# Solution to Problem No. 4 ## Given Data - **Specified Compressive Strength of Concrete, \( f'_c \) = 400 psi** - **Test Results (psi):** 485, 499, 386, 438, 447, 485, 440, 395, 468, 413 --- ## Part (a): Satisfactory Quality Check ### Step 1: Calculate the Average Strength \[ \text{Average} = \frac{485 + 499 + 386 + 438 + 447 + 485 + 440 + 395 + 468 + 413}{10} \] \[ = \frac{43560}{10} = 4356 \text{ psi} \] ### Step 2: Calculate the Standard Deviation (\( s \)) Let \( x_i \) be each result, and \(\bar{x} = 4356\): \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \] \[ \begin{align*} & (485-4356)^2 = 245,016 \\ & (499-4356)^2 = 401,956 \\ & (386-4356)^2 = 246,016 \\ & (438-4356)^2 = 576 \\ & (447-4356)^2 = 12,996 \\ & (485-4356)^2 = 245,016 \\ & (440-4356)^2 = 1,936 \\ & (395-4356)^2 = 164,836 \\ & (468-4356)^2 = 104,976 \\ & (413-4356)^2 = 50,176 \\ \end{align*} \] Sum: \(245,016 + 401,956 + 246,016 + 576 + 12,996 + 245,016 + 1,936 + 164,836 + 104,976 + 50,176 = 1,473,500\) \[ s = \sqrt{\frac{1,473,500}{9}} \approx \sqrt{163,722.22} \approx 404.6 \text{ psi} \] ### Step 3: Criteria for Satisfactory Quality Per ACI 318, concrete is satisfactory if: - **Average strength of all tests ≥ \( f'_{cr} \) (overdesign strength)** - **No individual test < \( f'_c - 500 \) psi** - **No two consecutive tests < \( f'_c \) by more than 500 psi** Here, \( f'_c = 400 \) psi, so \( f'_c - 500 = 350 \) psi. - All results > 350 psi. *(Lowest value is 386 psi)* - Average (4356 psi) > 400 psi #### **Conclusion for (a):** **Yes, the strength results represent concrete of satisfactory quality. All individual test results meet the minimum requirements and the average strength exceeds the specified strength.** --- ## Part (b): Mixture Adjustment Based on Standard Deviation Given: - **Standard deviation from previous project, \( s_s = 590 \) psi** - **Number of previous tests = 30** ### Step 1: Find Required Average Strength (\( f'_{cr} \)) According to ACI 318: \[ f'_{cr} = f'_c + 1.34s \] Where \( s \) is the standard deviation (use higher of current or historical if available). \[ f'_{cr} = 400 + 1.34 \times 590 = 400 + 790.6 = 479.6 \text{ psi} \] ### Step 2: Compare Current Average to Required Average - **Current average = 4356 psi** - **Required average = 4791 psi** ### Step 3: Conclusion Since the current average strength (4356 psi) is **less than** the required average strength (4791 psi): #### **Conclusion for (b):** **Yes, the mixture proportions must be adjusted. The average strength does not meet the required overdesign strength based on the higher standard deviation from previous data.** --- ## **Summary Table** | Test No. | Strength (psi) | |----------|---------------| | 1 | 485 | | 2 | 499 | | 3 | 386 | | 4 | 438 | | 5 | 447 | | 6 | 485 | | 7 | 440 | | 8 | 395 | | 9 | 468 | | 10 | 413 | | **Avg** | **4356** | --- ## **Final Answers** ### (a) **The concrete is of satisfactory quality based on test results and ACI requirements.** ### (b) **The mixture proportions must be adjusted, as the average strength does not meet the required value based on the higher standard deviation.**

# Solution to Problem No. 4 ## Given Information - **Characteristic Compressive Strength of Concrete, \( f'_c = 4000 \, \text{psi} \)** - **Ten Test Results (psi):** 4850, 4990, 3860, 4380, 4470, 4850, 4400, 3950, 4680, 4130 - **Sample Standard Deviation from Previous Project, \( s = 590 \, \text{psi} \)** ## ACI Acceptance Criteria for Concrete Strength 1. **Criterion 1 (Average of Three Consecutive Tests):** \[ \text{Average of any 3 consecutive tests} \geq f'_c \] 2. **Criterion 2 (Individual Test Limit):** \[ \text{No individual test result should be less than } f'_c - 500 \] For \( f'_c = 4000 \): \[ 4000 - 500 = 3500 \, \text{psi} \] --- ## Part (a): Do the Strength Results Represent Concrete of Satisfactory Quality? ### Step 1: Check Criterion 2 (Individual Test Results) - **Smallest value in the data:** \[ \text{min} = 3860 \, \text{psi} \] - **Required minimum:** \[ 3500 \, \text{psi} \] Since \( 3860 > 3500 \), **Criterion 2 is satisfied.** ### Step 2: Check Criterion 1 (Average of Every 3 Consecutive Tests) #### Compute All Possible 3-Test Moving Averages 1. \( \bar{x}_1 = \frac{4850 + 4990 + 3860}{3} = 4566.67 \) 2. \( \bar{x}_2 = \frac{4990 + 3860 + 4380}{3} = 4410 \) 3. \( \bar{x}_3 = \frac{3860 + 4380 + 4470}{3} = 4236.67 \) 4. \( \bar{x}_4 = \frac{4380 + 4470 + 4850}{3} = 4566.67 \) 5. \( \bar{x}_5 = \frac{4470 + 4850 + 4400}{3} = 4573.33 \) 6. \( \bar{x}_6 = \frac{4850 + 4400 + 3950}{3} = 4400 \) 7. \( \bar{x}_7 = \frac{4400 + 3950 + 4680}{3} = 4343.33 \) 8. \( \bar{x}_8 = \frac{3950 + 4680 + 4130}{3} = 4253.33 \) **Decision Rule:** Each average must be \( \geq f'_c = 4000 \). - All averages computed are greater than 4000. **Criterion 1 is satisfied.** ### Conclusion for Part (a) **Both ACI criteria are satisfied. The concrete strength results represent concrete of satisfactory quality.** --- ## Part (b): Must the Mix Proportions be Adjusted? ### Step 1: Compute \( f'_{cr1} \) \[ f'_{cr1} = f'_c + 1.34s \] \[ f'_{cr1} = 4000 + 1.34(590) = 4000 + 790.6 = 4790.6 \, \text{psi} \] ### Step 2: Compute \( f'_{cr2} \) \[ f'_{cr2} = f'_c + 2.33s - 500 \] \[ f'_{cr2} = 4000 + 2.33(590) - 500 = 4000 + 1374.7 - 500 = 4874.7 \, \text{psi} \] ### Step 3: Required Average Strength \[ f'_{cr} = \max(4790.6, 4874.7) = 4874.7 \, \text{psi} \] ### Step 4: Compute Actual Average of the 10 Tests \[ \bar{x} = \frac{4850 + 4990 + 3860 + 4380 + 4470 + 4850 + 4400 + 3950 + 4680 + 4130}{10} = 4456 \, \text{psi} \] ### Step 5: Comparison - **Required average:** \( 4874.7 \, \text{psi} \) - **Actual average:** \( 4456 \, \text{psi} \) Since \( 4456 < 4874.7 \): ### Conclusion for Part (b) **Yes, the mix proportions must be adjusted to increase the average strength.** --- ## Final Answers - **(a)** Yes, the concrete is of satisfactory quality based on ACI acceptance criteria. - **(b)** Yes, the mix proportions must be adjusted because the average strength is below the required average strength.

# Problem: Probability of Choosing a Double-Headed Coin ## Given Information - **Types of Coins:** - 3 fair coins - 2 double-headed coins - **Total Coins:** 5 (3 fair + 2 double-headed) - **Tossing:** A chosen coin is tossed 4 times, resulting in heads all 4 times. ## Definitions - Let \( D \) be the event that a double-headed coin is chosen. - Let \( H \) be the event that heads are obtained on all 4 tosses. ## Objective We want to find \( P(D | H) \), the probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Using Bayes' Theorem \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step 1: Calculate \( P(D) \) The prior probability of choosing a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] ### Step 2: Calculate \( P(H | D) \) If a double-headed coin is chosen, the probability of tossing heads 4 times is: \[ P(H | D) = 1 \] ### Step 3: Calculate \( P(H | F) \) Let \( F \) be the event of choosing a fair coin. The probability of tossing heads 4 times with a fair coin is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Step 4: Calculate \( P(F) \) The prior probability of choosing a fair coin: \[ P(F) = \frac{3}{5} \] ### Step 5: Calculate \( P(H) \) Using the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating \( P(H) \): \[ P(H) = \frac{2}{5} + \frac{3}{80} \] Finding a common denominator (80): \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Step 6: Substitute into Bayes' Theorem Now we can substitute back into Bayes' theorem: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] Calculating: \[ P(D | H) = \frac{2}{5} \cdot \frac{16}{7} = \frac{32}{35} \] ## Final Answer The probability that the chosen coin is double-headed given that all tosses resulted in heads is: \[ \boxed{\frac{32}{35}} \]

# Problem: Probability of Choosing a Double-Headed Coin ## Given Information - **Coins in the Box:** - 3 fair coins - 2 double-headed coins - **Total Coins:** 5 (3 fair + 2 double-headed) - **Tosses:** A chosen coin is tossed 4 times, resulting in heads in all tosses. ## Definitions - Let \( D \): event that a double-headed coin is chosen. - Let \( H \): event that heads are obtained in all 4 tosses. ## Objective We want to find \( P(D | H) \), the probability of having chosen a double-headed coin given that all tosses resulted in heads. ## Bayes' Theorem Application Using Bayes' theorem, we have: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Prior Probabilities 1. **Probability of choosing a double-headed coin:** \[ P(D) = \frac{2}{5} \] 2. **Probability of choosing a fair coin \( F \):** \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - If a double-headed coin is chosen, the probability of getting heads in all 4 tosses is: \[ P(H | D) = 1 \] - If a fair coin is chosen, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads Using the law of total probability, we calculate \( P(H) \): \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting our values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] Finding a common denominator (80): \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Now substituting back into Bayes' theorem: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] Calculating: \[ P(D | H) = \frac{2}{5} \cdot \frac{16}{7} = \frac{32}{35} \] ## Conclusion The probability that the chosen coin is double-headed given that all tosses resulted in heads is: \[ \boxed{\frac{32}{35}} \]

# Problem: Probability of Choosing a Double-Headed Coin ## Given Information - Coins in the box include: - 3 fair coins - 2 double-headed coins - Total coins: 5 (3 fair + 2 double-headed) - A chosen coin is tossed 4 times, and all results are heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. ## Objective We aim to determine \( P(D | H) \), the probability that a double-headed coin was chosen given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem states: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Prior Probabilities 1. The probability of picking a double-headed coin is calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. The probability of selecting a fair coin \( F \) is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - If a double-headed coin is chosen, the probability of getting heads 4 times is: \[ P(H | D) = 1 \] - If a fair coin is selected, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads Using the law of total probability, we compute \( P(H) \): \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires finding a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, adding both terms: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] Converting \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] Calculating this expression: 1. Rewrite: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] 2. Performing the multiplication: \[ P(D | H) = \frac{32}{35} \] ## Conclusion The probability that the chosen coin is double-headed, given that all tosses resulted in heads, is: \[ \boxed{\frac{32}{35}} \]

# Problem: Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, yielding heads on all tosses. ## Definitions and Notation - Let \( D \) denote the event that a double-headed coin is chosen. - Let \( H \) denote the event that heads appears in all 4 tosses. We seek to find \( P(D | H) \), which is the conditional probability that the chosen coin is a double-headed coin given that all tosses resulted in heads. ## Bayes' Theorem Overview Bayes' theorem relates the conditional and marginal probabilities of random events. It is mathematically expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Components of the Formula 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of initially selecting a double-headed coin from the box is given by: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of choosing a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Obtaining Heads with a Double-Headed Coin:** If a double-headed coin is chosen, every toss will result in heads. Thus: \[ P(H | D) = 1 \] - **Probability of Obtaining Heads with a Fair Coin:** For a fair coin, the probability of getting heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of getting heads in all 4 tosses is calculated as follows: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we use the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting in the known values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires finding a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we add both components: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] Converting \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem We substitute our derived values back into Bayes' theorem: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This can be simplified as follows: 1. Rewrite the equation: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] 2. Perform the multiplication: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After analyzing the probabilities and applying Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is: \[ \boxed{\frac{32}{35}} \]

# Problem: Determining the Probability of Picking a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins Totaling to 5 coins. A coin is randomly selected and tossed 4 times, resulting in heads for all tosses. ## Definitions - Let \( D \) represent the event of choosing a double-headed coin. - Let \( H \) represent the event of obtaining heads on all 4 tosses. Our goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Bayes' Theorem Bayes' theorem provides a way to update our estimate of the probability of an event based on new evidence. It is given by the formula: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Prior Probabilities 1. **Probability of Choosing a Double-Headed Coin:** The proportion of double-headed coins in the box is: \[ P(D) = \frac{2}{5} \] 2. **Probability of Choosing a Fair Coin:** Conversely, the probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **If a Double-Headed Coin is Selected:** The outcome of all tosses will certainly be heads. Therefore: \[ P(H | D) = 1 \] - **If a Fair Coin is Selected:** The probability of obtaining heads on one toss of a fair coin is \( \frac{1}{2} \). Thus, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To compute \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each component: 1. The first term yields: \[ P(H | D) \cdot P(D) = 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. For the second term, we find a common denominator: \[ P(H | F) \cdot P(F) = \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we combine both terms: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] Converting \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting the values back into Bayes' theorem, we compute: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After evaluating the probabilities through Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is: \[ \boxed{\frac{32}{35}} \]

# Problem: Calculating the Probability of Picking a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is selected at random and tossed 4 times, with each toss resulting in heads. ## Definitions - Let \( D \) denote the event of selecting a double-headed coin. - Let \( H \) denote the event of obtaining heads in all 4 tosses. We need to find \( P(D | H) \), the conditional probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Bayes' Theorem Bayes' theorem provides a way to update the probability of an event based on new evidence. The theorem is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Breakdown of Each Term 1. **Prior Probability of Choosing a Double-Headed Coin:** The probability of selecting a double-headed coin from the box is: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Choosing a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is chosen, every toss will definitely result in heads. Thus: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** If a fair coin is selected, the probability of getting heads on any single toss is \( \frac{1}{2} \). Therefore, the probability of getting heads in 4 consecutive tosses is calculated as follows: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To determine \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term involves finding a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we sum both terms: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To combine these, we convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting the derived values back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This can be simplified as follows: 1. Rewrite the expression: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] 2. Perform the multiplication: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After applying Bayes' theorem and evaluating the probabilities, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is: \[ \boxed{\frac{32}{35}} \]

# Problem: Finding the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is chosen at random and tossed 4 times, with all tosses showing heads. ## Definitions - Let \( D \) represent the event of selecting a double-headed coin. - Let \( H \) represent the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem allows us to update the probability of a hypothesis based on new evidence. The formula is: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Determining Each Component 1. **Prior Probability of Choosing a Double-Headed Coin:** The probability of selecting a double-headed coin is given by: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Choosing a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads. Therefore: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** If a fair coin is chosen, the chance of heads on one toss is \( \frac{1}{2} \). The probability of getting heads in 4 consecutive tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting in the values calculated: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the parts: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we add both terms together: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To combine these fractions, convert \( \frac{2}{5} \) to have a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Now, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now we substitute these values back into Bayes' theorem: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After analyzing the probabilities using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is: \[ \boxed{\frac{32}{35}} \]

# Problem: Assessing the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins Thus, the total number of coins is 5. A coin is randomly selected and tossed 4 times, resulting in heads on all tosses. ## Definitions - Let \( D \) denote the event of selecting a double-headed coin. - Let \( H \) denote the event of obtaining heads on all 4 tosses. We aim to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Bayes' Theorem Bayes' theorem provides a method for updating the probability of a hypothesis based on new evidence. The formula is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Components 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of initially choosing a double-headed coin is: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability for a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Getting Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always show heads in all tosses. Thus: \[ P(H | D) = 1 \] - **Probability of Getting Heads with a Fair Coin:** For a fair coin, the chance of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires finding a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, summing both terms: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To combine these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] So we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting the calculated values back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating the final result: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After analyzing the probabilities through the application of Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is: \[ \boxed{\frac{32}{35}} \]

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, and all tosses resulted in heads. ## Definitions - Let \( D \) represent the event of selecting a double-headed coin. - Let \( H \) represent the event of obtaining heads in all 4 tosses. Our goal is to determine \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Bayes' Theorem Bayes' theorem allows us to update the probability of a hypothesis based on new evidence. The formula is given by: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Breakdown of Components 1. **Prior Probability of Choosing a Double-Headed Coin:** The probability of selecting a double-headed coin is calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Choosing a Fair Coin:** Thus, the probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is chosen, it will yield heads on every toss. Therefore: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the chance of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we utilize the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires finding a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we add the two terms together: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To combine these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation with Bayes' Theorem Substituting the derived values back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After applying Bayes' theorem and evaluating the probabilities, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is: \[ \boxed{\frac{32}{35}} \]

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information The box contains: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, resulting in heads each time. ## Definitions - Let \( D \) represent the event of selecting a double-headed coin. - Let \( H \) represent the event of obtaining heads in all 4 tosses. We want to find \( P(D | H) \), which is the conditional probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem is a mathematical formula used to update probabilities based on new evidence. It states: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculation of Probabilities 1. **Prior Probability of Choosing a Double-Headed Coin:** The probability of selecting a double-headed coin is calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Choosing a Fair Coin:** The probability of choosing a fair coin is then: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is chosen, it will always show heads. Thus: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the chance of heads on each toss is \( \frac{1}{2} \). Therefore, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we utilize the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term involves finding a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, summing both terms: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To combine these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting the calculated values back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After performing the calculations and applying Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is: \[ \boxed{\frac{32}{35}} \]

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information Inside a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with all tosses resulting in heads. ## Definitions - Let \( D \) be the event of selecting a double-headed coin. - Let \( H \) be the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem provides a way to compute conditional probabilities by relating them to prior probabilities and likelihoods. It can be expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step 1: Calculate the Prior Probabilities 1. **Probability of Choosing a Double-Headed Coin:** The probability of selecting a double-headed coin from the box is: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Probability of Choosing a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Step 2: Calculate the Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always show heads in all tosses. Thus: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of getting heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Step 3: Calculate the Total Probability of Heads To find \( P(H) \), we use the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting in the known values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. For the second term, we calculate: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we sum these fractions: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] Converting \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Step 4: Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating the final result: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After evaluating the probabilities using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is: \[ \boxed{\frac{32}{35}} \]

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This totals to 5 coins. A coin is chosen at random and tossed 4 times, yielding heads on all tosses. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. We aim to determine P(D | H), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem allows us to update the probability of an event based on new evidence. The formula is represented as follows: P(D | H) = (P(H | D) * P(D)) / P(H) ### Component Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** This can be calculated as: P(D) = Number of double-headed coins / Total number of coins = 2 / 5 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: P(F) = 3 / 5 ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always result in heads. Therefore: P(H | D) = 1 - **Probability of Heads with a Fair Coin:** The chance of getting heads on a single toss for a fair coin is 1/2. Thus, the probability of getting heads in 4 tosses is: P(H | F) = (1/2)^4 = 1/16 ### Total Probability of Heads To find P(H), we apply the law of total probability: P(H) = P(H | D) * P(D) + P(H | F) * P(F) Substituting the known values gives: P(H) = 1 * (2 / 5) + (1/16) * (3 / 5) Calculating the terms yields: 1. The first term becomes: 1 * (2 / 5) = 2 / 5 2. The second term, requiring a common denominator, is: (1/16) * (3/5) = 3 / 80 Combining these, we find: P(H) = (2 / 5) + (3 / 80) To add these fractions, convert 2/5 into a fraction with a denominator of 80: 2/5 = 32/80 Thus, we have: P(H) = (32 / 80) + (3 / 80) = 35 / 80 = 7 / 16 ### Final Calculation Using Bayes' Theorem Substituting the calculated values back into Bayes' theorem provides: P(D | H) = (P(H | D) * P(D)) / P(H) = (1 * (2 / 5)) / (7 / 16) This simplifies to: P(D | H) = (2 / 5) / (7 / 16) = (2 / 5) * (16 / 7) Calculating this gives: P(D | H) = 32 / 35 ## Conclusion After analyzing the probabilities through the application of Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is 32/35.

# Problem: Determining the Probability of Choosing a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This leads to a total of 5 coins. A coin is randomly selected and tossed 4 times, resulting in heads on each toss. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. We are tasked with calculating \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem enables us to update the probability of an event based on new evidence. The theorem is expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step 1: Calculate Prior Probabilities 1. **Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin is calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Step 2: Calculate Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is chosen, it will always result in heads. Thus: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on one toss is \( \frac{1}{2} \). Therefore, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Step 3: Calculate the Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term results in: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Adding these together: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To combine these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Step 4: Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After evaluating the probabilities through Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Assessing the Probability of Selecting a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is selected at random and tossed 4 times, with all tosses resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Bayes' Theorem Application Bayes' theorem allows us to update our estimate of the probability of an event based on new evidence. The formula is expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculation of Probabilities The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] Conversely, the probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] For the conditional probabilities, if a double-headed coin is chosen, every toss will yield heads. Therefore: \[ P(H | D) = 1 \] For a fair coin, the probability of getting heads on a single toss is \( \frac{1}{2} \). Consequently, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] To find the total probability of heads, we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term becomes: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term can be computed as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, both terms are combined: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] Substituting all calculated values back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After analyzing the situation using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Calculating the Probability of Choosing a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins This means there are a total of 5 coins. A coin is selected at random and tossed 4 times, resulting in heads on every toss. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Bayes' Theorem Bayes' theorem allows us to compute conditional probabilities based on prior probabilities and likelihoods. The formula is given by: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Calculations The prior probability of selecting a double-headed coin can be determined as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we consider the conditional probabilities. If a double-headed coin is chosen, it will always result in heads: \[ P(H | D) = 1 \] For a fair coin, the probability of obtaining heads in one toss is \( \frac{1}{2} \). Therefore, the probability of getting heads in all 4 tosses is calculated as: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] To find the total probability of heads \( P(H) \), we utilize the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term results in: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we can combine these probabilities: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] So, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] Finally, substituting back into Bayes' theorem gives us: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon analyzing the situation using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This gives a total of 5 coins. A coin is randomly selected and tossed 4 times, resulting in heads each time. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Bayes' theorem is used to update the probability of an event based on new evidence. The formula is expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculation of Probabilities The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] Conversely, the probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] For the conditional probabilities, if a double-headed coin is chosen, every toss will yield heads: \[ P(H | D) = 1 \] For a fair coin, the probability of getting heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To determine \( P(H) \), we utilize the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term results in: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] We can now combine these probabilities: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] Finally, substituting back into Bayes' theorem gives us: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon analyzing the probabilities using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Finding the Probability of Choosing a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let \( D \) denote the event of selecting a double-headed coin. - Let \( H \) denote the event of obtaining heads in all 4 tosses. We seek to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Bayes' theorem allows for the updating of probabilities based on new evidence. The formula is as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations To find the prior probability of choosing a double-headed coin, we have: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] For selecting a fair coin, the probability is: \[ P(F) = \frac{3}{5} \] Next, we determine the conditional probabilities. If a double-headed coin is chosen, it will always result in heads: \[ P(H | D) = 1 \] For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is calculated as: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values, we get: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Combining these, we find: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting the computed values back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After evaluating the probabilities through the application of Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Assessing the Probability of Choosing a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is selected at random and tossed 4 times, yielding heads on all occasions. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The aim is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Bayes' theorem provides a method for updating the probability of an event based on new evidence. The equation is expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculation of Probabilities To calculate the prior probability of selecting a double-headed coin, we have: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we assess the conditional probabilities. If a double-headed coin is chosen, it will always show heads: \[ P(H | D) = 1 \] For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we employ the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting in the known values results in: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Combining these yields: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon reviewing the calculations through the application of Bayes' theorem, we determine that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins Thus, there are a total of 5 coins. A coin is randomly selected and tossed 4 times, with all tosses resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. We aim to find \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Using Bayes' theorem, we can express the conditional probability as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations To calculate the prior probability of selecting a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we consider the conditional probabilities. If a double-headed coin is selected, it will always produce heads: \[ P(H | D) = 1 \] For a fair coin, the probability of getting heads on one toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we utilize the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting in the known values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term can be computed as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we combine both components: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating the final result yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After carefully analyzing the problem using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Analyzing the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is selected at random and tossed 4 times, with all tosses resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem allows us to update the probability of an event based on new evidence. The formula is given by: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we analyze the conditional probabilities. If a double-headed coin is chosen, it will always produce heads: \[ P(H | D) = 1 \] For a fair coin, the probability of getting heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values results in: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term yields: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term, requiring a common denominator, is: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now we can combine both components: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After performing the calculations using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This gives a total of 5 coins. A coin is selected at random and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Bayes' theorem is a powerful tool that allows us to update the probability of an event based on new evidence. The theorem is mathematically represented as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations To find the prior probability of selecting a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we assess the conditional probabilities. If a double-headed coin is chosen, it will always yield heads: \[ P(H | D) = 1 \] For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term yields: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires finding a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Combining these yields: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting the values back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After reviewing the probabilities utilizing Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Analyzing the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let \( D \) represent the event of selecting a double-headed coin. - Let \( H \) denote the event of obtaining heads on all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Using Bayes' Theorem Bayes' theorem provides a method for updating probabilities based on new evidence. The theorem can be expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities First, we calculate the prior probability of selecting a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] Next, the probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we find the conditional probabilities. If a double-headed coin is chosen, it will always result in heads: \[ P(H | D) = 1 \] For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we use the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting in the values we have: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the terms yields: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term can be computed as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Next, we add both components: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To combine these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Now, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives us: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating the final result yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After evaluating the situation using Bayes' theorem, we determine that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This gives a total of 5 coins. A coin is randomly selected and tossed 4 times, resulting in heads on each toss. ## Definitions - Let \( D \) denote the event of selecting a double-headed coin. - Let \( H \) denote the event of obtaining heads in all 4 tosses. We want to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Bayes' theorem enables the calculation of conditional probabilities and is expressed mathematically as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Calculations First, we calculate the prior probability of choosing a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we find the conditional probabilities. If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] For a fair coin, the chance of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in 4 consecutive tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires finding a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now we can combine both components to find \( P(H) \): \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon evaluating the situation using Bayes' theorem, we conclude that the probability that the selected coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Determining the Probability of Choosing a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with heads appearing in every toss. ## Definitions - Let D be the event of selecting a double-headed coin. - Let H be the event of getting heads in all 4 tosses. We are tasked with calculating \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Bayes' theorem helps us update the probability of an event based on new evidence. The formula is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities To compute the prior probability of selecting a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we consider the conditional probabilities. If a double-headed coin is chosen, it will always result in heads: \[ P(H | D) = 1 \] For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we utilize the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting in the values calculated: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components yields: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term, after finding a common denominator, is: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now we can combine the two components: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting the calculated values back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After analyzing the scenario using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Finding the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with every toss resulting in heads. ## Definitions - Let D represent the event of choosing a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. Our goal is to calculate \( P(D | H) \), the conditional probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Bayes' theorem allows us to update the probability of an event based on new evidence. The formula is represented as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities First, we compute the prior probability of selecting a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we determine the conditional probabilities. If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we utilize the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting in the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term results in: 1. The first term: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires finding a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now we can combine both components: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After evaluating the probabilities using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins Thus, there are a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss yielding heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to find \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem provides a framework for updating probabilities based on new evidence. The formula is: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations First, we calculate the prior probability of selecting a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we assess the conditional probabilities. If a double-headed coin is chosen, it will always show heads: \[ P(H | D) = 1 \] For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now we can combine both components: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem gives us: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon evaluating the probabilities using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly chosen and tossed 4 times, with heads appearing in each toss. ## Definitions - Let D denote the event of choosing a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The task is to calculate \( P(D | H) \), the conditional probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem provides a method for updating the probability of an event based on new evidence. The formula is as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations To determine the prior probability of selecting a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we evaluate the conditional probabilities. If a double-headed coin is selected, it will always result in heads: \[ P(H | D) = 1 \] For a fair coin, the probability of getting heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To compute \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each component results in: 1. The first term is: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we can combine both terms: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After analyzing the scenario using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Evaluating the Probability of Choosing a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This totals to 5 coins. A coin is selected at random and tossed 4 times, producing heads in every toss. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads on all 4 tosses. Our objective is to calculate \( P(D | H) \), the conditional probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem provides a systematic way to revise probabilities based on new evidence. The formula can be expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations First, we calculate the prior probability of selecting a double-headed coin: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] Next, we assess the conditional probabilities. If a double-headed coin is chosen, it will always show heads: \[ P(H | D) = 1 \] For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each term results in: 1. The first term: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we combine these probabilities: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After evaluating the probabilities using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Finding the Probability of Choosing a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Bayes' theorem allows us to update probabilities based on new evidence. The formula is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculation of Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin is: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term requires a common denominator: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we combine both components to find \( P(H) \): \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting the values back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After evaluating the probabilities using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with heads appearing in each toss. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem provides a method for calculating conditional probabilities based on prior probabilities and new evidence. The formula is: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin is determined as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is chosen, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the chance of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of getting heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we can combine both terms: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After carefully analyzing the probabilities through Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Expected Number of Tosses to Get the Pattern HTH ## Given Information A fair coin is tossed repeatedly until the pattern HTH appears for the first time. We need to find the expected number of tosses required to achieve this. ## Definitions - Let E be the expected number of tosses needed to obtain the pattern HTH. - Let E_H be the expected number of additional tosses needed given that the last toss was H. - Let E_HT be the expected number of additional tosses needed given that the last tosses were HT. ## States and Expected Values 1. **State S0**: No part of the pattern HTH has been matched. The expected number of tosses needed from this state is E. 2. **State S1**: The last toss resulted in H. The expected number of tosses needed from this state is E_H. 3. **State S2**: The last two tosses resulted in HT. The expected number of tosses needed from this state is E_HT. ## Transition Probabilities - From state S0: - With a probability of 1/2, we toss H and move to state S1. - With a probability of 1/2, we toss T and remain in state S0. Thus, the equation for E is: E = 1 + (1/2)E_H + (1/2)E - From state S1: - With a probability of 1/2, we toss H again and remain in state S1. - With a probability of 1/2, we toss T and move back to state S0. Therefore, the equation for E_H is: E_H = 1 + (1/2)E + (1/2)E_H - From state S2: - With a probability of 1/2, we toss H and reach the final pattern HTH (no more tosses needed). - With a probability of 1/2, we toss T and revert to state S0. Thus, the equation for E_HT is: E_HT = 1 + (1/2) * 0 + (1/2)E ## Solving the Equations 1. Rearranging the equation for E: E = 1 + (1/2)E_H + (1/2)E E - (1/2)E = 1 + (1/2)E_H (1/2)E = 1 + (1/2)E_H E = 2 + E_H 2. Rearranging the equation for E_H: E_H = 1 + (1/2)E + (1/2)E_H E_H - (1/2)E_H = 1 + (1/2)E (1/2)E_H = 1 + (1/2)E E_H = 2 + E 3. Rearranging the equation for E_HT: E_HT = 1 + (1/2) * 0 + (1/2)E E_HT = 1 + (1/2)E ## Substituting and Solving Substituting E_H in the equation for E: E_H = 2 + E E = 2 + (2 + E) E = 2 + 2 + E E = 4 + E This provides an infinite loop, indicating a need to simplify. We now have: E_H = 2 + E E = 2 + (1 + (1/2)E) E_H = 2 + (2 + (1 + (1/2)E)) E_H = 4 + (1/2)E Now substituting E_H in E: E = 4 + (2 + (1/2)E) Simplifying gives us: E = 6 + (1/2)E (1/2)E = 6 E = 12 ## Conclusion The expected number of tosses required to first obtain the pattern HTH is 12.

# Problem: Expected Number of Tosses to Get the Pattern HTH ## Given Information A fair coin is tossed repeatedly until the pattern HTH appears for the first time. We want to determine the expected number of tosses required to achieve this. ## Definitions - Let E0 be the expected number of tosses starting from a state where no part of the pattern has been matched. - Let E1 be the expected number of additional tosses needed given that the last toss was H (first character matched). - Let E2 be the expected number of additional tosses needed given that the last two tosses were HT (first two characters matched). The process stops when the pattern HTH is completed, meaning there are no additional tosses needed (0 more tosses). ## Equations Based on States From state E0: - The first toss can result in: - With probability 1/2, H is tossed → transition to state E1. - With probability 1/2, T is tossed → remain in state E0. This leads to the equation: E0 = 1 + (1/2)E1 + (1/2)E0 Rearranging gives: E0 - (1/2)E0 = 1 + (1/2)E1 (1/2)E0 = 1 + (1/2)E1 E0 = 2 + E1 From state E1: - The next toss can result in: - With probability 1/2, H is tossed again → remain in state E1. - With probability 1/2, T is tossed → transition back to state E0. This leads to the equation: E1 = 1 + (1/2)E1 + (1/2)E0 Rearranging gives: E1 - (1/2)E1 = 1 + (1/2)E0 (1/2)E1 = 1 + (1/2)E0 E1 = 2 + E0 From state E2: - The next toss can result in: - With probability 1/2, H is tossed → pattern HTH is completed (0 more tosses needed). - With probability 1/2, T is tossed → transition back to state E0. This leads to the equation: E2 = 1 + (1/2)(0) + (1/2)E0 Simplifying gives: E2 = 1 + (1/2)E0 ## Solving the System of Equations We now have a system of three equations: 1. E0 = 2 + E1 2. E1 = 2 + E0 3. E2 = 1 + (1/2)E0 Substituting E1 from the first equation into the second equation: E1 = 2 + (2 + E1) E1 = 4 + E1 This reveals a need to substitute back into E0. Using the first equation: E0 = 2 + (2 + E0) E0 = 4 + E0 This indicates a simplification issue. Instead, we should use the relationships derived from each state without substituting too early. From E0: E0 = 2 + E1 E1 = 2 + (2 + E1) Now substituting E2 into E1: From E2 = 1 + (1/2)E0 into E0: We can now express E1 in terms of E0: 1. E1 = 2 + E0 Substituting this into the first equation gives: E0 = 2 + (2 + E0) E0 - E0 = 4 This leads to a contradiction; thus we need to express everything in terms of E0: E2 = 1 + (1/2)E0 E1 = 2 + E0 Substituting E0 into E1: E0 = 2 + (2 + E0) This again leads to infinite loops; thus we need to re-evaluate our states and probabilities. Following through with the E2 calculations leads us to: E2 = 1 + (1/2)E0 Substituting in the equations: 1. From E1 = 2 + E2 gives: E1 = 2 + 1 + (1/2)E0 This can be simplified to: E1 = 3 + (1/2)E0 Returning to E0: E0 = 2 + (3 + (1/2)E0) E0 = 5 + (1/2)E0 (1/2)E0 = 5 E0 = 10 ## Final Answer The expected number of tosses required to first obtain the pattern HTH is 10.

# Problem: Expected Number of Tosses to Get the Pattern HTH ## Given Information A fair coin is tossed repeatedly until the pattern HTH appears for the first time. We aim to determine the expected number of tosses required to achieve this. ## Definitions - Let \(E_0\) represent the expected number of tosses starting from a state where no part of the pattern has been matched. - Let \(E_1\) represent the expected number of additional tosses needed given that the last toss was H (first character matched). - Let \(E_2\) represent the expected number of additional tosses needed given that the last two tosses were HT (first two characters matched). The process stops when the complete pattern HTH is matched. ## Equations Based on States From state \(E_0\): - The first toss can result in: - With a probability of \( \frac{1}{2} \), H is tossed → transition to state \(E_1\). - With a probability of \( \frac{1}{2} \), T is tossed → remain in state \(E_0\). This leads to the equation: \[ E_0 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging gives: \[ E_0 - \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ E_0 = 2 + E_1 \] From state \(E_1\): - The next toss can result in: - With a probability of \( \frac{1}{2} \), we toss H again → remain in state \(E_1\). - With a probability of \( \frac{1}{2} \), we toss T → transition back to state \(E_0\). This leads to the equation: \[ E_1 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging gives: \[ E_1 - \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ E_1 = 2 + E_0 \] From state \(E_2\): - The next toss can result in: - With a probability of \( \frac{1}{2} \), we toss H and complete the pattern HTH (0 more tosses needed). - With a probability of \( \frac{1}{2} \), we toss T → transition back to state \(E_0\). This leads to the equation: \[ E_2 = 1 + \frac{1}{2} \cdot 0 + \frac{1}{2}E_0 \] Simplifying gives: \[ E_2 = 1 + \frac{1}{2}E_0 \] ## Solving the System of Equations Now we have a system of three equations: 1. \(E_0 = 2 + E_1\) 2. \(E_1 = 2 + E_0\) 3. \(E_2 = 1 + \frac{1}{2}E_0\) Substituting \(E_1\) from the first equation into the second equation: \[ E_1 = 2 + (2 + E_1) \] This leads to: \[ E_1 = 4 + E_1 \] This indicates a need to reorganize our approach. Instead, from \(E_1\): \[ E_1 = 2 + E_2 \] Substituting \(E_2\) into this equation: \[ E_2 = 1 + \frac{1}{2}E_0 \] We can express everything in terms of \(E_0\): From \(E_2\): \[ E_1 = 2 + \left(1 + \frac{1}{2}E_0\right) \] This simplifies to: \[ E_1 = 3 + \frac{1}{2}E_0 \] Substituting back into the first equation gives: \[ E_0 = 2 + \left(3 + \frac{1}{2}E_0\right) \] This results in: \[ E_0 = 5 + \frac{1}{2}E_0 \] Rearranging yields: \[ E_0 - \frac{1}{2}E_0 = 5 \] \[ \frac{1}{2}E_0 = 5 \] \[ E_0 = 10 \] ## Conclusion The expected number of tosses required to obtain the pattern HTH for the first time is 10.

# Problem: Expected Number of Tosses to Obtain the Pattern HTH ## Given Information A fair coin is tossed repeatedly until the sequence HTH appears for the first time. The objective is to find the expected number of tosses required to achieve this pattern. ## Definitions - Let \(E_0\) represent the expected number of tosses needed when starting from scratch (no part of the pattern matched). - Let \(E_1\) represent the expected number of additional tosses needed given that the last toss was H (first letter of the pattern matched). - Let \(E_2\) represent the expected number of additional tosses needed given that the last two tosses were HT (first two letters of the pattern matched). The process concludes when the complete pattern HTH is matched, resulting in zero additional tosses. ## Transition Equations From state \(E_0\): - The first toss can result in: - With probability \( \frac{1}{2} \), H is tossed → transition to state \(E_1\). - With probability \( \frac{1}{2} \), T is tossed → remain in state \(E_0\). This leads to the equation: \[ E_0 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging this, we have: \[ E_0 - \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ E_0 = 2 + E_1 \] From state \(E_1\): - The next toss can result in: - With probability \( \frac{1}{2} \), H is tossed again → remain in state \(E_1\). - With probability \( \frac{1}{2} \), T is tossed → transition back to state \(E_0\). Thus, we have: \[ E_1 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging this gives: \[ E_1 - \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ E_1 = 2 + E_0 \] From state \(E_2\): - The next toss can result in: - With probability \( \frac{1}{2} \), H is tossed → pattern HTH is completed (0 additional tosses needed). - With probability \( \frac{1}{2} \), T is tossed → transition back to state \(E_0\). Thus, we have: \[ E_2 = 1 + \frac{1}{2} \cdot 0 + \frac{1}{2}E_0 \] This simplifies to: \[ E_2 = 1 + \frac{1}{2}E_0 \] ## Solving the System of Equations We now have the following equations: 1. \(E_0 = 2 + E_1\) 2. \(E_1 = 2 + E_0\) 3. \(E_2 = 1 + \frac{1}{2}E_0\) Substituting the expression for \(E_1\) into the second equation: \[ E_1 = 2 + (2 + E_1) \] This leads to: \[ E_1 = 4 + E_1 \] This indicates a need to substitute back into \(E_0\). Instead, let's express everything in terms of \(E_0\): From \(E_1\): \[ E_1 = 2 + E_2 \] Substituting \(E_2\) back into this gives: \[ E_2 = 1 + \frac{1}{2}E_0 \] Thus: \[ E_1 = 2 + \left(1 + \frac{1}{2}E_0\right) \] This simplifies to: \[ E_1 = 3 + \frac{1}{2}E_0 \] Substituting this into the first equation gives: \[ E_0 = 2 + (3 + \frac{1}{2}E_0) \] This leads to: \[ E_0 = 5 + \frac{1}{2}E_0 \] Rearranging yields: \[ E_0 - \frac{1}{2}E_0 = 5 \] \[ \frac{1}{2}E_0 = 5 \] \[ E_0 = 10 \] ## Conclusion The expected number of tosses required to first obtain the pattern HTH is 10.

# Problem: Expected Number of Tosses to Obtain the Pattern HTH ## Given Information A fair coin is tossed repeatedly until the pattern HTH appears for the first time. The task is to determine the expected number of tosses needed to achieve this pattern. ## Definitions - Let \(E_0\) represent the expected number of tosses starting from a state where no part of the pattern has been matched. - Let \(E_1\) represent the expected number of additional tosses needed given that the last toss was H (the first character of the pattern matched). - Let \(E_2\) represent the expected number of additional tosses needed given that the last two tosses were HT (the first two characters matched). ## State Transition Equations From state \(E_0\): - The first toss can result in: - With probability \( \frac{1}{2} \), H is tossed → transition to state \(E_1\). - With probability \( \frac{1}{2} \), T is tossed → remain in state \(E_0\). This leads to the equation: \[ E_0 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging gives: \[ E_0 - \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ E_0 = 2 + E_1 \] From state \(E_1\): - The next toss can result in: - With probability \( \frac{1}{2} \), H is tossed again → remain in state \(E_1\). - With probability \( \frac{1}{2} \), T is tossed → transition back to state \(E_0\). This leads to the equation: \[ E_1 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging gives: \[ E_1 - \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ E_1 = 2 + E_0 \] From state \(E_2\): - The next toss can result in: - With probability \( \frac{1}{2} \), H is tossed → pattern HTH is completed (0 additional tosses needed). - With probability \( \frac{1}{2} \), T is tossed → transition back to state \(E_0\). This leads to the equation: \[ E_2 = 1 + \frac{1}{2} \cdot 0 + \frac{1}{2}E_0 \] Simplifying gives: \[ E_2 = 1 + \frac{1}{2}E_0 \] ## Solving the Equations We now have a system of three equations: 1. \(E_0 = 2 + E_1\) 2. \(E_1 = 2 + E_0\) 3. \(E_2 = 1 + \frac{1}{2}E_0\) Substituting \(E_1\) from the first equation into the second equation leads to: \[ E_1 = 2 + (2 + E_1) \] This simplifies to: \[ E_1 = 4 + E_1 \] This results in a contradiction, indicating an issue with substitution. Instead, we should express everything in terms of \(E_0\): Using \(E_2\): \[ E_1 = 2 + \left(1 + \frac{1}{2}E_0\right) \] This yields: \[ E_1 = 3 + \frac{1}{2}E_0 \] Substituting back into the first equation gives: \[ E_0 = 2 + (3 + \frac{1}{2}E_0) \] This leads to: \[ E_0 = 5 + \frac{1}{2}E_0 \] Rearranging results in: \[ E_0 - \frac{1}{2}E_0 = 5 \] \[ \frac{1}{2}E_0 = 5 \] \[ E_0 = 10 \] ## Conclusion The expected number of tosses required to first obtain the pattern HTH is 10.

# Problem: Expected Number of Tosses to Achieve the Pattern HTH ## Given Information A fair coin is tossed repeatedly until the pattern HTH appears for the first time. We want to calculate the expected number of tosses required to achieve this. ## Definitions - Let \(E_0\) represent the expected number of tosses starting from a state where no part of the pattern has been matched. - Let \(E_1\) represent the expected number of additional tosses needed given that the last toss was H (the first character of the pattern matched). - Let \(E_2\) represent the expected number of additional tosses needed given that the last two tosses were HT (the first two characters of the pattern matched). The process stops once the complete pattern HTH is matched, resulting in zero additional tosses. ## Transition Equations From state \(E_0\): - The first toss can result in: - With probability \( \frac{1}{2} \), H is tossed → transition to state \(E_1\). - With probability \( \frac{1}{2} \), T is tossed → remain in state \(E_0\). This leads to the equation: \[ E_0 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging gives: \[ E_0 - \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ E_0 = 2 + E_1 \] From state \(E_1\): - The next toss can result in: - With probability \( \frac{1}{2} \), H is tossed again → remain in state \(E_1\). - With probability \( \frac{1}{2} \), T is tossed → transition back to state \(E_0\). This leads to the equation: \[ E_1 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging gives: \[ E_1 - \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ E_1 = 2 + E_0 \] From state \(E_2\): - The next toss can result in: - With probability \( \frac{1}{2} \), H is tossed → the pattern HTH is completed (0 additional tosses needed). - With probability \( \frac{1}{2} \), T is tossed → transition back to state \(E_0\). This leads to the equation: \[ E_2 = 1 + \frac{1}{2} \cdot 0 + \frac{1}{2}E_0 \] Simplifying gives: \[ E_2 = 1 + \frac{1}{2}E_0 \] ## Solving the System of Equations We now have a system of three equations: 1. \(E_0 = 2 + E_1\) 2. \(E_1 = 2 + E_0\) 3. \(E_2 = 1 + \frac{1}{2}E_0\) Substituting \(E_1\) from the first equation into the second equation gives: \[ E_1 = 2 + (2 + E_1) \] This leads to: \[ E_1 = 4 + E_1 \] This situation reveals a circular dependency, indicating a need to express everything in terms of \(E_0\). Substituting \(E_1\) back into \(E_0\): Using \(E_2\) in the equation for \(E_1\): \[ E_1 = 2 + \left(1 + \frac{1}{2}E_0\right) \] This yields: \[ E_1 = 3 + \frac{1}{2}E_0 \] Substituting this back into the first equation yields: \[ E_0 = 2 + (3 + \frac{1}{2}E_0) \] Simplifying gives: \[ E_0 = 5 + \frac{1}{2}E_0 \] Rearranging leads us to: \[ E_0 - \frac{1}{2}E_0 = 5 \] \[ \frac{1}{2}E_0 = 5 \] Solving for \(E_0\): \[ E_0 = 10 \] ## Conclusion The expected number of tosses required to first obtain the pattern HTH is 10.

# Problem: Expected Number of Tosses to Obtain the Pattern HTH ## Given Information A fair coin is tossed repeatedly until the pattern HTH appears for the first time. The goal is to calculate the expected number of tosses needed to achieve this pattern. ## Definitions - Let \(E_0\) represent the expected number of tosses starting from a state where no part of the pattern has been matched. - Let \(E_1\) represent the expected number of additional tosses needed given that the last toss was H (the first character matched). - Let \(E_2\) represent the expected number of additional tosses needed given that the last two tosses were HT (the first two characters matched). The process stops when the complete pattern HTH is matched, meaning no additional tosses are needed at that point. ## Transition Equations From state \(E_0\): - The first toss can result in: - With probability \( \frac{1}{2} \), H is tossed → transition to state \(E_1\). - With probability \( \frac{1}{2} \), T is tossed → remain in state \(E_0\). This leads to the equation: \[ E_0 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging gives: \[ E_0 - \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ \frac{1}{2}E_0 = 1 + \frac{1}{2}E_1 \] \[ E_0 = 2 + E_1 \] From state \(E_1\): - The next toss can result in: - With probability \( \frac{1}{2} \), H is tossed again → remain in state \(E_1\). - With probability \( \frac{1}{2} \), T is tossed → transition back to state \(E_0\). This leads to the equation: \[ E_1 = 1 + \frac{1}{2}E_1 + \frac{1}{2}E_0 \] Rearranging gives: \[ E_1 - \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ \frac{1}{2}E_1 = 1 + \frac{1}{2}E_0 \] \[ E_1 = 2 + E_0 \] From state \(E_2\): - The next toss can result in: - With probability \( \frac{1}{2} \), H is tossed → pattern HTH is completed (0 more tosses needed). - With probability \( \frac{1}{2} \), T is tossed → transition back to state \(E_0\). This leads to the equation: \[ E_2 = 1 + \frac{1}{2} \cdot 0 + \frac{1}{2}E_0 \] This simplifies to: \[ E_2 = 1 + \frac{1}{2}E_0 \] ## Solving the Equations We now have a system of three equations: 1. \(E_0 = 2 + E_1\) 2. \(E_1 = 2 + E_0\) 3. \(E_2 = 1 + \frac{1}{2}E_0\) Substituting \(E_1\) from the first equation into the second equation gives: \[ E_1 = 2 + (2 + E_1) \] This leads to: \[ E_1 = 4 + E_1 \] This indicates a circular dependency, so let's express everything in terms of \(E_0\): From \(E_2\): \[ E_1 = 2 + \left(1 + \frac{1}{2}E_0\right) \] This simplifies to: \[ E_1 = 3 + \frac{1}{2}E_0 \] Now substituting this back into the first equation yields: \[ E_0 = 2 + (3 + \frac{1}{2}E_0) \] This leads to: \[ E_0 = 5 + \frac{1}{2}E_0 \] Rearranging gives: \[ E_0 - \frac{1}{2}E_0 = 5 \] \[ \frac{1}{2}E_0 = 5 \] Solving for \(E_0\): \[ E_0 = 10 \] ## Conclusion The expected number of tosses required to first obtain the pattern HTH is 10.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly chosen and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem allows us to calculate conditional probabilities, linking prior probabilities to new evidence. The formula is expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values provides: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term can be computed as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, adding both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After carefully analyzing the probabilities using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Assessing the Probability of Selecting a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with all tosses resulting in heads. ## Definitions - Let D be the event of selecting a double-headed coin. - Let H be the event of obtaining heads in all 4 tosses. Our goal is to determine \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Using Bayes' Theorem Bayes' theorem provides a method for calculating conditional probabilities by relating prior probabilities to observed evidence. The formula is given by: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To compute \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values leads to: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, adding both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To combine these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After analyzing the probabilities using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This means there are a total of 5 coins. A randomly selected coin is tossed 4 times, and all tosses result in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The goal is to find \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem allows us to calculate conditional probabilities by relating prior probabilities to observed evidence. The formula is: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin is: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To calculate \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After analyzing the probabilities using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Determining the Likelihood of Choosing a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, resulting in heads on each toss. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of achieving heads in all 4 tosses. The task is to compute \( P(D | H) \), which is the conditional probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is useful for calculating conditional probabilities and relates prior probabilities to new evidence. The theorem can be expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be expressed as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is chosen, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we utilize the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values leads to: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components yields: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, we combine both components: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to have a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation with Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon careful analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Calculating the Probability of Choosing a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins This gives a total of 5 coins. A coin is selected at random and tossed 4 times, with heads appearing in every toss. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. Our objective is to determine \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem is a fundamental tool for calculating conditional probabilities and is expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin is: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To determine \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating each component results in: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term is: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon careful analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly chosen and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The task is to calculate \( P(D | H) \), which is the conditional probability that the chosen coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a mathematical tool used to calculate conditional probabilities. It relates the probability of an event based on prior knowledge of conditions that might be related to the event. The formula is: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, adding both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To combine these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \).

# Problem: Assessing the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and all tosses yield heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Using Bayes' Theorem Bayes' theorem is a powerful statistical tool that allows us to update the probability of a hypothesis based on new evidence. The theorem can be mathematically expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] In this case, \(P(D | H)\) represents the probability we want to find, \(P(H | D)\) is the probability of getting heads in all tosses if a double-headed coin is chosen, \(P(D)\) is the prior probability of selecting a double-headed coin, and \(P(H)\) is the total probability of getting heads in all tosses. ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads in every toss. Therefore, the probability is: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Consequently, the probability of obtaining heads in all 4 tosses is calculated as follows: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To calculate the total probability \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components gives: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term can be computed as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, we convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we can express \( P(H) \) as: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now, we substitute back into Bayes' theorem: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a strong likelihood that the coin selected was double-headed, especially given the observed outcome of four consecutive heads.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, and each toss results in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem is a method for calculating conditional probabilities by linking prior probabilities to new evidence. It is represented mathematically as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is calculated as: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term can be computed as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, adding both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a high likelihood that the selected coin was indeed double-headed, especially considering the result of four consecutive heads.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with all tosses resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The goal is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a fundamental concept in probability that allows us to update the probability of a hypothesis based on new evidence. The theorem can be expressed mathematically as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the values we calculated: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates to: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Combining both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to have a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This suggests a strong likelihood that the selected coin was double-headed, especially given the outcome of four consecutive heads, which is a highly unlikely result with a fair coin.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information A box consists of: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly chosen coin is tossed 4 times, yielding heads every time. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of achieving heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem is a statistical method used to find conditional probabilities. It can be expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin is given by: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will certainly yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known probabilities gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates to: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, summing both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a high likelihood that the selected coin was double-headed, especially in light of the outcome of four consecutive heads, which is a rare event with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly chosen and tossed 4 times, with all tosses yielding heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to find \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Application of Bayes' Theorem Bayes' theorem is a foundational concept in probability theory used to find conditional probabilities. It can be articulated as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is chosen, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To calculate \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the values we calculated: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a strong likelihood that the selected coin was double-headed, particularly given the outcome of four consecutive heads, which is significantly improbable for a fair coin.

# Problem: Finding the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and each toss results in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The goal is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is an essential formula used to find conditional probabilities by linking prior knowledge with new evidence. The theorem can be expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components gives: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates to: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to have a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This substantial probability indicates that the selected coin is very likely to be double-headed, particularly given the outcome of four consecutive heads, which would be highly unlikely with a fair coin.

# Problem: Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, and each toss results in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The task is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Using Bayes' Theorem Bayes' theorem is a fundamental concept in probability that allows us to calculate conditional probabilities. It is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin is: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To calculate \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This result indicates a high likelihood that the selected coin was double-headed, particularly in light of the outcome of four consecutive heads, which would be improbable with a fair coin.

# Problem: Assessing the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with all tosses resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a fundamental principle in probability that allows for the calculation of conditional probabilities. The theorem is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin is calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To determine \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values provides: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon careful analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a strong likelihood that the selected coin was indeed double-headed, especially considering the result of four consecutive heads, which would be quite unlikely with a fair coin.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and each toss results in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The goal is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem is a statistical method used to compute conditional probabilities. It relates the conditional probability of an event to its prior probabilities and the likelihood of the observed evidence. The formula is given by: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components yields: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a strong likelihood that the selected coin was indeed double-headed, especially considering the outcome of four consecutive heads, which would be significantly improbable with a fair coin.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly chosen and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D be the event of selecting a double-headed coin. - Let H be the event of obtaining heads in all 4 tosses. The task is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a statistical method that allows us to determine the conditional probability of an event based on prior knowledge of conditions related to the event. The theorem is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be computed as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ P(H | D) \cdot P(D) = 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ P(H | F) \cdot P(F) = \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This suggests a high likelihood that the selected coin was double-headed, especially considering the outcome of four consecutive heads, which is very unlikely with a fair coin.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a powerful statistical tool that provides a way to update the probability of an event based on new evidence. The theorem is expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is calculated as: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this yields: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This suggests a strong likelihood that the selected coin was indeed double-headed, especially given the result of four consecutive heads, which would be highly improbable with a fair coin.

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, yielding heads on every toss. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a fundamental principle in probability that allows us to find conditional probabilities by linking prior knowledge with new evidence. The theorem can be expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on each toss is \( \frac{1}{2} \). Consequently, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term can be calculated as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a strong likelihood that the selected coin was indeed double-headed, particularly given the outcome of four consecutive heads, which is highly unlikely to occur with a fair coin.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let \( D \) represent the event of selecting a double-headed coin. - Let \( H \) represent the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem offers a way to compute conditional probabilities by relating prior probabilities to observed evidence. The formula is given by: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is calculated as: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components gives: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This result suggests a strong likelihood that the selected coin was indeed double-headed, especially given the outcome of four consecutive heads, which is highly unlikely to occur with a fair coin.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins Thus, there are a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The task is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a fundamental concept in probability that helps calculate conditional probabilities by relating prior probabilities to new evidence. The formula is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of getting heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a strong likelihood that the selected coin was indeed double-headed, particularly given the outcome of four consecutive heads, which is quite unlikely to occur with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The goal is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a fundamental principle in probability that helps calculate conditional probabilities by relating prior probabilities to new evidence. The theorem can be expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This suggests a significant likelihood that the selected coin was double-headed, especially considering the outcome of four consecutive heads, which is highly improbable with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, yielding heads on each toss. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem offers a way to calculate conditional probabilities by connecting prior probabilities with observed evidence. The formula is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we can express \( P(H) \) as: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This result indicates a strong likelihood that the selected coin was double-headed, particularly in view of the outcome of four consecutive heads, which is quite improbable with a fair coin.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss yielding heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The task is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a valuable tool for finding conditional probabilities by linking prior knowledge with new evidence. The formula is given by: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads, so: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known probabilities yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After careful analysis using Bayes' theorem, we conclude that the probability that the selected coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a significant likelihood that the chosen coin was indeed double-headed, particularly given the outcome of four consecutive heads, which would be highly improbable with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with all tosses resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem is a key method for determining conditional probabilities by connecting prior probabilities with observed evidence. It is expressed mathematically as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a significant likelihood that the selected coin was indeed double-headed, particularly given the outcome of four consecutive heads, which is highly improbable with a fair coin.

# Problem: Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and each toss results in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Using Bayes' Theorem Bayes' theorem is a statistical tool for calculating conditional probabilities, linking prior probabilities to new evidence. It is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting in the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This suggests a strong likelihood that the selected coin was double-headed, particularly given the outcome of four consecutive heads, which would be highly unlikely with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with every toss resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is an essential tool in probability that allows us to compute conditional probabilities by linking prior probabilities to new evidence. It is expressed mathematically as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is calculated as: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This assessment indicates a high likelihood that the selected coin was indeed double-headed, particularly in light of the outcome of four consecutive heads, which would be quite improbable with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and each toss results in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The goal is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a crucial method for determining conditional probabilities, connecting prior probabilities with observed evidence. The theorem can be expressed mathematically as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This conclusion suggests a significant likelihood that the selected coin was indeed double-headed, especially in light of the outcome of four consecutive heads, which is highly improbable with a fair coin.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss yielding heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The task is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is an important method for calculating conditional probabilities by connecting prior probabilities with observed evidence. The theorem can be expressed mathematically as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This suggests a high likelihood that the selected coin was indeed double-headed, particularly considering the outcome of four consecutive heads, which would be highly improbable with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and each toss results in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a statistical method for determining conditional probabilities, linking prior probabilities to observed evidence. The theorem can be expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a significant likelihood that the selected coin was indeed double-headed, particularly given the result of four consecutive heads, which would be highly unlikely with a fair coin.

# Problem: Finding the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, with each toss resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The task is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a statistical method used to compute conditional probabilities, relating prior knowledge to new evidence. The theorem can be expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a significant likelihood that the selected coin was indeed double-headed, particularly given the outcome of four consecutive heads, which would be highly improbable with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a statistical method that helps us determine conditional probabilities by linking prior knowledge with new evidence. The theorem can be represented mathematically as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is calculated as follows: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components gives: 1. The first term results in: \[ P(H | D) \cdot P(D) = 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ P(H | F) \cdot P(F) = \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a strong likelihood that the selected coin was indeed double-headed, especially given the outcome of four consecutive heads, which would be highly improbable with a fair coin.

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a statistical tool used to find conditional probabilities by linking prior probabilities to new evidence. The formula can be represented as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components gives: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components together gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a strong likelihood that the selected coin was indeed double-headed, particularly considering the outcome of four consecutive heads, which is highly unlikely with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and each toss results in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The objective is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem is a fundamental concept in probability that allows us to determine conditional probabilities by linking prior information with new evidence. The formula is given by: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Calculating the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is calculated as: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This suggests a strong likelihood that the selected coin was indeed double-headed, especially considering the outcome of four consecutive heads, which would be quite unlikely with a fair coin.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, with each toss yielding heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Using Bayes' Theorem Bayes' theorem is a statistical method used to calculate conditional probabilities, relating prior probabilities to new evidence. The formula is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components gives: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a significant likelihood that the selected coin was indeed double-headed, particularly given the outcome of four consecutive heads, which would be highly unlikely with a fair coin.

# Problem: Tim's Chances of Drawing to a Straight Flush and a Flush ## Given Information Tim's original 5 cards: - 4♣ (discarded) - 6♥ - 8♥ - 9♥ - K♦ (discarded) Tim keeps: 6♥, 8♥, 9♥ He draws 2 new cards from the remaining deck. ## Step 1: Cards Remaining in the Deck A standard deck consists of 52 cards. Since Tim has 5 known cards, the remaining cards in the deck are: \[ 52 - 5 = 47 \] Total ways to draw 2 cards from the remaining deck: \[ \binom{47}{2} = \frac{47 \times 46}{2} = 1081 \] ## Part A: Chance of a Straight Flush To achieve a straight flush in hearts, the final 5 cards must be consecutive hearts and include 6♥, 8♥, and 9♥. The only possible straight flush using these three cards is: 6♥, 7♥, 8♥, 9♥, 10♥ The two required cards to complete this straight flush are: - 7♥ - 10♥ These specific cards must be drawn. ### Probability of Drawing a Straight Flush Favorable ways to draw the required cards: 1 The probability of drawing a straight flush is: \[ P(\text{Straight Flush}) = \frac{1}{1081} \] ## Part B: Chance of a Flush A flush means that all 5 cards are hearts. Tim already has 3 hearts: - 6♥ - 8♥ - 9♥ Thus, both drawn cards must be hearts. ### Step 2: Count Remaining Hearts In a standard deck, there are 13 hearts. Since Tim already has 3 hearts, the remaining hearts are: \[ 13 - 3 = 10 \] ### Step 3: Ways to Draw 2 Hearts The number of ways to draw 2 hearts from the remaining 10 hearts is: \[ \binom{10}{2} = \frac{10 \times 9}{2} = 45 \] ### Step 4: Exclude Straight Flush Cases The straight flush (7♥ and 10♥) is included in the 45 cases, but we are counting flush-only cases. Therefore, we must exclude this specific case. Thus, the favorable cases for a flush (but not a straight flush) are: \[ 45 - 1 = 44 \] ### Step 5: Probability of a Flush The probability of drawing a flush (but not a straight flush) is: \[ P(\text{Flush}) = \frac{44}{1081} \] ## Final Answers - Probability of drawing a straight flush: \[ \frac{1}{1081} \] - Probability of drawing a flush (but not a straight flush): \[ \frac{44}{1081} \]

# Problem: Probability of Choosing a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins The total number of coins in the box is 5. A coin is selected at random and tossed 4 times, resulting in heads on all tosses. We need to determine the probability that the chosen coin is double-headed given this outcome. ## Definitions - Let \( D \) denote the event of selecting a double-headed coin. - Let \( H \) denote the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Using Bayes' Theorem Bayes' theorem provides a method for calculating conditional probabilities by relating prior probabilities to new evidence. The theorem can be expressed mathematically as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step 1: Calculate the Probabilities 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin is calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The prior probability of selecting a fair coin can be expressed as: \[ P(F) = \frac{3}{5} \] ### Step 2: Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads in all tosses. Therefore, the probability is: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of getting heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Step 3: Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components yields: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we get: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2 \cdot 16}{5 \cdot 7} = \frac{32}{35} \] ## Conclusion After performing the calculations using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a very high likelihood that the selected coin was double-headed, particularly considering the outcome of four consecutive heads, which is highly improbable if a fair coin were selected. This analysis illustrates the power of Bayes' theorem in updating our beliefs based on observed outcomes.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and each toss results in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a fundamental concept in probability that allows us to compute conditional probabilities by linking prior probabilities with new evidence. The theorem is expressed mathematically as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term gives: \[ P(H | D) \cdot P(D) = 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ P(H | F) \cdot P(F) = \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This result indicates a significant likelihood that the selected coin was indeed double-headed, especially considering the outcome of four consecutive heads, which would be highly improbable with a fair coin. This highlights the effectiveness of Bayes' theorem in updating our beliefs based on observed data.

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The task is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed, given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem is a foundational concept in probability theory that allows us to calculate conditional probabilities by relating prior probabilities to the evidence we observe. The theorem can be formulated as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is given by: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To calculate \( P(H) \), we use the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating this gives: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This implies a significant likelihood that the selected coin was double-headed, especially considering the highly unlikely event of obtaining four consecutive heads with a fair coin. This solution illustrates the power of Bayes' theorem in updating probabilities in light of new information.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This leads to a total of 5 coins. A coin is randomly chosen and tossed 4 times, yielding heads on every toss. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed, given that all tosses resulted in heads. ## Using Bayes' Theorem Bayes' theorem is a vital statistical tool for calculating conditional probabilities. The theorem can be stated as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components gives: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a very high likelihood that the selected coin was double-headed, particularly given the outcome of four consecutive heads, which would be highly improbable with a fair coin. This situation illustrates how Bayes' theorem effectively updates our understanding based on observed results.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and all tosses result in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is an essential statistical method for calculating conditional probabilities. It connects prior probabilities with observed evidence. The theorem can be expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This result indicates a significant likelihood that the selected coin was double-headed, especially considering the outcome of four consecutive heads, which is highly improbable if the coin were fair. This demonstrates the effectiveness of Bayes' theorem in updating our beliefs based on observed data.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, with all tosses resulting in heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a statistical method that allows us to calculate conditional probabilities by relating prior probabilities to observed evidence. The theorem can be expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion Upon thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a significant likelihood that the selected coin was indeed double-headed, particularly in light of the outcome of four consecutive heads, which is highly improbable with a fair coin. This scenario illustrates the power of Bayes' theorem in updating our beliefs based on observed data.

# Problem: Determining the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, yielding heads on every toss. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The task is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a statistical method for determining conditional probabilities by linking prior probabilities with observed evidence. The formula is expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components gives: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem yields: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This result highlights a strong likelihood that the selected coin was indeed double-headed, particularly given the outcome of four consecutive heads, which would be highly improbable with a fair coin. This exercise demonstrates the effectiveness of Bayes' theorem in updating our beliefs based on new information.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss resulting in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The goal is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a fundamental principle in probability that allows us to compute conditional probabilities by linking prior probabilities with observed data. The theorem can be expressed as follows: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values yields: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components gives: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This result indicates a very high likelihood that the selected coin was double-headed, especially considering the outcome of four consecutive heads, which would be highly improbable with a fair coin. This illustrates the effectiveness of Bayes' theorem in updating probabilities based on new evidence.

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A randomly selected coin is tossed 4 times, and each toss results in heads. ## Definitions - Let D denote the event of selecting a double-headed coin. - Let H denote the event of obtaining heads in all 4 tosses. The goal is to calculate \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Applying Bayes' Theorem Bayes' theorem is a crucial tool in probability theory that allows us to find conditional probabilities by linking prior probabilities with new evidence. The theorem is mathematically represented as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin is calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of getting heads on a single toss is \( \frac{1}{2} \). Thus, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a strong likelihood that the selected coin was indeed double-headed, especially considering the outcome of four consecutive heads, which would be quite improbable with a fair coin. This scenario highlights the effectiveness of Bayes' theorem in updating our beliefs based on observed outcomes.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This gives us a total of 5 coins. A randomly selected coin is tossed 4 times, and each toss results in heads. ## Definitions - Let \( D \) represent the event of selecting a double-headed coin. - Let \( H \) represent the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is a crucial tool in probability theory that allows us to update our beliefs about the likelihood of an event based on new evidence. The theorem can be mathematically expressed as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads on each toss: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** When a fair coin is selected, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we can use the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives us: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components results in: 1. The first term yields: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this gives: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After conducting a thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This result indicates a very high likelihood that the selected coin was indeed double-headed. This conclusion is particularly strong considering the outcome of four consecutive heads, which would be extremely unlikely with a fair coin. Such an analysis emphasizes the effectiveness of Bayes' theorem in updating our beliefs based on observed data, showcasing its application in real-world scenarios involving uncertainty and conditional probabilities.

# Problem: Calculating the Probability of Selecting a Double-Headed Coin ## Given Information A box contains: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with all results being heads. ## Definitions - Let D represent the event of selecting a double-headed coin. - Let H represent the event of obtaining heads in all 4 tosses. The objective is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem is an essential statistical tool that allows us to calculate conditional probabilities by connecting prior probabilities with observed evidence. The theorem can be expressed mathematically as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Step-by-Step Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin can be calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we apply the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term gives: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, convert \( \frac{2}{5} \) to a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we can express \( P(H) \) as: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Now substituting back into Bayes' theorem gives: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we find that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This suggests a very high likelihood that the selected coin is double-headed, particularly in light of the outcome of four consecutive heads, which would be exceedingly unlikely if the coin were fair. This example illustrates the practical application of Bayes' theorem in real-world scenarios, allowing us to update our understanding based on observed events.

# Problem: Evaluating the Probability of Selecting a Double-Headed Coin ## Given Information In a box, there are: - 3 fair coins - 2 double-headed coins This results in a total of 5 coins. A coin is randomly selected and tossed 4 times, with each toss yielding heads. ## Definitions - Let \( D \) represent the event of selecting a double-headed coin. - Let \( H \) represent the event of obtaining heads in all 4 tosses. The goal is to compute \( P(D | H) \), the conditional probability that the selected coin is double-headed given that all tosses resulted in heads. ## Utilizing Bayes' Theorem Bayes' theorem provides a systematic approach to calculating conditional probabilities by connecting prior knowledge with new evidence. The formula is stated as: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} \] ### Probability Calculations 1. **Prior Probability of Selecting a Double-Headed Coin:** The prior probability of selecting a double-headed coin is calculated as follows: \[ P(D) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{2}{5} \] 2. **Prior Probability of Selecting a Fair Coin:** The probability of selecting a fair coin is: \[ P(F) = \frac{3}{5} \] ### Conditional Probabilities - **Probability of Heads with a Double-Headed Coin:** If a double-headed coin is selected, it will always yield heads: \[ P(H | D) = 1 \] - **Probability of Heads with a Fair Coin:** For a fair coin, the probability of heads on a single toss is \( \frac{1}{2} \). Therefore, the probability of obtaining heads in all 4 tosses is: \[ P(H | F) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \] ### Total Probability of Heads To find \( P(H) \), we use the law of total probability: \[ P(H) = P(H | D) \cdot P(D) + P(H | F) \cdot P(F) \] Substituting the known values gives: \[ P(H) = 1 \cdot \frac{2}{5} + \frac{1}{16} \cdot \frac{3}{5} \] Calculating the components: 1. The first term results in: \[ 1 \cdot \frac{2}{5} = \frac{2}{5} \] 2. The second term calculates as: \[ \frac{1}{16} \cdot \frac{3}{5} = \frac{3}{80} \] Now, combining both components gives: \[ P(H) = \frac{2}{5} + \frac{3}{80} \] To add these fractions, we convert \( \frac{2}{5} \) into a fraction with a denominator of 80: \[ \frac{2}{5} = \frac{32}{80} \] Thus, we have: \[ P(H) = \frac{32}{80} + \frac{3}{80} = \frac{35}{80} = \frac{7}{16} \] ### Final Calculation Using Bayes' Theorem Substituting back into Bayes' theorem provides: \[ P(D | H) = \frac{P(H | D) \cdot P(D)}{P(H)} = \frac{1 \cdot \frac{2}{5}}{\frac{7}{16}} \] This simplifies to: \[ P(D | H) = \frac{\frac{2}{5}}{\frac{7}{16}} = \frac{2}{5} \cdot \frac{16}{7} \] Calculating this results in: \[ P(D | H) = \frac{32}{35} \] ## Conclusion After thorough analysis using Bayes' theorem, we conclude that the probability that the chosen coin is double-headed, given that all tosses resulted in heads, is \( \frac{32}{35} \). This indicates a high likelihood that the selected coin was indeed double-headed, especially given that four consecutive heads would be extremely improbable with a fair coin. This scenario highlights the utility of Bayes' theorem in updating our beliefs based on observed outcomes.