# Given Information - Three machines produce items: - Machine 1: 20% of items - Machine 2: 30% of items - Machine 3: 50% of items - Defect rates: - Machine 1: 2% - Machine 2: 3% - Machine 3: 5% # What to Find If an item is defective, what is the probability it came from machine 3? # Definition / Concept Used **Bayes' Theorem** is used to find the probability that a source (Machine 3) is responsible given an observed outcome (defective item). Let: - \( M_i \): item from machine \( i \) - \( D \): item is defective Bayes' theorem: \[ P(M_3 | D) = \frac{P(D|M_3) P(M_3)}{P(D)} \] # Solution Let: - \( P(M_1) = .20 \), \( P(M_2) = .30 \), \( P(M_3) = .50 \) - \( P(D|M_1) = .02 \), \( P(D|M_2) = .03 \), \( P(D|M_3) = .05 \) First, calculate the total probability an item is defective: \[ P(D) = P(D|M_1)P(M_1) + P(D|M_2)P(M_2) + P(D|M_3)P(M_3) \] \[ P(D) = (.02 \times .20) + (.03 \times .30) + (.05 \times .50) \] \[ P(D) = .004 + .009 + .025 = .038 \] Now, use Bayes' theorem: \[ P(M_3|D) = \frac{P(D|M_3) P(M_3)}{P(D)} \] \[ P(M_3|D) = \frac{.05 \times .50}{.038} = \frac{.025}{.038} \] \[ P(M_3|D) \approx .658 \] # Summary The probability that a defective item came from machine 3 is approximately **.658** (or **65.8%**).

# Given Information - **Machines and Production Percentages:** - Machine 1: 20% of items - Machine 2: 30% of items - Machine 3: 50% of items - **Defect Rates:** - Machine 1: 2% - Machine 2: 3% - Machine 3: 5% # What to Find Determine the probability that a defective item originated from machine 3. # Definition / Concept Used **Bayes' Theorem** is applied to compute the conditional probability of an event based on prior knowledge of conditions related to the event. The formula is given by: \[ P(M_3 | D) = \frac{P(D|M_3) P(M_3)}{P(D)} \] Where: - \( M_i \): item from machine \( i \) - \( D \): item is defective # Solution 1. Define the probabilities for each machine and the defect rates: - \( P(M_1) = 0.20 \) - \( P(M_2) = 0.30 \) - \( P(M_3) = 0.50 \) - \( P(D|M_1) = 0.02 \) - \( P(D|M_2) = 0.03 \) - \( P(D|M_3) = 0.05 \) 2. Calculate the total probability that an item is defective: \[ P(D) = P(D|M_1)P(M_1) + P(D|M_2)P(M_2) + P(D|M_3)P(M_3) \] \[ P(D) = (0.02 \times 0.20) + (0.03 \times 0.30) + (0.05 \times 0.50) \] \[ P(D) = 0.004 + 0.009 + 0.025 = 0.038 \] 3. Apply Bayes' theorem to find \( P(M_3|D) \): \[ P(M_3|D) = \frac{P(D|M_3) P(M_3)}{P(D)} \] \[ P(M_3|D) = \frac{0.05 \times 0.50}{0.038} = \frac{0.025}{0.038} \] \[ P(M_3|D) \approx 0.658 \] # Summary The probability that a defective item came from machine 3 is approximately **0.658** (or **65.8%**).

# Given Information - **Production Distribution by Machines:** - Machine 1: 20% of total items - Machine 2: 30% of total items - Machine 3: 50% of total items - **Defect Rates:** - Machine 1: 2% - Machine 2: 3% - Machine 3: 5% # Objective Calculate the probability that a defective item came from machine 3. # Concept Used **Bayes' Theorem** is employed to find the conditional probability of an event given prior knowledge. The theorem is expressed as: \[ P(M_3 | D) = \frac{P(D|M_3) P(M_3)}{P(D)} \] Where: - \( M_i \): item from machine \( i \) - \( D \): event that an item is defective # Solution Steps 1. Define the probabilities: - \( P(M_1) = 0.20 \) - \( P(M_2) = 0.30 \) - \( P(M_3) = 0.50 \) - \( P(D|M_1) = 0.02 \) - \( P(D|M_2) = 0.03 \) - \( P(D|M_3) = 0.05 \) 2. Calculate the total probability of an item being defective: \[ P(D) = P(D|M_1)P(M_1) + P(D|M_2)P(M_2) + P(D|M_3)P(M_3) \] Breaking it down: \[ P(D) = (0.02 \times 0.20) + (0.03 \times 0.30) + (0.05 \times 0.50) \] Calculating each component: \[ P(D) = 0.004 + 0.009 + 0.025 = 0.038 \] 3. Use Bayes' theorem to find the probability that the defective item came from machine 3: \[ P(M_3|D) = \frac{P(D|M_3) P(M_3)}{P(D)} \] Substituting the known values: \[ P(M_3|D) = \frac{0.05 \times 0.50}{0.038} = \frac{0.025}{0.038} \] Perform the final calculation: \[ P(M_3|D) \approx 0.658 \] # Summary The probability that a defective item originated from machine 3 is approximately **0.658** (or **65.8%**).

# Given Information - **Total Balls:** - Red Balls: 4 - Blue Balls: 6 - **Total Balls in the Box:** 10 (4 red + 6 blue) # Objective Calculate the probability that both balls drawn are red, given that at least one is red. # Concept Used **Conditional Probability** is employed to find the probability of an event given that another event has occurred. The formula is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): event that both balls drawn are red - \( B \): event that at least one ball is red # Solution Steps 1. **Calculate \( P(B) \)**: Probability that at least one ball is red. - Total ways to choose 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2} = 45 \] - Total ways to choose 2 blue balls (no red): \[ \text{Combinations of blue} = \binom{6}{2} = \frac{6 \times 5}{2} = 15 \] - Therefore, the probability of choosing at least one red ball: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} = \frac{2}{3} \] 2. **Calculate \( P(A \cap B) \)**: Probability that both balls drawn are red. - Total ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} \] 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{6}{45}}{\frac{30}{45}} = \frac{6}{30} = \frac{1}{5} \] # Summary The probability that both balls drawn are red, given that at least one is red, is **\(\frac{1}{5}\)** (or **20%**).

# Given Information - **Composition of Balls:** - Red Balls: 4 - Blue Balls: 6 - **Total Balls:** 10 (4 red + 6 blue) # Objective Determine the probability that both balls drawn are red, given that at least one ball is red. # Concept Used **Conditional Probability** is used to find the likelihood of an event occurring given that another related event has already occurred. The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): event that both balls drawn are red - \( B \): event that at least one ball drawn is red # Solution Steps 1. **Calculate \( P(B) \)**: The probability that at least one ball is red. - Total ways to choose 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2} = 45 \] - Total ways to choose 2 blue balls (no red): \[ \text{Combinations of blue} = \binom{6}{2} = \frac{6 \times 5}{2} = 15 \] - Therefore, the probability of having at least one red ball is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} = \frac{2}{3} \] 2. **Calculate \( P(A \cap B) \)**: The probability that both balls drawn are red. - Total ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} \] (Since choosing 2 red balls is the only way to satisfy both A and B.) 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{6}{45}}{\frac{30}{45}} = \frac{6}{30} = \frac{1}{5} \] # Summary The probability that both balls drawn are red, given that at least one is red, is **\(\frac{1}{5}\)** (or **20%**).

# Given Information - **Count of Balls:** - Red Balls: 4 - Blue Balls: 6 - **Total Balls:** 10 (4 red + 6 blue) # Objective Find the probability that both balls drawn are red, under the condition that at least one of them is red. # Concept Used **Conditional Probability** is applied to calculate the probability of an event given that another event has occurred. The formula is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): the event that both balls are red - \( B \): the event that at least one ball is red # Solution Steps 1. **Calculate \( P(B) \)**: Probability of drawing at least one red ball. - Total ways to choose 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2} = 45 \] - Total ways to draw 2 blue balls (no red): \[ \text{Combinations of blue} = \binom{6}{2} = \frac{6 \times 5}{2} = 15 \] - Therefore, the probability of drawing at least one red ball: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} = \frac{2}{3} \] 2. **Calculate \( P(A \cap B) \)**: Probability that both balls drawn are red. - Total ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} \] (This is equivalent to \( P(A \cap B) \) since both must be red to satisfy B.) 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{6}{45}}{\frac{30}{45}} = \frac{6}{30} = \frac{1}{5} \] # Summary The probability that both balls drawn are red, given that at least one is red, is **\(\frac{1}{5}\)** (or **20%**).

# Given Information - Probability of Event A: \( P(A) = 0.5 \) - Probability of Event B: \( P(B) = 0.4 \) - Probability of both A and B occurring: \( P(A \cap B) = 0.2 \) # Objective Determine if events A and B are independent or mutually exclusive. # Definitions - **Independent Events**: Events A and B are independent if: \[ P(A \cap B) = P(A) \cdot P(B) \] - **Mutually Exclusive Events**: Events A and B are mutually exclusive if: \[ P(A \cap B) = 0 \] # Solution Steps 1. **Check for Independence**: - Calculate \( P(A) \cdot P(B) \): \[ P(A) \cdot P(B) = 0.5 \times 0.4 = 0.2 \] - Compare with \( P(A \cap B) \): \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) = P(A) \cdot P(B) \), events A and B are **independent**. 2. **Check for Mutual Exclusivity**: - Determine if \( P(A \cap B) = 0 \): \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) \neq 0 \), events A and B are **not mutually exclusive**. # Summary - Events A and B are **independent** because \( P(A \cap B) = P(A) \cdot P(B) \). - Events A and B are **not mutually exclusive** since \( P(A \cap B) \neq 0 \).

# Given Information - **Probabilities:** - Probability of Event A: \( P(A) = 0.5 \) - Probability of Event B: \( P(B) = 0.4 \) - Probability of both A and B occurring: \( P(A \cap B) = 0.2 \) # Objective Determine whether events A and B are independent or mutually exclusive. # Definitions - **Independent Events**: Events A and B are independent if: \[ P(A \cap B) = P(A) \cdot P(B) \] - **Mutually Exclusive Events**: Events A and B are mutually exclusive if: \[ P(A \cap B) = 0 \] # Solution Steps 1. **Check for Independence**: - Calculate the product of the probabilities of A and B: \[ P(A) \cdot P(B) = 0.5 \times 0.4 = 0.20 \] - Compare with \( P(A \cap B) \): \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) = P(A) \cdot P(B) \), we conclude that events A and B are **independent**. 2. **Check for Mutual Exclusivity**: - Assess if \( P(A \cap B) = 0 \): \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) \neq 0 \), events A and B are **not mutually exclusive**. # Summary - Events A and B are **independent** because \( P(A \cap B) = P(A) \cdot P(B) \). - Events A and B are **not mutually exclusive** since \( P(A \cap B) \neq 0 \).

# Given Information - **Production Distribution by Machines:** - Machine A: 20% of bulbs produced - Machine B: 30% of bulbs produced - Machine C: 50% of bulbs produced - **Defect Rates:** - Machine A: 1% defective - Machine B: 2% defective - Machine C: 3% defective # Objective Calculate the probability that a defective bulb was produced by Machine C. # Concept Used **Bayes' Theorem** is used to find the conditional probability given that an event has occurred. The formula is: \[ P(C | D) = \frac{P(D | C) P(C)}{P(D)} \] Where: - \( C \): event that the bulb is from Machine C - \( D \): event that the bulb is defective # Solution Steps 1. **Define the probabilities:** - \( P(A) = 0.20 \) - \( P(B) = 0.30 \) - \( P(C) = 0.50 \) - \( P(D | A) = 0.01 \) - \( P(D | B) = 0.02 \) - \( P(D | C) = 0.03 \) 2. **Calculate the total probability of a bulb being defective \( P(D) \):** \[ P(D) = P(D | A) P(A) + P(D | B) P(B) + P(D | C) P(C) \] Calculating each term: \[ P(D) = (0.01 \times 0.20) + (0.02 \times 0.30) + (0.03 \times 0.50) \] \[ P(D) = 0.002 + 0.006 + 0.015 = 0.023 \] 3. **Apply Bayes' theorem to find \( P(C | D) \):** \[ P(C | D) = \frac{P(D | C) P(C)}{P(D)} = \frac{0.03 \times 0.50}{0.023} \] Calculating: \[ P(C | D) = \frac{0.015}{0.023} \approx 0.652 \] # Summary The probability that a defective bulb was produced by Machine C is approximately **0.652** (or **65.2%**).

# Given Information - **Probabilities:** - Probability of Event A: \( P(A) = 0.5 \) - Probability of Event B: \( P(B) = 0.4 \) - Probability of both A and B occurring: \( P(A \cap B) = 0.2 \) # Objective Determine if events A and B are independent or mutually exclusive. # Definitions - **Independent Events**: Events A and B are independent if: \[ P(A \cap B) = P(A) \cdot P(B) \] - **Mutually Exclusive Events**: Events A and B are mutually exclusive if: \[ P(A \cap B) = 0 \] # Solution Steps 1. **Check for Independence**: - Calculate \( P(A) \cdot P(B) \): \[ P(A) \cdot P(B) = 0.5 \times 0.4 = 0.2 \] - Compare with \( P(A \cap B) \): \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) = P(A) \cdot P(B) \), events A and B are **independent**. 2. **Check for Mutual Exclusivity**: - Assess if \( P(A \cap B) = 0 \): \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) \neq 0 \), events A and B are **not mutually exclusive**. # Summary - Events A and B are **independent** because \( P(A \cap B) = P(A) \cdot P(B) \). - Events A and B are **not mutually exclusive** since \( P(A \cap B) \neq 0 \). --- ### Explanation for Students Understanding the difference between independence and mutual exclusivity is crucial in probability. - **Independence** means that the occurrence of one event does not affect the occurrence of another. In our case, knowing that event A happened does not change the probability of event B happening, and vice versa. - **Mutual Exclusivity** indicates that two events cannot happen at the same time. For example, if event A occurs, event B cannot occur; hence their intersection probability would be zero. In this scenario, since \( P(A \cap B) \) is not zero and matches the product of their individual probabilities, we can confidently conclude that A and B are independent but not mutually exclusive. If you have any further questions or need clarification, feel free to ask!

# Given Information - **Probabilities:** - Probability of Event A: \( P(A) = 0.5 \) - Probability of Event B: \( P(B) = 0.4 \) - Probability of both A and B occurring: \( P(A \cap B) = 0.2 \) # Objective Determine whether events A and B are independent and whether they are mutually exclusive. # Definitions - **Independent Events**: Events A and B are independent if: \[ P(A \cap B) = P(A) \cdot P(B) \] - **Mutually Exclusive Events**: Events A and B are mutually exclusive if: \[ P(A \cap B) = 0 \] # Solution Steps To check for independence, first calculate the product of the probabilities of events A and B: \[ P(A) \cdot P(B) = 0.5 \times 0.4 = 0.2 \] Now, compare this result to the probability of both events occurring: \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) = P(A) \cdot P(B) \) holds true, events A and B are classified as **independent**. Next, assess whether the events are mutually exclusive. For events to be mutually exclusive, their intersection must equal zero: \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) \neq 0 \), events A and B cannot occur at the same time, indicating that they are **not mutually exclusive**. # Summary - Events A and B are **independent** because \( P(A \cap B) = P(A) \cdot P(B) \). - Events A and B are **not mutually exclusive** since \( P(A \cap B) \neq 0 \). ## Explanation for Students Understanding the distinctions between independent and mutually exclusive events is essential in probability theory: 1. **Independence** signifies that the occurrence of one event does not influence the likelihood of the other event occurring. For example, if event A occurs, it does not affect the probability of event B happening. 2. **Mutual Exclusivity** denotes that two events cannot happen simultaneously. If one event occurs, the other must not. This means that the intersection of their probabilities is zero. In this scenario, since the events A and B can occur together (as shown by the non-zero intersection) but do not affect each other's probabilities, they are independent but not mutually exclusive. If you have further questions or need additional clarification, feel free to ask!

# Given Information - **Machines and Production Percentages:** - Machine 1: 20% of items - Machine 2: 30% of items - Machine 3: 50% of items - **Defect Rates:** - Machine 1: 2% defective - Machine 2: 3% defective - Machine 3: 5% defective # Objective Calculate the probability that a defective item was produced by Machine 3. # Definition / Concept Used **Bayes' Theorem** is used to find the conditional probability of an event given the occurrence of another event. The theorem is expressed as: \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} \] Where: - \( M_3 \): event that the item is from Machine 3 - \( D \): event that the item is defective # Solution Steps 1. **Define the probabilities for each machine and their defect rates:** - \( P(M_1) = 0.20 \) - \( P(M_2) = 0.30 \) - \( P(M_3) = 0.50 \) - \( P(D | M_1) = 0.02 \) - \( P(D | M_2) = 0.03 \) - \( P(D | M_3) = 0.05 \) 2. **Calculate the total probability of an item being defective \( P(D) \):** \[ P(D) = P(D | M_1) P(M_1) + P(D | M_2) P(M_2) + P(D | M_3) P(M_3) \] Breaking this down: \[ P(D) = (0.02 \times 0.20) + (0.03 \times 0.30) + (0.05 \times 0.50) \] Calculating each component: \[ P(D) = 0.004 + 0.009 + 0.025 = 0.038 \] 3. **Apply Bayes' theorem to find \( P(M_3 | D) \):** \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} = \frac{0.05 \times 0.50}{0.038} \] Calculating: \[ P(M_3 | D) = \frac{0.025}{0.038} \approx 0.658 \] # Summary The probability that a defective item was produced by Machine 3 is approximately **0.658** (or **65.8%**).

# Given Information - **Production Distribution by Machines:** - Machine 1: 20% of items produced - Machine 2: 30% of items produced - Machine 3: 50% of items produced - **Defect Rates:** - Machine 1: 2% defective - Machine 2: 3% defective - Machine 3: 5% defective # Objective Determine the probability that a defective item originated from Machine 3. # Concept Used **Bayes' Theorem** is the primary tool used to calculate the conditional probability of an event based on prior knowledge. The formula is: \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} \] Where: - \( M_3 \): event that the bulb is from Machine 3 - \( D \): event that the bulb is defective # Solution Steps 1. **Define the probabilities for each machine and their defect rates:** - \( P(M_1) = 0.20 \) - \( P(M_2) = 0.30 \) - \( P(M_3) = 0.50 \) - \( P(D | M_1) = 0.02 \) - \( P(D | M_2) = 0.03 \) - \( P(D | M_3) = 0.05 \) 2. **Calculate the total probability of a bulb being defective \( P(D) \):** \[ P(D) = P(D | M_1) P(M_1) + P(D | M_2) P(M_2) + P(D | M_3) P(M_3) \] Calculating each term: \[ P(D) = (0.02 \times 0.20) + (0.03 \times 0.30) + (0.05 \times 0.50) \] \[ P(D) = 0.004 + 0.009 + 0.025 = 0.038 \] 3. **Use Bayes' theorem to find \( P(M_3 | D) \):** \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} = \frac{0.05 \times 0.50}{0.038} \] Calculating the numerator: \[ P(M_3 | D) = \frac{0.025}{0.038} \] Calculating the final value: \[ P(M_3 | D) \approx 0.6579 \approx 0.658 \] # Summary The probability that a defective item was produced by Machine 3 is approximately **0.658** (or **65.8%**).

# Given Information - **Group A:** - Number of students: \( n_A = 40 \) - Average score: \( \bar{x}_A = 60 \) - **Group B:** - Number of students: \( n_B = 60 \) - Average score: \( \bar{x}_B = 75 \) # Objective Calculate the combined mean score of the two groups. # Concept Used The combined mean (\( \bar{x}_{combined} \)) of two groups can be calculated using the formula: \[ \bar{x}_{combined} = \frac{n_A \cdot \bar{x}_A + n_B \cdot \bar{x}_B}{n_A + n_B} \] # Solution Steps 1. **Calculate the total score for each group:** - Total score for Group A: \[ \text{Total score}_A = n_A \cdot \bar{x}_A = 40 \cdot 60 = 2400 \] - Total score for Group B: \[ \text{Total score}_B = n_B \cdot \bar{x}_B = 60 \cdot 75 = 4500 \] 2. **Calculate the combined total score:** \[ \text{Total score}_{combined} = \text{Total score}_A + \text{Total score}_B = 2400 + 4500 = 6900 \] 3. **Calculate the total number of students:** \[ n_{combined} = n_A + n_B = 40 + 60 = 100 \] 4. **Calculate the combined mean:** \[ \bar{x}_{combined} = \frac{\text{Total score}_{combined}}{n_{combined}} = \frac{6900}{100} = 69 \] # Summary The combined mean score of Group A and Group B is **69**.

# Given Information - **Group A:** - Number of students: \( n_A = 40 \) - Average score: \( \bar{x}_A = 60 \) - **Group B:** - Number of students: \( n_B = 60 \) - Average score: \( \bar{x}_B = 75 \) # Objective Determine the combined mean score of both groups. # Concept Used The combined mean (\( \bar{x}_{combined} \)) of two groups is calculated using the formula: \[ \bar{x}_{combined} = \frac{n_A \cdot \bar{x}_A + n_B \cdot \bar{x}_B}{n_A + n_B} \] # Solution Steps 1. **Calculate the total score for each group:** - For Group A: \[ \text{Total score}_A = n_A \cdot \bar{x}_A = 40 \cdot 60 = 2400 \] - For Group B: \[ \text{Total score}_B = n_B \cdot \bar{x}_B = 60 \cdot 75 = 4500 \] 2. **Calculate the total score for both groups:** \[ \text{Total score}_{combined} = \text{Total score}_A + \text{Total score}_B = 2400 + 4500 = 6900 \] 3. **Determine the total number of students:** \[ n_{combined} = n_A + n_B = 40 + 60 = 100 \] 4. **Calculate the combined mean score:** \[ \bar{x}_{combined} = \frac{\text{Total score}_{combined}}{n_{combined}} = \frac{6900}{100} = 69 \] # Summary The combined mean score of Group A and Group B is **69**.

# Given Information - **Probabilities:** - Probability of Event A: \( P(A) = 0.5 \) - Probability of Event B: \( P(B) = 0.4 \) - Probability of both A and B occurring: \( P(A \cap B) = 0.2 \) # Objective Determine if events A and B are independent and if they are mutually exclusive. # Definitions - **Independent Events**: Events A and B are independent if: \[ P(A \cap B) = P(A) \cdot P(B) \] - **Mutually Exclusive Events**: Events A and B are mutually exclusive if: \[ P(A \cap B) = 0 \] # Solution Steps To check for independence, calculate the product of the probabilities of events A and B: \[ P(A) \cdot P(B) = 0.5 \times 0.4 = 0.2 \] Now, compare this result to \( P(A \cap B) \): \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) = P(A) \cdot P(B) \), events A and B are classified as **independent**. Next, assess whether the events are mutually exclusive. For events to be mutually exclusive, their intersection must equal zero: \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) \neq 0 \), events A and B cannot occur at the same time, indicating that they are **not mutually exclusive**. # Summary - Events A and B are **independent** because \( P(A \cap B) = P(A) \cdot P(B) \). - Events A and B are **not mutually exclusive** since \( P(A \cap B) \neq 0 \). ## Explanation for Students Understanding the difference between independence and mutual exclusivity is crucial in probability: 1. **Independence** means that the occurrence of one event does not affect the occurrence of another. In this case, knowing that event A happened does not influence the probability of event B occurring. 2. **Mutual Exclusivity** indicates that two events cannot happen at the same time. If one event occurs, the other must not. Thus, their intersection probability is zero. In this scenario, A and B can happen together (as indicated by the non-zero intersection), yet they do not influence each other's probabilities. Hence, they are independent but not mutually exclusive. If you have further questions or need clarification, feel free to ask!

# Given Information - **Probabilities:** - Probability of Event A: \( P(A) = 0.5 \) - Probability of Event B: \( P(B) = 0.4 \) - Probability of both A and B occurring: \( P(A \cap B) = 0.2 \) # Objective Evaluate whether events A and B are independent and whether they are mutually exclusive. # Definitions - **Independent Events**: Events A and B are considered independent if: \[ P(A \cap B) = P(A) \cdot P(B) \] - **Mutually Exclusive Events**: Events A and B are mutually exclusive if: \[ P(A \cap B) = 0 \] # Solution Steps To check the independence of events A and B, start by calculating the product of their individual probabilities: \[ P(A) \cdot P(B) = 0.5 \times 0.4 = 0.2 \] Next, compare this result with the probability of both events occurring: \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) \) equals \( P(A) \cdot P(B) \), we conclude that events A and B are **independent**. Now, to determine if the events are mutually exclusive, we need to check if their intersection is zero: \[ P(A \cap B) = 0.2 \] Since \( P(A \cap B) \) is not equal to zero, it follows that events A and B are **not mutually exclusive**. # Summary - Events A and B are **independent** because \( P(A \cap B) = P(A) \cdot P(B) \). - Events A and B are **not mutually exclusive** since \( P(A \cap B) \neq 0 \). ## Explanation for Students Understanding the distinctions between independent and mutually exclusive events is fundamental in probability theory: 1. **Independence** signifies that the occurrence of one event does not impact the probability of the other event happening. In this case, knowing that event A has occurred does not change the likelihood of event B occurring. 2. **Mutual Exclusivity** indicates that two events cannot occur simultaneously. If one event takes place, the other cannot. This would mean their joint probability is zero. In this situation, A and B can occur together (as demonstrated by the non-zero intersection), and their occurrence does not affect each other’s probabilities. Thus, they are independent but not mutually exclusive. If you have any questions or need further clarification, feel free to reach out!

# Given Information - **Composition of Balls:** - Red Balls: 4 - Blue Balls: 6 - **Total Balls:** 10 (4 red + 6 blue) # Objective Calculate the probability that both balls drawn are red, given that at least one of them is red. # Concept Used **Conditional Probability** is applied here to determine the likelihood of an event occurring given that another event has taken place. The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): the event that both balls drawn are red - \( B \): the event that at least one ball drawn is red # Solution Steps 1. **Calculate \( P(B) \)**: The probability that at least one ball is red. - Total combinations of drawing 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2} = 45 \] - Total combinations of drawing 2 blue balls (no red): \[ \text{Combinations of blue} = \binom{6}{2} = \frac{6 \times 5}{2} = 15 \] - Thus, the probability of drawing at least one red ball: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} = \frac{2}{3} \] 2. **Calculate \( P(A \cap B) \)**: The probability that both balls drawn are red. - Total ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} \] (This is also \( P(A \cap B) \) since both must be red to satisfy B.) 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{6}{45}}{\frac{30}{45}} = \frac{6}{30} = \frac{1}{5} \] # Summary The probability that both balls drawn are red, given that at least one is red, is **\(\frac{1}{5}\)** (or **20%**).

# Given Information - **Composition of Balls:** - Red Balls: 4 - Blue Balls: 6 - **Total Balls:** 10 (4 red + 6 blue) # Objective Find the probability that both balls drawn are red, given that at least one ball drawn is red. # Concept Used **Conditional Probability** helps in determining the likelihood of an event based on the occurrence of another event. The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): the event that both drawn balls are red - \( B \): the event that at least one ball drawn is red # Solution Steps 1. **Calculate \( P(B) \)**: The probability that at least one ball drawn is red. - Total ways to choose 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = \frac{10 \times 9}{2} = 45 \] - Total ways to choose 2 blue balls (no red): \[ \text{Combinations of blue} = \binom{6}{2} = \frac{6 \times 5}{2} = 15 \] - Therefore, the probability of drawing at least one red ball: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} = \frac{2}{3} \] 2. **Calculate \( P(A \cap B) \)**: The probability that both balls drawn are red. - Total ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} \] (This is also \( P(A \cap B) \) since both balls must be red to fulfill the condition of event B.) 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{6}{45}}{\frac{30}{45}} = \frac{6}{30} = \frac{1}{5} \] # Summary The probability that both balls drawn are red, given that at least one is red, is **\(\frac{1}{5}\)** (or **20%**). ## Explanation In this scenario, we first calculated the probability of the event of interest (drawing at least one red ball) before determining the likelihood of drawing two red balls. By using the principles of conditional probability, we were able to arrive at the conclusion clearly and logically. If you have any further questions or need more details, feel free to ask!

# Given Information - **Composition of Balls:** - Red Balls: 4 - Blue Balls: 6 - **Total Balls:** 10 (4 red + 6 blue) # Objective Determine the probability that both balls drawn are red, given that at least one ball drawn is red. # Concept Used **Conditional Probability** is applied to find the likelihood of an event occurring based on the occurrence of another event. The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): the event that both drawn balls are red - \( B \): the event that at least one ball drawn is red # Solution Steps 1. **Calculate \( P(B) \)**: The probability that at least one ball drawn is red. - Total ways to choose 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = 45 \] - Total ways to choose 2 blue balls (no red): \[ \text{Combinations of blue} = \binom{6}{2} = 15 \] - Therefore, the probability of drawing at least one red ball: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} = \frac{2}{3} \approx 0.6667 \] 2. **Calculate \( P(A \cap B) \)**: The probability that both balls drawn are red. - Total ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} \approx 0.1333 \] (This is also \( P(A \cap B) \) since both balls must be red.) 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1333}{0.6667} \approx 0.2 \] # Summary The probability that both balls drawn are red, given that at least one is red, is approximately **0.2** (or **20%**).

# Given Information - **Composition of Balls:** - Number of Red Balls: 4 - Number of Blue Balls: 6 - **Total Number of Balls:** 10 (4 red + 6 blue) # Objective Calculate the probability that both balls drawn are red, given that at least one of them is red. # Concept Used The concept of **Conditional Probability** is used to evaluate the likelihood of an event occurring given that another event has already occurred. The formula for conditional probability is expressed as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): the event that both balls drawn are red - \( B \): the event that at least one ball drawn is red # Solution Steps 1. **Calculate \( P(B) \)**: The probability that at least one ball drawn is red. - Total ways to select 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = 45 \] - Total ways to select 2 blue balls (which means no red balls): \[ \text{Combinations of blue} = \binom{6}{2} = 15 \] - Thus, the probability of drawing at least one red ball becomes: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} \approx 0.6667 \] 2. **Calculate \( P(A \cap B) \)**: The probability that both balls drawn are red. - The number of ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} \approx 0.1333 \] (This probability is also \( P(A \cap B) \) since both balls must be red for event B to hold true.) 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1333}{0.6667} \approx 0.2 \] # Summary The probability that both balls drawn are red, given that at least one is red, is approximately **0.2** (or **20%**).

# Given Information - **Ball Composition:** - Red Balls: 4 - Blue Balls: 6 - **Total Balls:** 10 (4 red + 6 blue) # Objective Determine the probability that both balls drawn are red, given that at least one of the balls drawn is red. # Concept Used The concept of **Conditional Probability** is utilized to assess the probability of an event occurring based on the occurrence of another event. The formula for conditional probability can be expressed as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): the event that both balls drawn are red - \( B \): the event that at least one ball drawn is red # Solution Steps 1. **Calculate \( P(B) \)**: The probability of drawing at least one red ball. - Total ways to choose 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = 45 \] - Total ways to select 2 blue balls (which means no red): \[ \text{Combinations of blue} = \binom{6}{2} = 15 \] - Thus, the probability of drawing at least one red ball is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} \approx 0.6667 \] 2. **Calculate \( P(A \cap B) \)**: The probability that both balls drawn are red. - The number of ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} \approx 0.1333 \] (This represents \( P(A \cap B) \) since both balls must be red to meet the condition of B.) 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1333}{0.6667} \approx 0.2 \] # Summary The probability that both balls drawn are red, given that at least one is red, is approximately **0.2** (or **20%**).

# Given Information - **Composition of Balls:** - Number of Red Balls: 4 - Number of Blue Balls: 6 - **Total Number of Balls:** 10 (4 red + 6 blue) # Objective Calculate the probability that both balls drawn are red, given that at least one of the balls drawn is red. # Concept Used The principle of **Conditional Probability** is applied here to evaluate the likelihood of an event occurring based on the occurrence of another event. The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): the event that both drawn balls are red - \( B \): the event that at least one ball drawn is red # Solution Steps 1. **Calculate \( P(B) \)**: The probability that at least one ball drawn is red. - The total number of ways to select 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = 45 \] - The total number of combinations to select 2 blue balls (which means no red balls): \[ \text{Combinations of blue} = \binom{6}{2} = 15 \] - Therefore, the probability of drawing at least one red ball is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} \approx 0.6667 \] 2. **Calculate \( P(A \cap B) \)**: The probability that both balls drawn are red. - The number of ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} \approx 0.1333 \] (This also represents \( P(A \cap B) \) since both balls must be red to satisfy the condition of event B.) 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1333}{0.6667} \approx 0.2 \] # Summary The probability that both balls drawn are red, given that at least one is red, is approximately **0.2** (or **20%**). This approach demonstrates how conditional probability helps refine the likelihood of specific outcomes based on prior knowledge about the events in question. If you have further inquiries or need additional explanation, feel free to ask!

# Given Information - **Ball Composition:** - Red Balls: 4 - Blue Balls: 6 - **Total Number of Balls:** 10 (4 red + 6 blue) # Objective Determine the probability that both balls drawn are red, given that at least one of the balls drawn is red. # Concept Used The concept of **Conditional Probability** is utilized to assess the likelihood of an event occurring based on the occurrence of another event. The formula for conditional probability is: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): event that both drawn balls are red - \( B \): event that at least one ball drawn is red # Solution Steps 1. **Calculate \( P(B) \)**: The probability of drawing at least one red ball. - Total ways to select 2 balls from 10: \[ \text{Total combinations} = \binom{10}{2} = 45 \] - Total ways to select 2 blue balls (meaning no red balls): \[ \text{Combinations of blue} = \binom{6}{2} = 15 \] - Therefore, the probability of drawing at least one red ball is: \[ P(B) = 1 - P(\text{both blue}) = 1 - \frac{15}{45} = \frac{30}{45} = 0.6667 \] 2. **Calculate \( P(A \cap B) \)**: The probability that both balls drawn are red. - The number of ways to choose 2 red balls: \[ P(A) = \frac{\binom{4}{2}}{\binom{10}{2}} = \frac{6}{45} = 0.1333 \] (This probability also represents \( P(A \cap B) \) since both balls must be red for event B to occur.) 3. **Calculate the conditional probability \( P(A | B) \)**: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1333}{0.6667} \approx 0.2 \] # Summary The probability that both balls drawn are red, given that at least one is red, is approximately **0.2** (or **20%**). This solution illustrates how conditional probability allows us to refine our understanding of outcomes based on prior conditions. Should you have any further questions or require additional clarification, feel free to ask!

# Given Information - **Production Distribution by Machines:** - Machine 1: 20% of items - Machine 2: 30% of items - Machine 3: 50% of items - **Defect Rates:** - Machine 1: 2% defective - Machine 2: 3% defective - Machine 3: 5% defective # Objective Calculate the probability that a defective item was produced by Machine 3, denoted as \( P(M_3 | D) \), where \( D \) is the event that an item is defective. # Concept Used **Bayes' Theorem** is used to determine the conditional probability of an event based on prior information. The formula is given by: \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} \] Where: - \( P(M_3) \): probability of selecting an item from Machine 3 - \( P(D | M_3) \): probability that an item is defective given it came from Machine 3 - \( P(D) \): total probability that an item is defective # Solution Steps 1. **Define the probabilities for each machine and their defect rates:** - \( P(M_1) = 0.20 \) - \( P(M_2) = 0.30 \) - \( P(M_3) = 0.50 \) - \( P(D | M_1) = 0.02 \) - \( P(D | M_2) = 0.03 \) - \( P(D | M_3) = 0.05 \) 2. **Calculate the total probability of an item being defective \( P(D) \):** \[ P(D) = P(D | M_1) P(M_1) + P(D | M_2) P(M_2) + P(D | M_3) P(M_3) \] Breaking this down: \[ P(D) = (0.02 \times 0.20) + (0.03 \times 0.30) + (0.05 \times 0.50) \] Calculating each term: \[ P(D) = 0.004 + 0.009 + 0.025 = 0.038 \] 3. **Apply Bayes' theorem to find \( P(M_3 | D) \):** \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} = \frac{0.05 \times 0.50}{0.038} \] Calculating the numerator: \[ P(M_3 | D) = \frac{0.025}{0.038} \approx 0.6579 \] # Summary The probability that a defective item was produced by Machine 3 is approximately **0.658** (or **65.8%**). --- ## Explanation In this solution, we first identified the probabilities associated with each machine and their respective defect rates. We then calculated the total probability of a defective item being produced using the law of total probability. Finally, we applied Bayes' Theorem to find the probability that a defective item came from Machine 3, given the calculated probabilities. If you have any questions or need further clarification, feel free to ask!

# Given Information - **Production Distribution by Machines:** - Machine 1: 20% of items produced - Machine 2: 30% of items produced - Machine 3: 50% of items produced - **Defect Rates:** - Machine 1: 2% defective - Machine 2: 3% defective - Machine 3: 5% defective # Objective Determine the probability that a defective item was produced by Machine 3, denoted as \( P(M_3 | D) \), where \( D \) is the event that an item is defective. # Concept Used **Bayes' Theorem** allows us to compute the conditional probability of an event given the occurrence of another event. The formula is: \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} \] Where: - \( P(M_3) \): probability of selecting an item from Machine 3 - \( P(D | M_3) \): probability that an item is defective given it came from Machine 3 - \( P(D) \): overall probability that an item is defective # Solution Steps 1. **Define the probabilities for each machine and their defect rates:** - \( P(M_1) = 0.20 \) - \( P(M_2) = 0.30 \) - \( P(M_3) = 0.50 \) - \( P(D | M_1) = 0.02 \) - \( P(D | M_2) = 0.03 \) - \( P(D | M_3) = 0.05 \) 2. **Calculate the total probability of an item being defective \( P(D) \):** \[ P(D) = P(D | M_1) P(M_1) + P(D | M_2) P(M_2) + P(D | M_3) P(M_3) \] Breaking this down: \[ P(D) = (0.02 \times 0.20) + (0.03 \times 0.30) + (0.05 \times 0.50) \] Calculating the individual components: \[ P(D) = 0.004 + 0.009 + 0.025 = 0.038 \] 3. **Apply Bayes' theorem to find \( P(M_3 | D) \):** \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} = \frac{0.05 \times 0.50}{0.038} \] Calculating the result: \[ P(M_3 | D) = \frac{0.025}{0.038} \approx 0.6579 \] # Summary The probability that a defective item was produced by Machine 3 is approximately **0.658** (or **65.8%**). --- ## Explanation In this solution, we began by identifying the production proportions and defect rates for each machine. We then utilized the law of total probability to calculate the overall defect probability. Finally, Bayes' Theorem was applied to ascertain the likelihood that a defective item originated from Machine 3 given the probabilities calculated. If you require further clarification or have any questions, please don't hesitate to ask!

# Given Information - **Production Distribution by Machines:** - Machine 1: 20% of items produced - Machine 2: 30% of items produced - Machine 3: 50% of items produced - **Defect Rates:** - Machine 1: 2% defective - Machine 2: 3% defective - Machine 3: 5% defective # Objective Calculate the probability that a defective item was produced by Machine 3, denoted as \( P(M_3 | D) \), where \( D \) is the event that an item is defective. # Concept Used **Bayes' Theorem** is used to find the conditional probability of an event based on prior information. The formula is: \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} \] Where: - \( P(M_3) \): probability of selecting an item from Machine 3. - \( P(D | M_3) \): probability that an item is defective given it came from Machine 3. - \( P(D) \): total probability that an item is defective. # Solution Steps 1. **Define the probabilities for each machine and their defect rates:** - \( P(M_1) = 0.20 \) - \( P(M_2) = 0.30 \) - \( P(M_3) = 0.50 \) - \( P(D | M_1) = 0.02 \) - \( P(D | M_2) = 0.03 \) - \( P(D | M_3) = 0.05 \) 2. **Calculate the total probability of an item being defective \( P(D) \):** \[ P(D) = P(D | M_1) P(M_1) + P(D | M_2) P(M_2) + P(D | M_3) P(M_3) \] Breaking this down: \[ P(D) = (0.02 \times 0.20) + (0.03 \times 0.30) + (0.05 \times 0.50) \] Calculating the individual components: \[ P(D) = 0.004 + 0.009 + 0.025 = 0.038 \] 3. **Apply Bayes' theorem to find \( P(M_3 | D) \):** \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} = \frac{0.05 \times 0.50}{0.038} \] Calculating the result: \[ P(M_3 | D) = \frac{0.025}{0.038} \approx 0.6579 \] # Summary The probability that a defective item was produced by Machine 3 is approximately **0.658** (or **65.8%**). --- ## Explanation In this solution, we started by outlining the production percentages and defect rates for each machine. We then calculated the overall probability of an item being defective using the law of total probability. Finally, Bayes' Theorem was applied to determine the likelihood that a defective item originated from Machine 3 based on the calculated probabilities. If you have any questions or need further elaboration, feel free to ask!

# Given Information - **Production Distribution by Machines:** - Machine 1: 20% of total items produced - Machine 2: 30% of total items produced - Machine 3: 50% of total items produced - **Defect Rates:** - Machine 1: 2% defect rate - Machine 2: 3% defect rate - Machine 3: 5% defect rate # Objective Determine the probability that a defective item came from Machine 3, represented as \( P(M_3 | D) \), where \( D \) is the event that an item is defective. # Concept Used **Bayes' Theorem** provides a way to calculate conditional probabilities based on prior information. The relevant formula is: \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} \] Where: - \( P(M_3) \): probability of an item being produced by Machine 3 - \( P(D | M_3) \): probability that an item is defective given it was made by Machine 3 - \( P(D) \): overall probability that an item is defective # Solution Steps 1. **Identify the probabilities for each machine and their defect rates:** - \( P(M_1) = 0.20 \) - \( P(M_2) = 0.30 \) - \( P(M_3) = 0.50 \) - \( P(D | M_1) = 0.02 \) - \( P(D | M_2) = 0.03 \) - \( P(D | M_3) = 0.05 \) 2. **Calculate the total probability of a defective item \( P(D) \):** \[ P(D) = P(D | M_1) P(M_1) + P(D | M_2) P(M_2) + P(D | M_3) P(M_3) \] Breaking this down: \[ P(D) = (0.02 \times 0.20) + (0.03 \times 0.30) + (0.05 \times 0.50) \] Calculating each component gives: \[ P(D) = 0.004 + 0.009 + 0.025 = 0.038 \] 3. **Utilize Bayes' theorem to find \( P(M_3 | D) \):** \[ P(M_3 | D) = \frac{P(D | M_3) P(M_3)}{P(D)} = \frac{0.05 \times 0.50}{0.038} \] Calculating the numerator: \[ P(M_3 | D) = \frac{0.025}{0.038} \approx 0.6579 \] # Summary The probability that a defective item was produced by Machine 3 is approximately **0.658** (or **65.8%**). --- ## Explanation In this analysis, we began by specifying the production percentages and defect rates for each machine. We then calculated the overall probability of a defective item using the law of total probability. Finally, we applied Bayes' Theorem to ascertain the likelihood that a defective item originated from Machine 3 based on the previously calculated probabilities. If you have any further questions or require clarification, please don't hesitate to ask!

# Given Information - **Production Distribution by Machines:** - Machine A: 20% of bulbs produced - Machine B: 30% of bulbs produced - Machine C: 50% of bulbs produced - **Defect Rates:** - Machine A: 1% defective - Machine B: 2% defective - Machine C: 3% defective # Objective Calculate the probability that a defective bulb was produced by Machine C, denoted as \( P(C | D) \), where \( D \) is the event that a bulb is defective. # Concept Used **Bayes' Theorem** is used to compute the conditional probability of an event based on prior probabilities. The formula is: \[ P(C | D) = \frac{P(D | C) P(C)}{P(D)} \] Where: - \( P(C) \): probability of selecting a bulb from Machine C - \( P(D | C) \): probability that a bulb is defective given it came from Machine C - \( P(D) \): overall probability that a bulb is defective # Solution Steps 1. **Define the probabilities for each machine and their defect rates:** - \( P(A) = 0.20 \) - \( P(B) = 0.30 \) - \( P(C) = 0.50 \) - \( P(D | A) = 0.01 \) - \( P(D | B) = 0.02 \) - \( P(D | C) = 0.03 \) 2. **Calculate the total probability of a bulb being defective \( P(D) \):** \[ P(D) = P(D | A) P(A) + P(D | B) P(B) + P(D | C) P(C) \] Breaking it down: \[ P(D) = (0.01 \times 0.20) + (0.02 \times 0.30) + (0.03 \times 0.50) \] Calculating each term: \[ P(D) = 0.002 + 0.006 + 0.015 = 0.023 \] 3. **Apply Bayes' theorem to find \( P(C | D) \):** \[ P(C | D) = \frac{P(D | C) P(C)}{P(D)} = \frac{0.03 \times 0.50}{0.023} \] Calculating the numerator: \[ P(C | D) = \frac{0.015}{0.023} \approx 0.6522 \] # Summary The probability that a defective bulb was produced by Machine C is approximately **0.652** (or **65.2%**). --- ## Explanation This solution outlines the steps taken to determine the probability that a defective bulb originated from Machine C. By using Bayes' Theorem, we first calculated the total probability of a bulb being defective based on the production rates and defect rates of each machine. We then applied the theorem to find the desired conditional probability. If you have any questions or need further clarification, feel free to ask!

# Given Information - **Diameter of Glass Sphere:** 30 mm - **Radius of Glass Sphere:** \( r = \frac{30}{2} = 15 \) mm = 0.015 m - **Force Applied:** \( F = 5 \) N - **Material Properties:** - Glass: - Young's Modulus, \( E_G = 46.2 \) GPa = \( 46.2 \times 10^9 \) Pa - Poisson's Ratio, \( v_G = 0.245 \) - Steel: - Young's Modulus, \( E_S = 207 \) GPa = \( 207 \times 10^9 \) Pa - Poisson's Ratio, \( v_S = 0.292 \) # Objective (a) Determine the radius of the contact surface. (b) Determine the maximum pressure at the contact surface. (c) Calculate the principal stresses \( \sigma_1, \sigma_2, \) and \( \sigma_3 \) in the glass sphere at the depth \( z = 0.037 \) mm. (d) Draw the Mohr circles for the stresses and show the point corresponding to maximum shear stress. --- ## Solution Steps ### (a) Determine the Radius of the Contact Surface The radius of the contact surface \( a \) can be determined using the following formula for a sphere pressed against a flat surface: \[ a = \left( \frac{3F}{2 \pi E_G} \right)^{1/3} \] Substituting the values: \[ a = \left( \frac{3 \times 5}{2 \pi \times 46.2 \times 10^9} \right)^{1/3} \] Calculating: \[ a = \left( \frac{15}{2 \pi \times 46.2 \times 10^9} \right)^{1/3} \approx 0.000162 \, m = 0.162 \, mm \] ### (b) Determine the Maximum Pressure at the Contact Surface The maximum pressure \( P \) at the contact surface can be calculated using: \[ P = \frac{F}{\pi a^2} \] Substituting the values: \[ P = \frac{5}{\pi (0.000162)^2} \] Calculating: \[ P \approx \frac{5}{\pi \times 2.6244 \times 10^{-8}} \approx 6.031 \times 10^5 \, Pa = 603.1 \, kPa \] ### (c) Calculate the Principal Stresses \( \sigma_1, \sigma_2, \) and \( \sigma_3 \) The principal stresses in the glass sphere can be calculated using the following formulas: 1. **Radial Stress \( \sigma_r \)**: \[ \sigma_r = -P \] \[ \sigma_r = -603.1 \times 10^3 \, Pa \] 2. **Circumferential Stress \( \sigma_\theta \)**: \[ \sigma_\theta = \frac{P}{2} \left( \frac{1 + v_G}{1 - v_G} \right) \] \[ \sigma_\theta = \frac{603.1 \times 10^3}{2} \left( \frac{1 + 0.245}{1 - 0.245} \right) \approx \frac{603.1 \times 10^3}{2} \times 1.319 \approx 398.0 \times 10^3 \, Pa \] 3. **Using the formula for the third principal stress \( \sigma_3 \)**: \[ \sigma_3 = \sigma_r + \frac{2\sigma_\theta}{3} \] \[ \sigma_3 = -603.1 \times 10^3 + \frac{2 \times 398.0 \times 10^3}{3} \approx -603.1 \times 10^3 + 265.3 \times 10^3 \approx -337.8 \times 10^3 \, Pa \] ### (d) Draw the Mohr Circles for the Stresses To sketch the Mohr's Circle, we plot the normal stresses on the x-axis and shear stresses on the y-axis. The points representing the principal stresses are: 1. Point A for \( \sigma_1 \): - \( \sigma_1 = 398.0 \times 10^3 \, Pa \) (Circumferential stress) 2. Point B for \( \sigma_3 \): - \( \sigma_3 = -337.8 \times 10^3 \, Pa \) (Radial stress) **Maximum Shear Stress \( \tau_{max} \)**: The maximum shear stress can be calculated as: \[ \tau_{max} = \frac{\sigma_1 - \sigma_3}{2} \] \[ \tau_{max} = \frac{398.0 \times 10^3 - (-337.8 \times 10^3)}{2} \approx 367.9 \times 10^3 \, Pa \] ### Summary - (a) The radius of the contact surface is approximately **0.162 mm**. - (b) The maximum pressure at the contact surface is approximately **603.1 kPa**. - (c) The principal stresses in the glass sphere at the depth \( z = 0.037 \, mm \) are: - \( \sigma_1 \approx 398.0 kPa \) - \( \sigma_2 \approx -603.1 kPa \) (Radial) - \( \sigma_3 \approx -337.8 kPa \) - (d) The Mohr's Circle can be drawn based on the principal stresses, with the maximum shear stress point plotted accordingly. If you have any questions or need clarification on any part of the solution, feel free to ask!

# Given Information - **Diameter of Glass Sphere:** 30 mm - **Radius of Glass Sphere:** \( r = \frac{30 \, \text{mm}}{2} = 15 \, \text{mm} = 0.015 \, \text{m} \) - **Force Applied:** \( F = 5 \, \text{N} \) - **Material Properties:** - Glass: - Young's Modulus, \( E_G = 46.2 \, \text{GPa} = 46.2 \times 10^9 \, \text{Pa} \) - Poisson's Ratio, \( v_G = 0.245 \) - Steel: - Young's Modulus, \( E_S = 207 \, \text{GPa} = 207 \times 10^9 \, \text{Pa} \) - Poisson's Ratio, \( v_S = 0.292 \) # Objective (a) Determine the radius of the contact surface. (b) Determine the maximum pressure at the contact surface. (c) Calculate the principal stresses \( \sigma_1, \sigma_2, \) and \( \sigma_3 \) in the glass sphere at the depth \( z = 0.037 \, \text{mm} \). (d) Draw the Mohr circles for the stresses and show the point corresponding to maximum shear stress. --- ## Solution Steps ### (a) Determine the Radius of the Contact Surface The radius of the contact area \( a \) can be calculated using the formula derived from Hertzian contact mechanics: \[ a = \left( \frac{3F}{2\pi E_G} \right)^{1/3} \] Substituting the values: \[ a = \left( \frac{3 \times 5 \, \text{N}}{2 \pi \times (46.2 \times 10^9 \, \text{Pa})} \right)^{1/3} \] Calculating: \[ a = \left( \frac{15}{2 \pi \times 46.2 \times 10^9} \right)^{1/3} \approx \left( \frac{15}{289.453 \times 10^9} \right)^{1/3} \approx \left( 5.18 \times 10^{-8} \right)^{1/3} \] \[ a \approx 0.000368 \, \text{m} = 0.368 \, \text{mm} \] ### (b) Determine the Maximum Pressure at the Contact Surface The maximum pressure \( P \) at the contact surface can be calculated using: \[ P = \frac{F}{\pi a^2} \] Substituting the calculated radius: \[ P = \frac{5}{\pi (0.000368)^2} \] Calculating: \[ P \approx \frac{5}{\pi \times 1.356224 \times 10^{-7}} \approx \frac{5}{4.255 \times 10^{-7}} \approx 1.176 \times 10^7 \, \text{Pa} = 11760 \, \text{kPa} \] ### (c) Calculate the Principal Stresses \( \sigma_1, \sigma_2, \) and \( \sigma_3 \) 1. **Radial Stress \( \sigma_r \)**: \[ \sigma_r = -P \] \[ \sigma_r = -11760 \, \text{kPa} \] 2. **Circumferential Stress \( \sigma_\theta \)**: Using the formula: \[ \sigma_\theta = \frac{P}{2} \left( \frac{1 + v_G}{1 - v_G} \right) \] \[ \sigma_\theta = \frac{11760}{2} \left( \frac{1 + 0.245}{1 - 0.245} \right) \] \[ \sigma_\theta = 5880 \left( \frac{1.245}{0.755} \right) \approx 5880 \times 1.646 = 9671.88 \, \text{kPa} \] 3. **Using the formula for the third principal stress \( \sigma_3 \)**: \[ \sigma_3 = \sigma_r + \frac{2\sigma_\theta}{3} \] \[ \sigma_3 = -11760 + \frac{2 \cdot 9671.88}{3} \approx -11760 + 6450.59 \approx -5309.41 \, \text{kPa} \] ### (d) Draw the Mohr Circles for the Stresses To draw Mohr's Circle: 1. The x-axis represents normal stress and the y-axis represents shear stress. 2. The points corresponding to the principal stresses are: - Point A for \( \sigma_1 = 9671.88 \, \text{kPa} \) - Point B for \( \sigma_3 = -5309.41 \, \text{kPa} \) **Maximum Shear Stress \( \tau_{max} \)** can be found using: \[ \tau_{max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{9671.88 - (-5309.41)}{2} \approx \frac{15081.29}{2} \approx 7540.65 \, \text{kPa} \] ### Summary - (a) The radius of the contact surface is approximately **0.368 mm**. - (b) The maximum pressure at the contact surface is approximately **11760 kPa**. - (c) The principal stresses in the glass sphere at the depth \( z = 0.037 \, mm \) are: - \( \sigma_1 \approx 9671.88 \, kPa \) - \( \sigma_2 \approx -11760 \, kPa \) - \( \sigma_3 \approx -5309.41 \, kPa \) - (d) The Mohr's Circle can be drawn using the principal stresses, showing maximum shear stress. If you have any further questions or need clarification, feel free to ask!