# Given Information - There are 3 independent models: \(M_1\), \(M_2\), \(M_3\). - Probabilities of assignment: - \(P(M_1) = .40\) - \(P(M_2) = .35\) - \(P(M_3) = .25\) - Each model applies two independent tests: - **Test A (Credit Score Filter) pass probabilities:** - \(M_1:\ .85\) - \(M_2:\ .80\) - \(M_3:\ .75\) - **Test B (Income Verification) pass probabilities:** - \(M_1:\ .90\) - \(M_2:\ .70\) - \(M_3:\ .60\) - An application is approved only if it passes both tests. # What to Find Find the probability that an application is approved, denoted as \(P(A)\). # Definition or Concept Used - **Law of Total Probability:** If events \(M_1, M_2, M_3\) partition the sample space, then: \[ P(A) = \sum_{i=1}^3 P(A|M_i)P(M_i) \] - **Independence of Tests:** The probability of passing both tests for a model is the product of individual pass probabilities: \[ P(\text{Approved}|M_i) = P(\text{Pass A}|M_i) \times P(\text{Pass B}|M_i) \] # Solution For \(M_1\): - \(P(\text{Approved}|M_1) = .85 \times .90 = .765\) - Contribution: \(.765 \times .40 = .306\) For \(M_2\): - \(P(\text{Approved}|M_2) = .80 \times .70 = .56\) - Contribution: \(.56 \times .35 = .196\) For \(M_3\): - \(P(\text{Approved}|M_3) = .75 \times .60 = .45\) - Contribution: \(.45 \times .25 = .1125\) Add all contributions: \[ P(A) = .306 + .196 + .1125 = .6145 \] # Summary The probability that an application is approved, \(P(A)\), is **.6145**.

# Given Information - **Models**: Three independent evaluation models: \(M_1\), \(M_2\), \(M_3\). - **Probabilities of Assignment**: - \(P(M_1) = 0.40\) - \(P(M_2) = 0.35\) - \(P(M_3) = 0.25\) - **Test A (Credit Score Filter) Pass Probabilities**: - \(P(\text{Pass A}|M_1) = 0.85\) - \(P(\text{Pass A}|M_2) = 0.80\) - \(P(\text{Pass A}|M_3) = 0.75\) - **Test B (Income Verification) Pass Probabilities**: - \(P(\text{Pass B}|M_1) = 0.90\) - \(P(\text{Pass B}|M_2) = 0.70\) - \(P(\text{Pass B}|M_3) = 0.60\) An application is approved only if it passes both tests. # What to Find Calculate the probability that an application is approved, denoted as \(P(A)\). # Definition or Concept Used - **Law of Total Probability**: If events \(M_1, M_2, M_3\) partition the sample space: \[ P(A) = \sum_{i=1}^3 P(A|M_i) P(M_i) \] - **Independence of Tests**: The probability of passing both tests for a model is the product of individual pass probabilities: \[ P(\text{Approved}|M_i) = P(\text{Pass A}|M_i) \times P(\text{Pass B}|M_i) \] # Solution For each model, calculate the probability of approval: - **Model \(M_1\)**: - \(P(\text{Approved}|M_1) = P(\text{Pass A}|M_1) \times P(\text{Pass B}|M_1) = 0.85 \times 0.90 = 0.765\) - Contribution to \(P(A)\): \[ 0.765 \times 0.40 = 0.306 \] - **Model \(M_2\)**: - \(P(\text{Approved}|M_2) = P(\text{Pass A}|M_2) \times P(\text{Pass B}|M_2) = 0.80 \times 0.70 = 0.56\) - Contribution to \(P(A)\): \[ 0.56 \times 0.35 = 0.196 \] - **Model \(M_3\)**: - \(P(\text{Approved}|M_3) = P(\text{Pass A}|M_3) \times P(\text{Pass B}|M_3) = 0.75 \times 0.60 = 0.45\) - Contribution to \(P(A)\): \[ 0.45 \times 0.25 = 0.1125 \] Add all contributions to find \(P(A)\): \[ P(A) = 0.306 + 0.196 + 0.1125 = 0.6145 \] # Summary The probability that an application is approved, \(P(A)\), is **0.6145**.

# Given Information - **Total Number of People**: - Democrats: \(D = 13\) - Republicans: \(R = 12\) - Independents: \(I = 8\) - **Total**: \[ T = D + R + I = 13 + 12 + 8 = 33 \] - **Committee Size**: 8 members - **Desired Outcome**: Exactly 5 members must be Democrats. # What to Find Find the probability that exactly 5 of the committee members will be Democrats. # Definition or Concept Used - **Hypergeometric Distribution**: Used for sampling without replacement from a finite population. The probability of getting exactly \(k\) successes (Democrats) in \(n\) draws (committee members) is given by: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] Where: - \(K\) = total number of successes in the population (Democrats) - \(N\) = total population size - \(n\) = number of draws (committee members) - \(k\) = number of observed successes (Democrats in the committee) # Solution Set parameters for the hypergeometric distribution: - \(K = 13\) (total Democrats) - \(N = 33\) (total people) - \(n = 8\) (committee members) - \(k = 5\) (Democrats in the committee) 1. Calculate the number of ways to choose 5 Democrats from 13: \[ \binom{K}{k} = \binom{13}{5} \] 2. Calculate the number of ways to choose the remaining 3 members from the 20 non-Democrats (12 Republicans + 8 Independents): \[ \binom{N-K}{n-k} = \binom{20}{3} \] 3. Calculate the total number of ways to select any 8 members from the total of 33: \[ \binom{N}{n} = \binom{33}{8} \] Now, compute these values: - \(\binom{13}{5} = \frac{13!}{5!(13-5)!} = 1287\) - \(\binom{20}{3} = \frac{20!}{3!(20-3)!} = 1140\) - \(\binom{33}{8} = \frac{33!}{8!(33-8)!} = 105625\) Now, substitute these values into the hypergeometric probability formula: \[ P(X = 5) = \frac{\binom{13}{5} \cdot \binom{20}{3}}{\binom{33}{8}} = \frac{1287 \cdot 1140}{105625} \] Calculate the result: - Numerator: \(1287 \times 1140 = 1468290\) - Denominator: \(105625\) Thus: \[ P(X = 5) = \frac{1468290}{105625} \approx 0.1389 \] # Summary The probability that exactly five of the committee members will be Democrats is approximately **0.1389**.

# Given Information - **Total Number of People**: - Democrats: \(D = 13\) - Republicans: \(R = 12\) - Independents: \(I = 8\) - **Total**: \[ T = D + R + I = 13 + 12 + 8 = 33 \] - **Committee Size**: 8 members - **Desired Outcome**: Exactly 5 members must be Democrats. # What to Find Find the probability that exactly 5 of the committee members will be Democrats. # Definition or Concept Used - **Hypergeometric Distribution**: Used for sampling without replacement from a finite population. The probability of getting exactly \(k\) successes (Democrats) in \(n\) draws (committee members) is given by: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] Where: - \(K\) = total number of successes in the population (Democrats) - \(N\) = total population size - \(n\) = number of draws (committee members) - \(k\) = number of observed successes (Democrats in the committee) # Solution Set parameters for the hypergeometric distribution: - \(K = 13\) (total Democrats) - \(N = 33\) (total people) - \(n = 8\) (committee members) - \(k = 5\) (Democrats in the committee) 1. Calculate the number of ways to choose 5 Democrats from 13: \[ \binom{K}{k} = \binom{13}{5} \] 2. Calculate the number of ways to choose the remaining 3 members from the 20 non-Democrats (12 Republicans + 8 Independents): \[ \binom{N-K}{n-k} = \binom{20}{3} \] 3. Calculate the total number of ways to select any 8 members from the total of 33: \[ \binom{N}{n} = \binom{33}{8} \] Now, compute these values: - **Calculate \(\binom{13}{5}\)**: \[ \binom{13}{5} = \frac{13!}{5!(13-5)!} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287 \] - **Calculate \(\binom{20}{3}\)**: \[ \binom{20}{3} = \frac{20!}{3!(20-3)!} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140 \] - **Calculate \(\binom{33}{8}\)**: \[ \binom{33}{8} = \frac{33!}{8!(33-8)!} = \frac{33 \times 32 \times 31 \times 30 \times 29 \times 28 \times 27 \times 26}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 105625 \] Now, substitute these values into the hypergeometric probability formula: \[ P(X = 5) = \frac{\binom{13}{5} \cdot \binom{20}{3}}{\binom{33}{8}} = \frac{1287 \cdot 1140}{105625} \] Calculate the result: - **Numerator**: \[ 1287 \times 1140 = 1468290 \] - **Denominator**: \[ 105625 \] Thus: \[ P(X = 5) = \frac{1468290}{105625} \approx 0.1389 \] # Summary The probability that exactly five of the committee members will be Democrats is approximately **0.1389**.

# Given Information - **Total Number of People**: - Democrats: \(D = 13\) - Republicans: \(R = 12\) - Independents: \(I = 8\) - **Total**: \[ T = D + R + I = 13 + 12 + 8 = 33 \] - **Committee Size**: 8 members - **Desired Outcome**: Exactly 5 members must be Democrats. # What to Find Find the probability that exactly 5 of the committee members will be Democrats. # Definition or Concept Used - **Hypergeometric Distribution**: The probability of getting exactly \(k\) successes (Democrats) in \(n\) draws (committee members) is given by: \[ P(X = k) = \frac{\binom{K}{k} \cdot \binom{N-K}{n-k}}{\binom{N}{n}} \] Where: - \(K = 13\) (total Democrats) - \(N = 33\) (total people) - \(n = 8\) (committee members) - \(k = 5\) (Democrats in the committee) # Solution ### Use the Hypergeometric Formula \[ P(X = 5) = \frac{\binom{13}{5} \cdot \binom{20}{3}}{\binom{33}{8}} \] ### Compute Each Term 1. **Calculate \(\binom{13}{5}\)**: \[ \binom{13}{5} = \frac{13!}{5!(13-5)!} = 1287 \] 2. **Calculate \(\binom{20}{3}\)**: \[ \binom{20}{3} = \frac{20!}{3!(20-3)!} = 1140 \] 3. **Calculate \(\binom{33}{8}\)**: \[ \binom{33}{8} = \frac{33!}{8!(33-8)!} = 13,884,156 \] ### Compute Probability Substituting the computed values into the formula: \[ P(X = 5) = \frac{1287 \times 1140}{13,884,156} \] Calculating the numerator: \[ 1287 \times 1140 = 1,467,180 \] Thus, \[ P(X = 5) = \frac{1,467,180}{13,884,156} \approx 0.10567 \] # Summary The probability that exactly five of the committee members will be Democrats is approximately **0.10567**.

# Given Information - **Weight Distribution**: The weights of jars of mustard are normally distributed. - **Mean (\(\mu\))**: \(137.2\) ounces - **Standard Deviation (\(\sigma\))**: \(1.6\) ounces - **Label Weight**: \(135\) ounces # What to Find a. The probability that a randomly chosen jar contains more than the stated contents. b. The probability that among ten randomly chosen jars, at least eight contain more than the stated contents. c. The new standard deviation required so that \(95\%\) of all jars contain more than the stated contents. # Definitions or Concepts Used - **Normal Distribution**: A continuous probability distribution characterized by the mean and standard deviation. - **Z-Score Formula**: \[ Z = \frac{X - \mu}{\sigma} \] Where \(X\) is the value of interest, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. - **Cumulative Distribution Function (CDF)**: The probability that a random variable takes a value less than or equal to a specified value. # Solution ### Part a: Probability of More Than Stated Contents 1. Calculate the Z-score for \(X = 135\): \[ Z = \frac{135 - 137.2}{1.6} = \frac{-2.2}{1.6} = -1.375 \] 2. Use the Z-table or a calculator to find \(P(Z < -1.375)\): \[ P(Z < -1.375) \approx 0.0844 \] 3. Therefore, the probability that a jar contains more than 135 ounces: \[ P(X > 135) = 1 - P(Z < -1.375) = 1 - 0.0844 = 0.9156 \] ### Part b: Probability of At Least 8 Out of 10 Jars 1. Let \(p\) be the probability that a single jar contains more than 135 ounces, \(p = 0.9156\). 2. The number of jars containing more than 135 ounces follows a binomial distribution \(B(n=10, p)\). 3. We need to calculate \(P(X \geq 8)\): \[ P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) \] Using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] 4. Compute each term: - \(P(X = 8)\): \[ P(X = 8) = \binom{10}{8} (0.9156)^8 (0.0844)^2 \approx 0.1937 \] - \(P(X = 9)\): \[ P(X = 9) = \binom{10}{9} (0.9156)^9 (0.0844)^1 \approx 0.1414 \] - \(P(X = 10)\): \[ P(X = 10) = \binom{10}{10} (0.9156)^{10} (0.0844)^0 \approx 0.0939 \] 5. Summing these probabilities: \[ P(X \geq 8) = 0.1937 + 0.1414 + 0.0939 \approx 0.4290 \] ### Part c: New Standard Deviation for 95% Probability 1. To find the required standard deviation \(\sigma'\), we need \(P(X > 135) = 0.95\). 2. This implies \(P(Z < z) = 0.05\) for a standard normal variable, which gives \(z \approx -1.645\). 3. Set up the equation: \[ -1.645 = \frac{135 - 137.2}{\sigma'} \] 4. Solving for \(\sigma'\): \[ \sigma' = \frac{2.2}{1.645} \approx 1.34 \] # Summary - a. The probability that a randomly chosen jar contains more than the stated contents is approximately **0.9156**. - b. The probability that at least eight of ten randomly chosen jars contain more than the stated contents is approximately **0.4290**. - c. The new standard deviation required for 95% of jars to contain more than the stated contents is approximately **1.34 ounces**.

# Given Information - **Random Variable**: Let \(x\) be the number of heads in 5 tosses of a fair coin. - **Distribution**: \(x\) follows a Binomial distribution \(B(n=5, p=0.5)\). # What to Find 1. \(E(x^2)\) - The expected value of the square of \(x\). 2. \(\text{Var}(2x + 3)\) - The variance of \(2x + 3\). # Definitions or Concepts Used - **Expected Value**: For a Binomial distribution \(B(n, p)\): \[ E(x) = n \cdot p \] - **Variance**: For a Binomial distribution \(B(n, p)\): \[ \text{Var}(x) = n \cdot p \cdot (1 - p) \] - **Properties of Expectation**: \[ E(aX + b) = aE(X) + b \] - **Variance of a Linear Transformation**: \[ \text{Var}(aX + b) = a^2 \cdot \text{Var}(X) \] # Solution ### Step 1: Calculate \(E(x)\) and \(\text{Var}(x)\) - **Parameters**: - \(n = 5\) - \(p = 0.5\) 1. **Expected Value**: \[ E(x) = n \cdot p = 5 \cdot 0.5 = 2.5 \] 2. **Variance**: \[ \text{Var}(x) = n \cdot p \cdot (1 - p) = 5 \cdot 0.5 \cdot 0.5 = 1.25 \] ### Step 2: Calculate \(E(x^2)\) Using the formula: \[ E(x^2) = \text{Var}(x) + (E(x))^2 \] Substituting values: \[ E(x^2) = 1.25 + (2.5)^2 = 1.25 + 6.25 = 7.5 \] ### Step 3: Calculate \(\text{Var}(2x + 3)\) Using the variance property: \[ \text{Var}(2x + 3) = 2^2 \cdot \text{Var}(x) = 4 \cdot 1.25 = 5 \] # Summary - \(E(x^2) = 7.5\) - \(\text{Var}(2x + 3) = 5\)

# Given Information - **Random Variable**: Let \(x\) represent the number of heads obtained from 5 tosses of a fair coin. - **Distribution**: The random variable \(x\) follows a Binomial distribution \(B(n=5, p=0.5)\). # Objective 1. Determine \(E(x^2)\) - the expected value of the square of \(x\). 2. Calculate \(\text{Var}(2x + 3)\) - the variance of the expression \(2x + 3\). # Definitions and Concepts Utilized - **Expected Value**: For a Binomial distribution \(B(n, p)\), the expected value is calculated as: \[ E(x) = n \cdot p \] - **Variance**: The variance for a Binomial distribution is given by: \[ \text{Var}(x) = n \cdot p \cdot (1 - p) \] - **Expectation Property**: The expected value of a linear transformation is: \[ E(aX + b) = aE(X) + b \] - **Variance Property**: The variance of a linear transformation is expressed as: \[ \text{Var}(aX + b) = a^2 \cdot \text{Var}(X) \] # Solution Steps ### Step 1: Calculate \(E(x)\) and \(\text{Var}(x)\) - **Parameters**: - \(n = 5\) (number of trials) - \(p = 0.5\) (probability of success) 1. **Calculating the Expected Value**: \[ E(x) = n \cdot p = 5 \cdot 0.5 = 2.5 \] 2. **Calculating the Variance**: \[ \text{Var}(x) = n \cdot p \cdot (1 - p) = 5 \cdot 0.5 \cdot 0.5 = 1.25 \] ### Step 2: Derive \(E(x^2)\) Utilizing the relationship between variance and expected value: \[ E(x^2) = \text{Var}(x) + (E(x))^2 \] Substituting previously derived values: \[ E(x^2) = 1.25 + (2.5)^2 = 1.25 + 6.25 = 7.5 \] ### Step 3: Compute \(\text{Var}(2x + 3)\) Applying the variance property for linear transformations: \[ \text{Var}(2x + 3) = 2^2 \cdot \text{Var}(x) = 4 \cdot 1.25 = 5 \] # Summary of Results - The expected value \(E(x^2)\) is **7.5**. - The variance \(\text{Var}(2x + 3)\) is **5**.

# Given Information - **Random Variable**: Let \(x\) denote the number of heads obtained when tossing a fair coin 5 times. - **Distribution**: The variable \(x\) follows a Binomial distribution denoted as \(B(n=5, p=0.5)\). # Objective The goals are to compute: 1. \(E(x^2)\) - the expected value of the square of \(x\). 2. \(\text{Var}(2x + 3)\) - the variance of the linear expression \(2x + 3\). # Definitions and Concepts - **Expected Value**: For a Binomial distribution \(B(n, p)\), the expected value is expressed as: \[ E(x) = n \cdot p \] - **Variance**: The variance for a Binomial distribution is defined as: \[ \text{Var}(x) = n \cdot p \cdot (1 - p) \] - **Expectation Property**: The expected value of a linear transformation can be calculated using: \[ E(aX + b) = aE(X) + b \] - **Variance Property**: The variance of a linear transformation is given by: \[ \text{Var}(aX + b) = a^2 \cdot \text{Var}(X) \] # Solution Process First, establish the parameters for the binomial distribution: - Let \(n = 5\) (total number of trials). - Let \(p = 0.5\) (probability of getting heads). To find the expected value of \(x\): \[ E(x) = n \cdot p = 5 \cdot 0.5 = 2.5 \] Next, calculate the variance of \(x\): \[ \text{Var}(x) = n \cdot p \cdot (1 - p) = 5 \cdot 0.5 \cdot 0.5 = 1.25 \] Now, to derive \(E(x^2)\), utilize the relationship between variance and expected value: \[ E(x^2) = \text{Var}(x) + (E(x))^2 \] Substituting the calculated values: \[ E(x^2) = 1.25 + (2.5)^2 = 1.25 + 6.25 = 7.5 \] For the variance of the expression \(2x + 3\), apply the variance property: \[ \text{Var}(2x + 3) = 2^2 \cdot \text{Var}(x) = 4 \cdot 1.25 = 5 \] # Summary of Results - The expected value \(E(x^2)\) is **7.5**. - The variance \(\text{Var}(2x + 3)\) amounts to **5**.

# Given Information - **Random Variable**: Define \(x\) as the count of heads observed in 5 tosses of a fair coin. - **Distribution**: The variable \(x\) follows a Binomial distribution, specifically \(B(n=5, p=0.5)\). # Objective The tasks are to evaluate: 1. The expected value \(E(x^2)\) - the expectation of the square of \(x\). 2. The variance \(\text{Var}(2x + 3)\) - the variance of the expression \(2x + 3\). # Definitions and Concepts - **Expected Value**: For a Binomial distribution \(B(n, p)\), the expected value is calculated as: \[ E(x) = n \cdot p \] - **Variance**: The variance for a Binomial distribution is defined as: \[ \text{Var}(x) = n \cdot p \cdot (1 - p) \] - **Expectation Property**: The expected value of a linear transformation follows the rule: \[ E(aX + b) = aE(X) + b \] - **Variance Property**: The variance of a linear transformation can be determined by: \[ \text{Var}(aX + b) = a^2 \cdot \text{Var}(X) \] # Solution Process To begin, identify the parameters of the binomial distribution: - \(n = 5\) (number of trials). - \(p = 0.5\) (probability of heads). The expected value \(E(x)\) is computed as: \[ E(x) = n \cdot p = 5 \cdot 0.5 = 2.5 \] Next, calculate the variance of \(x\): \[ \text{Var}(x) = n \cdot p \cdot (1 - p) = 5 \cdot 0.5 \cdot 0.5 = 1.25 \] To find \(E(x^2)\), utilize the relationship between variance and expected value: \[ E(x^2) = \text{Var}(x) + (E(x))^2 \] Substituting the known values: \[ E(x^2) = 1.25 + (2.5)^2 = 1.25 + 6.25 = 7.5 \] For the variance of the linear expression \(2x + 3\), apply the variance property: \[ \text{Var}(2x + 3) = 2^2 \cdot \text{Var}(x) = 4 \cdot 1.25 = 5 \] # Summary of Findings - The expected value \(E(x^2)\) is determined to be **7.5**. - The variance \(\text{Var}(2x + 3)\) is calculated to be **5**.

# Given Information - **Joint Probability Density Function (PDF)**: \[ f(x, y) = 6xy \quad \text{for } 0 < x < 1, \, 0 < y < 1 \] - **Support**: The variables \(X\) and \(Y\) are both bounded between 0 and 1. # What to Find 1. Determine if \(X\) and \(Y\) are independent. 2. Calculate the probability \(P(X + Y < 1)\). # Definitions and Concepts - **Independence of Random Variables**: Two random variables \(X\) and \(Y\) are independent if: \[ f(x, y) = f_X(x) \cdot f_Y(y) \] where \(f_X(x)\) and \(f_Y(y)\) are the marginal PDFs of \(X\) and \(Y\) respectively. - **Marginal Probability Density Functions**: - For \(X\): \[ f_X(x) = \int_0^1 f(x, y) \, dy \] - For \(Y\): \[ f_Y(y) = \int_0^1 f(x, y) \, dx \] - **Probability Calculation**: \[ P(X + Y < 1) = \int_0^1 \int_0^{1-x} f(x, y) \, dy \, dx \] # Solution ### Step 1: Calculate Marginal PDFs #### Marginal PDF of \(X\): \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 6xy \, dy = 6x \left(\int_0^1 y \, dy\right) = 6x \cdot \frac{1}{2} = 3x \] #### Marginal PDF of \(Y\): \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 6xy \, dx = 6y \left(\int_0^1 x \, dx\right) = 6y \cdot \frac{1}{2} = 3y \] ### Step 2: Check Independence To check if \(X\) and \(Y\) are independent, compare \(f(x, y)\) with \(f_X(x) \cdot f_Y(y)\): \[ f_X(x) \cdot f_Y(y) = (3x)(3y) = 9xy \] Since \(f(x, y) = 6xy\) does not equal \(9xy\), we conclude that \(X\) and \(Y\) are not independent. ### Step 3: Calculate \(P(X + Y < 1)\) Set up the double integral: \[ P(X + Y < 1) = \int_0^1 \int_0^{1-x} f(x, y) \, dy \, dx \] Substituting \(f(x, y)\): \[ P(X + Y < 1) = \int_0^1 \int_0^{1-x} 6xy \, dy \, dx \] #### Evaluate the inner integral: \[ \int_0^{1-x} 6xy \, dy = 6x \left(\int_0^{1-x} y \, dy\right) = 6x \cdot \left[\frac{y^2}{2}\right]_0^{1-x} = 6x \cdot \frac{(1-x)^2}{2} = 3x(1 - 2x + x^2) = 3x(1 - 2x + x^2) \] #### Now evaluate the outer integral: \[ P(X + Y < 1) = \int_0^1 3x(1 - 2x + x^2) \, dx = 3 \left(\int_0^1 x \, dx - 2\int_0^1 x^2 \, dx + \int_0^1 x^3 \, dx\right) \] Calculating each integral: - \(\int_0^1 x \, dx = \frac{1}{2}\) - \(\int_0^1 x^2 \, dx = \frac{1}{3}\) - \(\int_0^1 x^3 \, dx = \frac{1}{4}\) Now substituting these values: \[ P(X + Y < 1) = 3 \left(\frac{1}{2} - 2 \cdot \frac{1}{3} + \frac{1}{4}\right) = 3 \left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right) \] Finding a common denominator (12): \[ = 3 \left(\frac{6}{12} - \frac{8}{12} + \frac{3}{12}\right) = 3 \left(\frac{1}{12}\right) = \frac{1}{4} \] # Summary - \(X\) and \(Y\) are **not independent**. - The probability \(P(X + Y < 1)\) is **\(\frac{1}{4}\)**.

# Given Information - **Joint Probability Density Function (PDF)**: \[ f(x, y) = c(x + y) \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 1 \] - The PDF is zero outside the specified range. # What to Find 1. Determine the value of \(c\). 2. Find the marginal PDF of \(X\). 3. Calculate \(E[X]\). 4. Assess the independence of \(X\) and \(Y\). 5. Compute the covariance \(Cov(X, Y)\). # Definitions and Concepts - **Normalization Condition**: The total probability must equal 1: \[ \int_0^1 \int_0^1 f(x, y) \, dy \, dx = 1 \] - **Marginal PDFs**: - For \(X\): \[ f_X(x) = \int_0^1 f(x, y) \, dy \] - **Expected Value**: \[ E[X] = \int_0^1 x f_X(x) \, dx \] - **Independence**: \(X\) and \(Y\) are independent if: \[ f(x, y) = f_X(x) \cdot f_Y(y) \] - **Covariance**: \[ Cov(X, Y) = E[XY] - E[X]E[Y] \] # Solution ### Finding the Value of \(c\) To find \(c\), use the normalization condition: \[ \int_0^1 \int_0^1 c(x + y) \, dy \, dx = 1 \] Calculating the integral: 1. Evaluate the inner integral: \[ \int_0^1 (x + y) \, dy = \int_0^1 x \, dy + \int_0^1 y \, dy = x \cdot 1 + \left[\frac{y^2}{2}\right]_0^1 = x + \frac{1}{2} \] 2. Now, evaluate the outer integral: \[ \int_0^1 \left(x + \frac{1}{2}\right) \, dx = \left[\frac{x^2}{2} + \frac{x}{2}\right]_0^1 = \frac{1}{2} + \frac{1}{2} = 1 \] Combining these gives: \[ c \int_0^1 \left(x + \frac{1}{2}\right) \, dx = c \left(1\right) = 1 \] Thus, \(c = 1\). ### Finding the Marginal PDF of \(X\) Now calculate the marginal PDF of \(X\): \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 (x + y) \, dy = x \cdot 1 + \left[\frac{y^2}{2}\right]_0^1 = x + \frac{1}{2} \] ### Calculating \(E[X]\) Now, compute \(E[X]\): \[ E[X] = \int_0^1 x f_X(x) \, dx = \int_0^1 x \left(x + \frac{1}{2}\right) \, dx = \int_0^1 \left(x^2 + \frac{x}{2}\right) \, dx \] Calculating each term: \[ \int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} \] \[ \int_0^1 \frac{x}{2} \, dx = \frac{1}{2} \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{4} \] Thus: \[ E[X] = \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \] ### Assessing Independence of \(X\) and \(Y\) To check if \(X\) and \(Y\) are independent, compare: \[ f(x, y) = c(x + y) \quad \text{with } f_X(x) \cdot f_Y(y) \] First, find \(f_Y(y)\): \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 (x + y) \, dx = \left[\frac{x^2}{2} + yx\right]_0^1 = \frac{1}{2} + y \] Now, \(f_X(x) \cdot f_Y(y)\): \[ f_X(x) \cdot f_Y(y) = (x + \frac{1}{2})\left(y + \frac{1}{2}\right) \] Since \(f(x, y) = c(x + y)\) does not equal \(f_X(x) \cdot f_Y(y)\), \(X\) and \(Y\) are not independent. ### Calculating Covariance \(Cov(X, Y)\) To compute \(Cov(X, Y)\): 1. Find \(E[XY]\): \[ E[XY] = \int_0^1 \int_0^1 xy f(x, y) \, dy \, dx = \int_0^1 \int_0^1 xy (x + y) \, dy \, dx \] This expands to: \[ E[XY] = \int_0^1 \left[x \int_0^1 (xy + y^2) \, dy\right] \, dx \] Calculating the inner integral: \[ \int_0^1 (xy + y^2) \, dy = \frac{x}{2} + \frac{1}{3} \] Then: \[ E[XY] = \int_0^1 x\left(\frac{x}{2} + \frac{1}{3}\right) \, dx = \int_0^1 \left(\frac{x^2}{2} + \frac{x}{3}\right) \, dx \] Calculating this gives: \[ E[XY] = \left[\frac{x^3}{6} + \frac{x^2}{6}\right]_0^1 = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \] Finally, calculate the covariance: \[ Cov(X, Y) = E[XY] - E[X]E[Y] \] We already have: - \(E[X] = \frac{7}{12}\) - To find \(E[Y]\), it will be equal to \(E[X]\) due to symmetry, hence \(E[Y] = \frac{7}{12}\). Now substitute: \[ Cov(X, Y) = \frac{1}{3} - \left(\frac{7}{12} \cdot \frac{7}{12}\right) = \frac{1}{3} - \frac{49}{144} \] Finding a common denominator: \[ Cov(X, Y) = \frac{48}{144} - \frac{49}{144} = -\frac{1}{144} \] # Summary - The value of \(c\) is **1**. - The marginal PDF of \(X\) is \(f_X(x) = x + \frac{1}{2}\). - The expected value \(E[X]\) is \(\frac{7}{12}\). - \(X\) and \(Y\) are **not independent**. - The covariance \(Cov(X, Y)\) is \(-\frac{1}{144}\).

# Given Information - **Total Balls**: - Red balls: \(5\) - Blue balls: \(5\) - Green balls: \(5\) - **Total number of balls**: \(15\) (5 red + 5 blue + 5 green) - **Drawing Method**: Three balls are drawn without replacement. # What to Find Calculate the probability that at least two of the drawn balls are of the same color. # Definition or Concept Used - **Complement Rule**: To find \(P(A)\), where \(A\) is the event that at least two balls are of the same color, use: \[ P(A) = 1 - P(A^c) \] Here, \(P(A^c)\) is the probability that all three drawn balls are of different colors. # Solution ### Step 1: Calculate \(P(A^c)\) To find \(P(A^c)\) (the probability that all three balls are of different colors): 1. **Total ways to choose 3 balls from 15**: \[ \text{Total combinations} = \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] 2. **Ways to choose 1 ball of each color**: Since there are 3 colors (red, blue, green), we can choose 1 ball from each color: \[ \text{Ways to choose 1 ball from each color} = \binom{5}{1} \times \binom{5}{1} \times \binom{5}{1} = 5 \times 5 \times 5 = 125 \] 3. **Calculate \(P(A^c)\)**: \[ P(A^c) = \frac{\text{Ways to choose 3 different colors}}{\text{Total ways to choose 3 balls}} = \frac{125}{455} \] ### Step 2: Calculate \(P(A)\) Using the complement rule: \[ P(A) = 1 - P(A^c) = 1 - \frac{125}{455} = \frac{455 - 125}{455} = \frac{330}{455} \] ### Step 3: Simplify the Probability Simplifying \(\frac{330}{455}\): - Both numbers can be divided by 5: \[ P(A) = \frac{66}{91} \] # Summary The probability that at least two balls drawn are of the same color is **\(\frac{66}{91}\)**.

# Given Information - **Joint Probability Density Function (PDF)**: \[ f(x, y) = c(x + y) \quad \text{for } 0 < x < 1 \text{ and } 0 < y < 1 \] - The PDF is zero outside the specified range. # What to Find 1. The value of \(c\). 2. The marginal PDF of \(X\). 3. The expected value \(E[X]\). 4. The independence of \(X\) and \(Y\). 5. The covariance \(Cov(X, Y)\). # Definitions and Concepts - **Normalization Condition**: The total probability must equal 1: \[ \int_0^1 \int_0^1 f(x, y) \, dy \, dx = 1 \] - **Marginal PDFs**: - For \(X\): \[ f_X(x) = \int_0^1 f(x, y) \, dy \] - **Expected Value**: \[ E[X] = \int_0^1 x f_X(x) \, dx \] - **Independence**: \(X\) and \(Y\) are independent if: \[ f(x, y) = f_X(x) \cdot f_Y(y) \] - **Covariance**: \[ Cov(X, Y) = E[XY] - E[X]E[Y] \] # Solution ### Finding the Value of \(c\) To find \(c\), use the normalization condition: \[ \int_0^1 \int_0^1 c(x + y) \, dy \, dx = 1 \] Calculating the integral: 1. Evaluate the inner integral: \[ \int_0^1 (x + y) \, dy = \int_0^1 x \, dy + \int_0^1 y \, dy = x \cdot 1 + \left[\frac{y^2}{2}\right]_0^1 = x + \frac{1}{2} \] 2. Now, evaluate the outer integral: \[ \int_0^1 \left(x + \frac{1}{2}\right) \, dx = \left[\frac{x^2}{2} + \frac{x}{2}\right]_0^1 = \frac{1}{2} + \frac{1}{2} = 1 \] Combining these gives: \[ c \cdot 1 = 1 \] Thus, \(c = 1\). ### Finding the Marginal PDF of \(X\) Now calculate the marginal PDF of \(X\): \[ f_X(x) = \int_0^1 f(x, y) \, dy = \int_0^1 (x + y) \, dy = x \cdot 1 + \left[\frac{y^2}{2}\right]_0^1 = x + \frac{1}{2} \] ### Calculating \(E[X]\) Now, compute \(E[X]\): \[ E[X] = \int_0^1 x f_X(x) \, dx = \int_0^1 x \left(x + \frac{1}{2}\right) \, dx = \int_0^1 \left(x^2 + \frac{x}{2}\right) \, dx \] Calculating each term: \[ \int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} \] \[ \int_0^1 \frac{x}{2} \, dx = \frac{1}{2} \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{4} \] Thus: \[ E[X] = \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \] ### Assessing Independence of \(X\) and \(Y\) To check if \(X\) and \(Y\) are independent, compare: \[ f(x, y) = c(x + y) \quad \text{with } f_X(x) \cdot f_Y(y) \] First, find \(f_Y(y)\): \[ f_Y(y) = \int_0^1 f(x, y) \, dx = \int_0^1 (x + y) \, dx = \left[\frac{x^2}{2} + yx\right]_0^1 = \frac{1}{2} + y \] Now, \(f_X(x) \cdot f_Y(y)\): \[ f_X(x) \cdot f_Y(y) = \left(x + \frac{1}{2}\right)\left(y + \frac{1}{2}\right) \] Since \(f(x, y) = c(x + y)\) does not equal \(f_X(x) \cdot f_Y(y)\), \(X\) and \(Y\) are not independent. ### Calculating Covariance \(Cov(X, Y)\) To compute \(Cov(X, Y)\): 1. Find \(E[XY]\): \[ E[XY] = \int_0^1 \int_0^1 xy f(x, y) \, dy \, dx = \int_0^1 \int_0^1 xy (x + y) \, dy \, dx \] This expands to: \[ E[XY] = \int_0^1 \left[x \int_0^1 (xy + y^2) \, dy\right] \, dx \] Calculating the inner integral: \[ \int_0^1 (xy + y^2) \, dy = \frac{x}{2} + \frac{1}{3} \] Then: \[ E[XY] = \int_0^1 x\left(\frac{x}{2} + \frac{1}{3}\right) \, dx = \int_0^1 \left(\frac{x^2}{2} + \frac{x}{3}\right) \, dx \] Calculating this gives: \[ E[XY] = \left[\frac{x^3}{6} + \frac{x^2}{6}\right]_0^1 = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \] Finally, calculate the covariance: \[ Cov(X, Y) = E[XY] - E[X]E[Y] \] We already have: - \(E[X] = \frac{7}{12}\) - To find \(E[Y]\), it will be equal to \(E[X]\) due to symmetry, hence \(E[Y] = \frac{7}{12}\). Now substitute: \[ Cov(X, Y) = \frac{1}{3} - \left(\frac{7}{12} \cdot \frac{7}{12}\right) = \frac{1}{3} - \frac{49}{144} \] Finding a common denominator: \[ Cov(X, Y) = \frac{48}{144} - \frac{49}{144} = -\frac{1}{144} \] # Summary - The value of \(c\) is **1**. - The marginal PDF of \(X\) is \(f_X(x) = x + \frac{1}{2}\). - The expected value \(E[X]\) is \(\frac{7}{12}\). - \(X\) and \(Y\) are **not independent**. - The covariance \(Cov(X, Y)\) is \(-\frac{1}{144}\).