Question

The random valuable X follows binomial distribution B (n, p) for which the difference of the mean and the variance is 1. If 2P(X = 2) = 3P(X = 1), then n²P(X > 1) is equal to

• A. 15
• B. 16
• C. 12
• D. 11

Answer 1

The Correct Option is D

Given that \( 2P(x=2) = 3P(x=1) \), we can express the probabilities using the binomial distribution formula. \[ 2P(x=2) = 3P(x=1) \] \[ 2 \times \binom{n}{2} p^2 (1-p)^{n-2} = 3 \times \binom{n}{1} p (1-p)^{n-1} \] Simplifying: \[ 2 \times \frac{n(n-1)}{2} \times p^2 (1-p)^{n-2} = 3 \times n \times p \times (1-p)^{n-1} \] \[ \Rightarrow n(n-1)p^2 = 3n(1-p)p \] \[ \Rightarrow (n-1)p = 3(1-p) \quad \text{(Equation 1)} \] From here, we simplify further to find the value of \( p \) and \( n \). 
Step 1: Solve for \( p \) and \( n \). From Equation 1: \[ (n-1)p = 3(1-p) \] \[ np - p = 3 - 3p \] \[ np + 3p = 3 + p \] \[ p(n + 3) = 3 \] \[ p = \frac{3}{n+3} \] Substitute this value of \( p \) into the equation for the variance and mean difference. 
Step 2: Apply the condition for the mean and variance. The mean of a binomial distribution is \( \mu = np \), and the variance is \( \sigma^2 = np(1-p) \). Given that the difference between the mean and variance is 1: \[ np - np(1-p) = 1 \] Simplify this equation and solve for \( n \). 
Step 3: Solve for \( n \). We find that \( n = 4 \). 
Step 4: Find \( n^2 P(X > 1) \). Now, we compute \( P(X > 1) \) for \( n = 4 \) and \( p = \frac{1}{2} \). \[ P(X > 1) = 1 - P(X = 0) - P(X = 1) \] \[ P(X = 0) = \binom{4}{0} \left( \frac{1}{2} \right)^0 \left( \frac{1}{2} \right)^4 = \frac{1}{16} \] \[ P(X = 1) = \binom{4}{1} \left( \frac{1}{2} \right)^1 \left( \frac{1}{2} \right)^3 = \frac{4}{16} = \frac{1}{4} \] \[ P(X > 1) = 1 - \frac{1}{16} - \frac{4}{16} = \frac{11}{16} \] Now calculate \( n^2 P(X > 1) \): \[ n^2 P(X > 1) = 4^2 \times \frac{11}{16} = 16 \times \frac{11}{16} = 11 \]