# Given Information - \( X \): Daily change in stock market index (\(-1\), \(\), \(1\)) - \( Y \): Daily change in a related commodity price index (\(-1\), \(\), \(1\)) - Joint probability table is provided for \( X \) and \( Y \). # What Have to Find Find the **conditional probability** that the stock market index increases given that the commodity price index also increases, i.e., \[ P(X = 1 \mid Y = 1) \] # Definition or Concept Used Conditional probability is given by: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \] Here, \(A\) is the event \(X=1\), and \(B\) is the event \(Y=1\). # Step-by-Step Solution First, find \(P(X = 1 \cap Y = 1)\). From the table, this is the joint probability where \(X = 1\) and \(Y = 1\): - \(P(X = 1, Y = 1) = .20\) Next, find \(P(Y = 1)\). This is the sum of all joint probabilities where \(Y = 1\): \[ P(Y = 1) = P(X = -1, Y = 1) + P(X = , Y = 1) + P(X = 1, Y = 1) \] From the table: \[ P(X = -1, Y = 1) = .05 \\ P(X = , Y = 1) = .10 \\ P(X = 1, Y = 1) = .20 \\ P(Y = 1) = .05 + .10 + .20 = .35 \] Now, calculate the conditional probability: \[ P(X = 1 \mid Y = 1) = \frac{P(X = 1, Y = 1)}{P(Y = 1)} = \frac{.20}{.35} \] \[ P(X = 1 \mid Y = 1) = \frac{20}{35} = \frac{4}{7} \] # Summary The probability that the stock market index increases given that the commodity price index also increases is **\(\frac{4}{7}\)**.

# Given Information - **Indicators**: - \( X \): Daily change in stock market index (\(-1\): decrease, \(0\): no change, \(1\): increase) - \( Y \): Daily change in related commodity price index (\(-1\): decrease, \(0\): no change, \(1\): increase) - **Joint Probability Distribution Table**: | \(Y\) | \(-1\) | \(0\) | \(1\) | |:---------:|:--------:|:---------:|:---------:| | **\(-1\)**| 0.10 | 0.05 | 0.05 | | **\(0\)** | 0.10 | 0.20 | 0.10 | | **\(1\)** | 0.05 | 0.15 | 0.20 | # What We Have to Find Calculate the **conditional probability**: \[ P(X = 1 \mid Y = 1) \] # Definition or Concept Used The conditional probability formula is: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \] Where: - \( A \): Event \(X = 1\) - \( B \): Event \(Y = 1\) # Step-by-Step Solution **Find \(P(X = 1 \cap Y = 1)\)**: From the table, this is the joint probability where \(X = 1\) and \(Y = 1\): - \( P(X = 1, Y = 1) = 0.20 \) **Find \(P(Y = 1)\)**: Calculate \( P(Y = 1) \) by summing the joint probabilities where \(Y = 1\): \[ P(Y = 1) = P(X = -1, Y = 1) + P(X = 0, Y = 1) + P(X = 1, Y = 1) \] From the table: - \( P(X = -1, Y = 1) = 0.05 \) - \( P(X = 0, Y = 1) = 0.10 \) - \( P(X = 1, Y = 1) = 0.20 \) Thus, \[ P(Y = 1) = 0.05 + 0.10 + 0.20 = 0.35 \] **Calculate the Conditional Probability**: \[ P(X = 1 \mid Y = 1) = \frac{P(X = 1, Y = 1)}{P(Y = 1)} = \frac{0.20}{0.35} \] This simplifies to: \[ P(X = 1 \mid Y = 1) = \frac{20}{35} = \frac{4}{7} \] # Summary The probability that the stock market index increases given that the commodity price index also increases is **\(\frac{4}{7}\)**.

# Given Information - **Distribution**: Number of incoming calls follows a **Poisson distribution**. - **Average rate (\(\lambda\))**: 5 calls per 10-minute interval. - **Random Variable**: \( X \) represents the number of calls received in a 10-minute interval. # What We Have to Find Calculate the **probability** that exactly three calls are received during the interval: \[ P(X = 3) \] # Definition or Concept Used The probability mass function for a Poisson distribution is given by: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] Where: - \( k \): number of events (calls, in this case) - \( \lambda \): average rate of events - \( e \): Euler's number (approximately 2.71828) # Step-by-Step Solution Substituting the values into the formula for \( P(X = 3) \): 1. Set \( k = 3 \) and \( \lambda = 5 \): \[ P(X = 3) = \frac{5^3 e^{-5}}{3!} \] 2. Calculate \( 5^3 \): \[ 5^3 = 125 \] 3. Calculate \( 3! \): \[ 3! = 6 \] 4. Substitute these values: \[ P(X = 3) = \frac{125 e^{-5}}{6} \] 5. Calculate \( e^{-5} \) (using a calculator): \[ e^{-5} \approx 0.006737947 \] 6. Substitute \( e^{-5} \) into the equation: \[ P(X = 3) = \frac{125 \times 0.006737947}{6} \] 7. Perform the multiplication: \[ 125 \times 0.006737947 \approx 0.842368375 \] 8. Finally, divide by 6: \[ P(X = 3) \approx \frac{0.842368375}{6} \approx 0.140394729 \] # Summary The probability that exactly three calls are received during the 10-minute interval is approximately **0.1404**.

# Given Information - **Event Definitions**: - \( A \): The bag contains a prohibited item. - \( B \): The automated screening system flags the bag for inspection. - **Probabilities**: - \( P(A) = 0.02 \) (Probability that a bag contains a prohibited item) - \( P(A') = 0.98 \) (Probability that a bag does not contain a prohibited item) - \( P(B | A) = 0.94 \) (Probability that the system flags a bag when it contains a prohibited item) - \( P(B | A') = 0.08 \) (Probability that the system flags a bag when it does not contain a prohibited item) # What We Have to Find Calculate the **conditional probability**: \[ P(A \mid B) \] # Definition or Concept Used Using **Bayes' Theorem**: \[ P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)} \] **Total Probability \( P(B) \)** can be calculated as: \[ P(B) = P(B \mid A) P(A) + P(B \mid A') P(A') \] # Step-by-Step Solution **Calculate \( P(B) \)**: 1. Substitute values into the total probability formula: \[ P(B) = P(B \mid A) P(A) + P(B \mid A') P(A') \] \[ = (0.94 \times 0.02) + (0.08 \times 0.98) \] 2. Calculate each term: - For \( P(B \mid A) P(A) \): \[ 0.94 \times 0.02 = 0.0188 \] - For \( P(B \mid A') P(A') \): \[ 0.08 \times 0.98 = 0.0784 \] 3. Add the results: \[ P(B) = 0.0188 + 0.0784 = 0.0972 \] **Calculate \( P(A \mid B) \)** using Bayes' Theorem: \[ P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)} = \frac{0.94 \times 0.02}{0.0972} \] \[ = \frac{0.0188}{0.0972} \approx 0.193 \] # Summary The probability that a bag actually contains a prohibited item, given that the automated screening system has flagged it for inspection, is approximately **0.193**.

# Given Information - **Random Variable**: \( X \) represents the number of attempts required for the student to pass the practice test for the first time. - **Probability of passing on a single attempt**: \( P(\text{Pass}) = 0.30 \) - **Probability of failing on a single attempt**: \( P(\text{Fail}) = 1 - P(\text{Pass}) = 0.70 \) # What We Have to Find Calculate the **probability** that the student passes the test for the first time on the fourth attempt: \[ P(X = 4) \] # Definition or Concept Used The situation follows a **geometric distribution**, where the probability of the first success on the \( k \)-th trial is given by: \[ P(X = k) = (1 - p)^{k-1} p \] Where: - \( p \): probability of success (passing) - \( k \): number of trials until the first success # Step-by-Step Solution 1. Set \( p = 0.30 \) (probability of passing) and \( k = 4 \). 2. Substitute these values into the geometric distribution formula: \[ P(X = 4) = (0.70)^{4-1} \times 0.30 \] 3. Calculate \( (0.70)^{3} \): \[ (0.70)^{3} = 0.343 \] 4. Now substitute back into the equation: \[ P(X = 4) = 0.343 \times 0.30 \] 5. Perform the multiplication: \[ P(X = 4) = 0.1029 \] # Summary The probability that the student passes the test for the first time on the fourth attempt is approximately **0.1029**.

# Given Information - **Probability of choosing a blue chip**: \( P(B) = 0.27 \) # What We Have to Find Calculate the **probability of not choosing a blue chip**: \[ P(B') = ? \] # Definition or Concept Used The probability of the complementary event can be calculated as: \[ P(B') = 1 - P(B) \] # Step-by-Step Solution 1. Substitute the given probability into the equation: \[ P(B') = 1 - 0.27 \] 2. Perform the subtraction: \[ P(B') = 0.73 \] # Summary The probability of not choosing a blue chip is **0.73**.

# Given Information - **Random Variable**: \( X \) represents the fuel efficiency measurement. - **Mean (\(\mu\))**: 22 km per liter. - **Standard Deviation (\(\sigma\))**: 3 km per liter. - **Interval**: From 16 km per liter to 28 km per liter. # What We Have to Find Determine the minimum probability that the fuel efficiency measurement will lie within 6 km per liter of the mean: \[ P(16 \leq X \leq 28) \] # Definition or Concept Used Using **Chebyshev's Inequality**: \[ P(|X - \mu| < k\sigma) \geq 1 - \frac{1}{k^2} \] Where: - \( k \) is the number of standard deviations from the mean. # Step-by-Step Solution 1. **Calculate the distance from the mean**: - The interval is from 16 km to 28 km. - The distance from the mean (22 km) to the endpoints: - \( 22 - 16 = 6 \) km (lower limit) - \( 28 - 22 = 6 \) km (upper limit) 2. **Determine \( k \)**: - The distance of 6 km corresponds to the number of standard deviations: \[ k = \frac{6}{3} = 2 \] 3. **Apply Chebyshev's Inequality**: - Substitute \( k = 2 \) into the inequality: \[ P(|X - 22| < 6) \geq 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \] # Summary The minimum probability that the fuel efficiency measurement will lie within 6 km per liter of the mean is at least **0.75**.

# Given Information - **Total Books**: 60 - **Highly Fragile Books**: 15 - **Standard Rare Books**: 45 (i.e., \(60 - 15\)) - **Books Selected**: 8 - **Random Variable**: \( X \) represents the number of highly fragile books selected. # What We Have to Find Calculate the **probability** that exactly 3 of the selected books are highly fragile: \[ P(X = 3) \] # Definition or Concept Used The scenario follows a **hypergeometric distribution**, where: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] Where: - \( N \): Total number of items (60 books) - \( K \): Total number of success states in the population (15 highly fragile books) - \( n \): Number of draws (8 books selected) - \( k \): Number of observed successes (3 highly fragile books) # Step-by-Step Solution 1. **Identify Parameters**: - \( N = 60 \) - \( K = 15 \) - \( n = 8 \) - \( k = 3 \) 2. **Calculate Combinations**: - Compute \( \binom{K}{k} \): \[ \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] - Compute \( \binom{N-K}{n-k} \): \[ \binom{45}{5} = \frac{45!}{5!(45-5)!} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1} = 1221759 \] - Compute \( \binom{N}{n} \): \[ \binom{60}{8} = \frac{60!}{8!(60-8)!} = \frac{60 \times 59 \times 58 \times 57 \times 56 \times 55 \times 54 \times 53}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 750297520 \] 3. **Substitute into Hypergeometric Formula**: \[ P(X = 3) = \frac{\binom{15}{3} \binom{45}{5}}{\binom{60}{8}} = \frac{455 \times 1221759}{750297520} \] 4. **Calculate the Probability**: - First multiply: \[ 455 \times 1221759 = 555391845 \] - Then divide: \[ P(X = 3) = \frac{555391845}{750297520} \approx 0.739 \] # Summary The probability that exactly three of the selected books are highly fragile is approximately **0.739**.

# Given Information - **Total Books**: 60 - **Highly Fragile Books**: 15 - **Standard Rare Books**: 45 (i.e., \( 60 - 15 \)) - **Books Selected**: 8 - **Random Variable**: \( X \) represents the number of highly fragile books selected. # What We Have to Find Calculate the **probability** that exactly 3 of the selected books are highly fragile: \[ P(X = 3) \] # Definition or Concept Used The scenario follows a **hypergeometric distribution**. The probability mass function is given by: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] Where: - \( N \): Total number of items (60 books) - \( K \): Total number of success states in the population (15 highly fragile books) - \( n \): Number of draws (8 books selected) - \( k \): Number of observed successes (3 highly fragile books) # Step-by-Step Solution 1. **Identify Parameters**: - \( N = 60 \) - \( K = 15 \) - \( n = 8 \) - \( k = 3 \) 2. **Calculate Combinations**: - Compute \( \binom{K}{k} \): \[ \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] - Compute \( \binom{N-K}{n-k} \): \[ \binom{45}{5} = \frac{45!}{5!(45-5)!} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1} = 1221759 \] - Compute \( \binom{N}{n} \): \[ \binom{60}{8} = \frac{60!}{8!(60-8)!} = \frac{60 \times 59 \times 58 \times 57 \times 56 \times 55 \times 54 \times 53}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 750297520 \] 3. **Substitute into Hypergeometric Formula**: \[ P(X = 3) = \frac{\binom{15}{3} \binom{45}{5}}{\binom{60}{8}} = \frac{455 \times 1221759}{750297520} \] 4. **Calculate the Probability**: - First multiply: \[ 455 \times 1221759 = 555391845 \] - Then divide: \[ P(X = 3) = \frac{555391845}{750297520} \approx 0.739 \] # Final Verification Using a calculator or software, ensure the calculations for combinations and probabilities are accurate. # Summary The probability that exactly three of the selected books are highly fragile is approximately **0.739**.

# Given Information - **Total Books**: \( N = 60 \) - **Highly Fragile Books**: \( K = 15 \) - **Standard Rare Books**: \( N - K = 45 \) - **Books Selected**: \( n = 8 \) - **Fragile Books Wanted**: \( k = 3 \) # What We Have to Find Calculate the **probability** that exactly 3 of the selected books are highly fragile: \[ P(X = 3) \] # Definition or Concept Used The scenario follows a **hypergeometric distribution**: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] # Step-by-Step Solution **Step 1: Identify Parameters** - Total books \( N = 60 \) - Fragile books \( K = 15 \) - Standard books \( N - K = 45 \) - Books selected \( n = 8 \) - Fragile books wanted \( k = 3 \) **Step 2: Hypergeometric Probability Formula** Substituting values: \[ P(X = 3) = \frac{\binom{15}{3} \binom{45}{5}}{\binom{60}{8}} \] **Step 3: Compute Combinations** 1. Compute \( \binom{15}{3} \): \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] 2. Compute \( \binom{45}{5} \): \[ \binom{45}{5} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1} = 1,221,759 \] 3. Compute \( \binom{60}{8} \): \[ \binom{60}{8} = \frac{60 \times 59 \times 58 \times 57 \times 56 \times 55 \times 54 \times 53}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 2,558,620,845 \] **Step 4: Calculate Probability** Substituting the combination results: \[ P(X = 3) = \frac{455 \times 1,221,759}{2,558,620,845} \] Calculating the numerator: \[ 455 \times 1,221,759 = 555,900,345 \] Calculating the probability: \[ P(X = 3) = \frac{555,900,345}{2,558,620,845} \approx 0.21727 \] # Summary The probability that exactly three of the selected books are highly fragile is approximately **0.2173** (or **21.73%**).

# Given Information - **Total Digits**: Each code consists of 6 digits. - **Even Digits**: The even digits available are {0, 2, 4, 6, 8} (total of 5 even digits). - **Odd Digits**: The odd digits available are {1, 3, 5, 7, 9} (total of 5 odd digits). - **Total Possible Digits**: 10 digits (0-9). # What We Have to Find Calculate the probability that a randomly generated six-digit code contains **at least 4 even digits**. # Definition or Concept Used This scenario can be modeled using the **binomial distribution**, where: - \( n \): number of trials (digits) = 6 - \( p \): probability of success (choosing an even digit) = \(\frac{5}{10} = 0.5\) - \( k \): number of successes (even digits in the code) The probability mass function for a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \] To find the probability of at least 4 even digits: \[ P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) \] # Step-by-Step Solution **Step 1: Calculate Individual Probabilities** 1. **For \( k = 4 \)**: \[ P(X = 4) = \binom{6}{4} (0.5)^4 (0.5)^{6-4} = \binom{6}{4} (0.5)^6 \] \[ \binom{6}{4} = 15 \text{ (since \( \binom{6}{4} = \binom{6}{2} \))} \] \[ P(X = 4) = 15 \times (0.5)^6 = 15 \times \frac{1}{64} = \frac{15}{64} \] 2. **For \( k = 5 \)**: \[ P(X = 5) = \binom{6}{5} (0.5)^5 (0.5)^{6-5} = \binom{6}{5} (0.5)^6 \] \[ \binom{6}{5} = 6 \] \[ P(X = 5) = 6 \times (0.5)^6 = 6 \times \frac{1}{64} = \frac{6}{64} \] 3. **For \( k = 6 \)**: \[ P(X = 6) = \binom{6}{6} (0.5)^6 (0.5)^{6-6} = 1 \times (0.5)^6 = \frac{1}{64} \] **Step 2: Sum the Probabilities** Now, sum the probabilities for \( k = 4, 5, 6 \): \[ P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) \] \[ P(X \geq 4) = \frac{15}{64} + \frac{6}{64} + \frac{1}{64} = \frac{22}{64} \] \[ P(X \geq 4) = \frac{11}{32} \] # Summary The probability that a randomly generated six-digit code contains **at least four even digits** is approximately **\(\frac{11}{32}\)** or **0.34375**.

# Given Information - **Random Numbers Generated**: - \( A, B, C \) (each from 1 to 100) - **Balance Value**: \[ \text{Balance Value} = (A + B + C) - (A + B) = C \] - Each number is generated independently and uniformly. # What We Have to Find Calculate the **probability** that the balance value is greater than 50: \[ P(C > 50) \] # Definition or Concept Used Since \( C \) is uniformly distributed over the integers from 1 to 100, we can determine the probability by counting the favorable outcomes. **Uniform Distribution**: The probability of a specific outcome in a uniform distribution is: \[ P(X = x) = \frac{1}{N} \] Where \( N \) is the total number of possible outcomes. # Step-by-Step Solution 1. **Determine the Range of \( C \)**: - \( C \) can take values from 1 to 100. 2. **Count Favorable Outcomes**: - Favorable outcomes for \( C > 50 \) are the integers from 51 to 100. - The total number of favorable outcomes is: \[ 100 - 50 = 50 \] 3. **Total Possible Outcomes**: - The total number of possible outcomes for \( C \) is 100. 4. **Calculate the Probability**: \[ P(C > 50) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Possible Outcomes}} = \frac{50}{100} = 0.5 \] # Summary The probability that the balance value is greater than 50 is **0.5**.

# Given Information - **Random Variable**: \( X \) represents the number of alerts recorded in an hour. - **Expected Number of Alerts**: \( E(X) = 3 \). # What We Have to Find Calculate the **probability** that the deviation score is equal to zero: \[ P(\text{Deviation Score} = 0) \] # Definition or Concept Used The **deviation score** is defined as: \[ \text{Deviation Score} = (X - 3) + (3 - X) = 0 \] This implies that the deviation score is zero if and only if the number of alerts recorded is equal to the expected number. Therefore: \[ P(X - 3 + 3 - X = 0) = P(X = 3) \] The number of alerts \( X \) follows a **Poisson distribution** with parameter \( \lambda = 3 \). The probability mass function for a Poisson distribution is given by: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] # Step-by-Step Solution 1. **Identify Parameters**: - \( \lambda = 3 \) - We need to find \( P(X = 3) \). 2. **Substitute into the Poisson Formula**: \[ P(X = 3) = \frac{3^3 e^{-3}}{3!} \] 3. **Calculate \( 3^3 \)**: \[ 3^3 = 27 \] 4. **Calculate \( 3! \)**: \[ 3! = 6 \] 5. **Substitute Values**: \[ P(X = 3) = \frac{27 e^{-3}}{6} \] 6. **Calculate \( e^{-3} \)** (using a calculator): \[ e^{-3} \approx 0.049787 \] 7. **Substitute \( e^{-3} \) into the Probability**: \[ P(X = 3) = \frac{27 \times 0.049787}{6} \approx \frac{1.344429}{6} \approx 0.22407 \] # Summary The probability that the deviation score is equal to zero during a randomly observed hour of operation is approximately **0.2241**.

# Given Information - **Random Variable**: \( X \) represents the number of alerts recorded in an hour. - **Deviation Score**: \[ \text{Deviation Score} = (X - 3) + (3 - X) \] # What We Have to Find Calculate the **probability** that the deviation score is equal to zero: \[ P(\text{Deviation Score} = 0) \] # Step-by-Step Solution **Step 1: Simplify the Deviation Score Expression** The deviation score can be simplified as follows: \[ (X - 3) + (3 - X) = X - 3 + 3 - X = 0 \] **Step 2: Interpret the Result** The deviation score is always equal to zero regardless of the value of \( X \). Therefore: \[ P(\text{Deviation Score} = 0) = 1 \] **Step 3: Conclusion** Since the deviation score is always zero for any value of \( X \) (the number of alerts), the probability that it equals zero is: \[ P(\text{Deviation Score} = 0) = 1 \] # Summary The probability that the deviation score is equal to zero during a randomly observed hour of operation is **1** (or 100%).

# Given Information - **Random Variable**: \( X \) represents the number of alerts recorded in an hour. - **Deviation Score**: \[ \text{Deviation Score} = (X - 3) + (3 - X) \] # What We Have to Find Calculate the **probability** that the deviation score is equal to zero: \[ P(\text{Deviation Score} = 0) \] # Definition or Concept Used The deviation score can be simplified to show that it is always zero, regardless of the value of \( X \): \[ (X - 3) + (3 - X) = 0 \] # Step-by-Step Solution First, simplify the expression for the deviation score: - The deviation score can be expanded: \[ (X - 3) + (3 - X) = X - 3 + 3 - X \] - This simplifies to: \[ 0 \] Since the deviation score simplifies to \( 0 \) for any value of \( X \), it implies that the deviation score is always zero. Given that the deviation score is consistently equal to zero, we can conclude: \[ P(\text{Deviation Score} = 0) = 1 \] # Summary The probability that the deviation score is equal to zero during a randomly observed hour of operation is **1** (or 100%).

# Given Information - **Population Mean Score**: \( \mu = 68 \) marks - **Population Standard Deviation**: \( \sigma = 12 \) marks - **Sample Size**: \( n = 36 \) # What We Have to Find Evaluate the probability that the sample mean score \( \bar{X} \) is greater than a certain value (e.g., 70 marks): \[ P(\bar{X} > 70) \] # Definition or Concept Used According to the **Central Limit Theorem**, the distribution of the sample mean \( \bar{X} \) will approximate a normal distribution as the sample size increases, regardless of the population's distribution, provided \( n \) is sufficiently large. The mean and standard deviation of the sample mean are given by: - Mean of the sample mean: \[ E(\bar{X}) = \mu = 68 \] - Standard deviation of the sample mean (standard error): \[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2 \] # Step-by-Step Solution First, calculate the z-score for the sample mean when it exceeds 70 marks: \[ z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} \] Substituting in the values: \[ z = \frac{70 - 68}{2} = \frac{2}{2} = 1 \] Next, refer to the standard normal distribution table (or use a calculator) to find the probability associated with the z-score of 1: \[ P(Z < 1) \approx 0.8413 \] To find \( P(\bar{X} > 70) \), we subtract the cumulative probability from 1: \[ P(\bar{X} > 70) = 1 - P(Z < 1) = 1 - 0.8413 = 0.1587 \] # Summary The probability that a randomly selected group of students performs better than the overall average score of 68 marks, achieving a sample mean greater than 70 marks, is approximately **0.1587** (or **15.87%**).

# Given Information - **Weights of Tests**: - Test 1: 10% of final score - Test 2: 20% of final score - Test 3: 30% of final score - Test 4: 40% of final score - **Marks Obtained**: - Test 1: 80 marks - Test 2: 70 marks - Test 3: 60 marks - **Target Final Score**: 70 marks out of 100 # What We Have to Find Determine the score needed in Test 4 (\(X\)) to achieve a final weighted internal score of 70 marks: \[ \text{Final Score} = \frac{(\text{Weight}_1 \times \text{Score}_1) + (\text{Weight}_2 \times \text{Score}_2) + (\text{Weight}_3 \times \text{Score}_3) + (\text{Weight}_4 \times X)}{100} = 70 \] # Definition or Concept Used The final weighted score is calculated as: \[ \text{Final Score} = \frac{W_1 \cdot S_1 + W_2 \cdot S_2 + W_3 \cdot S_3 + W_4 \cdot S_4}{100} \] Where \(W_i\) is the weight assigned to each test and \(S_i\) is the score obtained in each test. # Step-by-Step Solution First, substitute the known values into the equation: \[ \frac{(10 \cdot 80) + (20 \cdot 70) + (30 \cdot 60) + (40 \cdot X)}{100} = 70 \] Calculate the contribution of the first three tests: - Test 1 contribution: \(10 \cdot 80 = 800\) - Test 2 contribution: \(20 \cdot 70 = 1400\) - Test 3 contribution: \(30 \cdot 60 = 1800\) Summing these contributions: \[ 800 + 1400 + 1800 = 4000 \] Now, substitute back into the equation: \[ \frac{4000 + 40X}{100} = 70 \] Multiply both sides by 100: \[ 4000 + 40X = 7000 \] Subtract 4000 from both sides: \[ 40X = 3000 \] Divide by 40: \[ X = \frac{3000}{40} = 75 \] # Summary The student must obtain **75 marks** in Test 4 in order to achieve a final weighted internal score of 70 marks.

# Given Information - **Random Digits**: A, B, C, D (each digit is independently chosen from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}). - **Difference Indicator**: \[ \text{Difference Indicator} = (A + B + C + D) - (A + B + C) = D \] - **Range of Generated Numbers**: Four-digit numbers from 0000 to 9999. # What We Have to Find Calculate the probability that the **difference indicator** (which is equal to \( D \)) is greater than 7: \[ P(D > 7) \] # Definition or Concept Used Since \( D \) can take on values from 0 to 9, the probability can be calculated by determining the number of favorable outcomes where \( D \) is any digit greater than 7 (i.e., 8 or 9). # Step-by-Step Solution 1. **Identify Possible Values for \( D \)**: - The possible values for \( D \) are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. 2. **Determine Favorable Outcomes**: - Favorable outcomes for \( D > 7 \) are the digits 8 and 9. - Hence, there are 2 favorable outcomes (8 and 9). 3. **Total Possible Outcomes**: - Since each digit can be any of the 10 digits (0 through 9), the total number of outcomes for one digit is 10. 4. **Calculate the Probability**: \[ P(D > 7) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Possible Outcomes}} = \frac{2}{10} = 0.2 \] # Summary The probability that the difference indicator is greater than 7 for a randomly generated four-digit number is **0.2** (or **20%**).

# Given Information - **Random Digits**: \( A, B, C, D \) (each digit is independently chosen from the set \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}). - **Difference Indicator**: \[ \text{Difference Indicator} = (A + B + C + D) - (A + B + C) = D \] - **Range of Generated Numbers**: Four-digit numbers from 0000 to 9999. # What We Have to Find Calculate the probability that the **difference indicator** (which is equal to \( D \)) is greater than 7: \[ P(D > 7) \] # Definition or Concept Used The difference indicator \( D \) can take on values from 0 to 9. The probability can be calculated by determining the number of favorable outcomes where \( D \) is any digit greater than 7 (i.e., 8 or 9). # Step-by-Step Solution **Identify Possible Values for \( D \)**: - The possible values for \( D \) are \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}. **Determine Favorable Outcomes**: - Favorable outcomes for \( D > 7 \) are the digits 8 and 9. - Hence, there are 2 favorable outcomes (8 and 9). **Total Possible Outcomes**: - Each digit can be any of the 10 digits (0 through 9), so the total number of outcomes for one digit is 10. **Calculate the Probability**: \[ P(D > 7) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Possible Outcomes}} = \frac{2}{10} = 0.2 \] # Summary The probability that the difference indicator is greater than 7 for a randomly generated four-digit number is **0.2** (or **20%**).

# Given Information - **Total Packages**: \( N = 100 \) - **Fragile Packages**: \( K = 15 \) - **Non-Fragile Packages**: \( N - K = 85 \) - **Packages Selected**: \( n = 10 \) - **Fragile Packages Wanted**: \( k = 3 \) # What We Have to Find Calculate the **probability** that exactly 3 of the selected packages are fragile: \[ P(X = 3) \] # Definition or Concept Used This scenario follows a **hypergeometric distribution**, where the probability mass function is given by: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] Where: - \( N \): Total number of items (100 packages) - \( K \): Total number of success states in the population (15 fragile packages) - \( n \): Number of draws (10 packages selected) - \( k \): Number of observed successes (3 fragile packages) # Step-by-Step Solution **Identify Parameters**: - \( N = 100 \) - \( K = 15 \) - \( n = 10 \) - \( k = 3 \) **Calculate Combinations**: 1. Compute \( \binom{K}{k} \): \[ \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] 2. Compute \( \binom{N-K}{n-k} \): \[ \binom{85}{7} = \frac{85!}{7!(85-7)!} = \frac{85 \times 84 \times 83 \times 82 \times 81 \times 80 \times 79}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 2280000 \] 3. Compute \( \binom{N}{n} \): \[ \binom{100}{10} = \frac{100!}{10!(100-10)!} = \frac{100 \times 99 \times 98 \times 97 \times 96 \times 95 \times 94 \times 93 \times 92 \times 91}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 17310309456440 \] **Substitute into Hypergeometric Formula**: \[ P(X = 3) = \frac{\binom{15}{3} \cdot \binom{85}{7}}{\binom{100}{10}} = \frac{455 \cdot 2280000}{17310309456440} \] **Calculate the Probability**: First multiply: \[ 455 \cdot 2280000 = 1037400000 \] Then divide: \[ P(X = 3) = \frac{1037400000}{17310309456440} \approx 0.05993 \] # Summary The probability that exactly 3 of the 10 selected packages are fragile is approximately **0.05993** (or **5.99%**).

# Given Information - **Total Packages**: \( N = 100 \) - **Fragile Packages**: \( K = 15 \) - **Non-Fragile Packages**: \( N - K = 85 \) - **Packages Selected**: \( n = 10 \) - **Fragile Packages Desired in Sample**: \( k = 3 \) # What We Have to Find Calculate the **probability** that exactly 3 of the selected packages are fragile: \[ P(X = 3) \] # Definition or Concept Used This scenario follows a **hypergeometric distribution**, where the probability mass function is given by: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] Where: - \( N \): Total number of items (100 packages) - \( K \): Total number of fragile packages (15) - \( n \): Number of packages selected (10) - \( k \): Number of fragile packages desired (3) # Step-by-Step Solution **Identify Parameters**: - Total Packages, \( N = 100 \) - Fragile Packages, \( K = 15 \) - Packages Selected, \( n = 10 \) - Fragile Packages Desired, \( k = 3 \) **Calculate Combinations**: 1. Compute \( \binom{K}{k} \): \[ \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] 2. Compute \( \binom{N-K}{n-k} \): \[ \binom{85}{7} = \frac{85!}{7!(85-7)!} = \frac{85 \times 84 \times 83 \times 82 \times 81 \times 80 \times 79}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \approx 2280000 \] 3. Compute \( \binom{N}{n} \): \[ \binom{100}{10} = \frac{100!}{10!(100-10)!} \approx 17310309456440 \] **Substitute into Hypergeometric Formula**: \[ P(X = 3) = \frac{\binom{15}{3} \cdot \binom{85}{7}}{\binom{100}{10}} = \frac{455 \cdot 2280000}{17310309456440} \] **Calculate the Probability**: 1. First, multiply: \[ 455 \cdot 2280000 = 1037400000 \] 2. Then divide: \[ P(X = 3) = \frac{1037400000}{17310309456440} \approx 0.05993 \] # Summary The probability that exactly 3 of the 10 selected packages are fragile is approximately **0.05993** (or **5.99%**).

# Given Information - **Mean Amount of Juice per Bottle**: \( \mu = 500 \) milliliters - **Standard Deviation of Juice Amounts**: \( \sigma = 8 \) milliliters - **Sample Size**: \( n = 36 \) - **Threshold Amount**: \( 497 \) milliliters # What We Have to Find Calculate the probability that the sample mean amount of juice \( \bar{X} \) is less than 497 milliliters: \[ P(\bar{X} < 497) \] # Definition or Concept Used According to the **Central Limit Theorem**, the distribution of the sample mean \( \bar{X} \) will approximate a normal distribution as the sample size increases, regardless of the population's distribution. The mean and standard deviation of the sample mean are given by: - Mean of the sample mean: \[ E(\bar{X}) = \mu = 500 \] - Standard deviation of the sample mean (standard error): \[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{36}} = \frac{8}{6} = \frac{4}{3} \approx 2.67 \] # Step-by-Step Solution First, calculate the z-score for the sample mean when it is less than 497 milliliters: \[ z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} \] Substituting in the values: \[ z = \frac{497 - 500}{\frac{4}{3}} = \frac{-3}{\frac{4}{3}} = -3 \times \frac{3}{4} = -2.25 \] Next, refer to the standard normal distribution table (or use a calculator) to find the probability associated with the z-score of -2.25: \[ P(Z < -2.25) \approx 0.0119 \] # Summary The probability that the average amount of juice in the sample of 36 bottles is less than 497 milliliters is approximately **0.0119** (or **1.19%**).

# Given Information - **Total Packages Selected**: \( n = 50 \) - **Probability of Misrouting**: \( p = 0.03 \) - **Random Variable**: \( X \) represents the number of misrouted packages. # What We Have to Find Calculate the probability that at most 2 packages are misrouted: \[ P(X \leq 2) \] # Definition or Concept Used The scenario follows a **binomial distribution**, where the probability mass function is given by: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] Where: - \( n \): number of trials (packages) - \( k \): number of successes (misrouted packages) - \( p \): probability of success (misrouting) # Step-by-Step Solution First, we need to calculate \( P(X \leq 2) \): \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \] **Calculating Each Probability**: 1. **For \( k = 0 \)**: \[ P(X = 0) = \binom{50}{0} (0.03)^0 (0.97)^{50} = 1 \cdot 1 \cdot (0.97)^{50} \] \[ P(X = 0) \approx (0.97)^{50} \approx 0.2161 \] 2. **For \( k = 1 \)**: \[ P(X = 1) = \binom{50}{1} (0.03)^1 (0.97)^{49} = 50 \cdot 0.03 \cdot (0.97)^{49} \] \[ P(X = 1) \approx 50 \cdot 0.03 \cdot 0.2233 \approx 0.3350 \] 3. **For \( k = 2 \)**: \[ P(X = 2) = \binom{50}{2} (0.03)^2 (0.97)^{48} = \frac{50 \cdot 49}{2} \cdot (0.03)^2 \cdot (0.97)^{48} \] \[ P(X = 2) \approx 1225 \cdot 0.0009 \cdot 0.2296 \approx 0.2536 \] **Summing the Probabilities**: \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \] \[ P(X \leq 2) \approx 0.2161 + 0.3350 + 0.2536 \approx 0.8047 \] # Summary The probability that at most 2 packages are misrouted among the 50 selected packages is approximately **0.8047** (or **80.47%**).

# Given Information - **Total Coupons**: Numbers from 1 to 100. - **Multiples of 10** in the range: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 (total of 10). - **Total Possible Numbers**: 100. - **Sample Size**: 5 coupons selected. - **Desired Multiples of 10 in Sample**: \( k = 2 \). # What We Have to Find Calculate the **probability** that exactly 2 of the selected coupons contain numbers that are multiples of 10: \[ P(X = 2) \] # Definition or Concept Used This scenario can be modeled using the **hypergeometric distribution**, where the probability mass function is given by: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] Where: - \( N \): Total number of items (100 coupons). - \( K \): Total number of success states in the population (10 multiples of 10). - \( n \): Number of draws (5 coupons selected). - \( k \): Number of observed successes (2 multiples of 10). # Step-by-Step Solution **Identify Parameters**: - Total Coupons, \( N = 100 \). - Total Multiples of 10, \( K = 10 \). - Total Coupons Selected, \( n = 5 \). - Desired Multiples of 10 in Sample, \( k = 2 \). **Calculate Combinations**: 1. Compute \( \binom{K}{k} \): \[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] 2. Compute \( \binom{N-K}{n-k} \): \[ \binom{90}{3} = \frac{90!}{3!(90-3)!} = \frac{90 \times 89 \times 88}{3 \times 2 \times 1} = 117480 \] 3. Compute \( \binom{N}{n} \): \[ \binom{100}{5} = \frac{100!}{5!(100-5)!} = \frac{100 \times 99 \times 98 \times 97 \times 96}{5 \times 4 \times 3 \times 2 \times 1} = 75287520 \] **Substitute into Hypergeometric Formula**: \[ P(X = 2) = \frac{\binom{10}{2} \cdot \binom{90}{3}}{\binom{100}{5}} = \frac{45 \cdot 117480}{75287520} \] **Calculate the Probability**: 1. First multiply: \[ 45 \cdot 117480 = 5286600 \] 2. Then divide: \[ P(X = 2) = \frac{5286600}{75287520} \approx 0.0702 \] # Summary The probability that exactly 2 out of the 5 selected coupons contain numbers that are multiples of 10 is approximately **0.0702** (or **7.02%**).

# Given Information - **Total Students**: \( N = 80 \) - **Initial Percentage of Students Passed**: \( 60\% \) - **Initial Number of Students Passed**: \[ P_{\text{initial}} = 0.60 \times 80 = 48 \] - **Grading Mistakes**: - One passing student incorrectly recorded as passing is actually failing. - One failing student incorrectly recorded as failing is actually passing. # What We Have to Find Calculate the final percentage of students who passed the exam after correcting the mistakes. # Step-by-Step Solution 1. **Adjust the Number of Students Who Passed**: - Correct the initial count: - One student who passed is now failing: \( 48 - 1 = 47 \) - One student who failed is now passing: \( 47 + 1 = 48 \) Therefore, the number of students who passed after corrections remains \( 48 \). 2. **Calculate the Final Percentage of Students Passed**: \[ P_{\text{final}} = \frac{48}{80} \times 100\% = 60\% \] # Summary The final percentage of students who passed the exam after the corrections is **60%**.

# Given Information - **Total Pills in Batch**: \( n = 200 \) - **Defective Rate**: \( p = 0.05 \) (5% of pills are defective) - **Sample Size**: \( n = 20 \) (pills selected for testing) - **Random Variable**: \( X \) represents the number of defective pills in the sample. # What We Have to Find Calculate the probability that exactly 2 pills in the sample are defective: \[ P(X = 2) \] # Definition or Concept Used This scenario follows a **binomial distribution**, where the probability mass function is given by: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] Where: - \( n \): number of trials (pills selected) - \( k \): number of successes (defective pills) - \( p \): probability of success (defective pill) # Step-by-Step Solution 1. **Identify Parameters**: - Total Pills Selected, \( n = 20 \) - Probability of Defective Pill, \( p = 0.05 \) - Desired Defective Pills in Sample, \( k = 2 \) 2. **Calculate the Probability**: \[ P(X = 2) = \binom{20}{2} (0.05)^2 (0.95)^{20-2} \] 3. **Calculate the Combinations**: \[ \binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20 \times 19}{2 \times 1} = 190 \] 4. **Substitute into the Probability Formula**: \[ P(X = 2) = 190 \cdot (0.05)^2 \cdot (0.95)^{18} \] 5. **Calculate Each Component**: - \( (0.05)^2 = 0.0025 \) - \( (0.95)^{18} \approx 0.407 \) (using a calculator) 6. **Combine the Results**: \[ P(X = 2) = 190 \cdot 0.0025 \cdot 0.407 \approx 190 \cdot 0.0010175 \approx 0.1933 \] # Summary The probability that exactly 2 pills in the sample are defective is approximately **0.1933** (or **19.33%**).

# Given Information - **Kellie's Investment**: - Amount: \( P = 6000 \) - Duration: \( t = 30 \) months - APR: \( r = 3.5\% = 0.035 \) - **Mariah's Investment**: - Amount: \( P = 9000 \) - Duration: \( t = 24 \) months - APR: \( r = 3.5\% = 0.035 \) # What We Have to Find Calculate the **APY (Annual Percentage Yield)** for both Kellie and Mariah. # Definition or Concept Used The formula for APY when interest is compounded monthly is: \[ APY = \left(1 + \frac{r}{n}\right)^{n} - 1 \] Where: - \( r \): Annual interest rate (as a decimal) - \( n \): Number of compounding periods per year (monthly compounding means \( n = 12 \)) # Step-by-Step Solution **Calculate Kellie's APY**: 1. Substitute values into the APY formula: \[ APY_{Kellie} = \left(1 + \frac{0.035}{12}\right)^{12} - 1 \] 2. Calculate \( \frac{0.035}{12} \): \[ \frac{0.035}{12} \approx 0.00291667 \] 3. Calculate \( \left(1 + 0.00291667\right)^{12} \): \[ \left(1.00291667\right)^{12} \approx 1.03557 \] 4. Subtract 1: \[ APY_{Kellie} \approx 1.03557 - 1 \approx 0.03557 \] 5. Convert to percentage: \[ APY_{Kellie} \approx 0.03557 \times 100 \approx 3.557\% \] **Calculate Mariah's APY**: 1. Substitute values into the APY formula: \[ APY_{Mariah} = \left(1 + \frac{0.035}{12}\right)^{12} - 1 \] 2. The calculation will be the same as Kellie's since they have the same APR and compounding frequency: \[ APY_{Mariah} = APY_{Kellie} \approx 3.557\% \] # Summary Both Kellie and Mariah's CDs have the same APY of approximately **3.557%**. The correct statement is: **Their CDs have the same APY because they have the same number of compoundings per year and the same APR. The amount invested and the time period of the investment don't matter because they aren't part of the APY formula.**

# Given Information - **Total Tablets in Batch**: \( n = 500 \) - **Defective Rate**: \( p = 0.04 \) (4% of tablets are defective) - **Sample Size**: \( n = 25 \) (tablets selected for testing) - **Random Variable**: \( X \) represents the number of defective tablets in the sample. # What We Have to Find Calculate the probability that exactly 2 tablets in the sample are defective: \[ P(X = 2) \] # Definition or Concept Used This scenario follows a **binomial distribution**, where the probability mass function is given by: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] Where: - \( n \): number of trials (tablets selected) - \( k \): number of successes (defective tablets) - \( p \): probability of success (defective tablet) # Step-by-Step Solution **Identify Parameters**: - Total Tablets Selected, \( n = 25 \) - Probability of Defective Tablet, \( p = 0.04 \) - Desired Defective Tablets in Sample, \( k = 2 \) **Calculate the Probability**: \[ P(X = 2) = \binom{25}{2} (0.04)^2 (0.96)^{25-2} \] **Calculate the Combinations**: \[ \binom{25}{2} = \frac{25!}{2!(25-2)!} = \frac{25 \times 24}{2 \times 1} = 300 \] **Substitute into the Probability Formula**: \[ P(X = 2) = 300 \cdot (0.04)^2 \cdot (0.96)^{23} \] **Calculate Each Component**: 1. \( (0.04)^2 = 0.0016 \) 2. \( (0.96)^{23} \approx 0.3677 \) (using a calculator) **Combine the Results**: \[ P(X = 2) = 300 \cdot 0.0016 \cdot 0.3677 \approx 300 \cdot 0.00058832 \approx 0.1765 \] # Summary The probability that exactly 2 tablets in the sample are defective is approximately **0.1765** (or **17.65%**).