Question

A random variable \(X\) follows a binomial distribution in which the difference between its mean and variance is 1. If \(2P(x=2) = 3P(x=1)\), then \(n^2 P(x>1)\) is:

• A. 13
• B. 11
• C. 15
• D. 12

Hint:

Use properties of binomial distribution and algebraic manipulation to find unknowns.

Answer 1

The Correct Option is B

Step 1: Binomial distribution parameters
\[ X \sim Binomial(n, p) \] Step 2: Use given conditions
\[ \text{Mean} = np, \quad \text{Variance} = np(1-p) \] \[ \text{Mean} - \text{Variance} = 1 \implies np - np(1-p) = 1 \implies np^2 = 1 \] Step 3: Given
\[ 2P(x=2) = 3P(x=1) \] \[ 2 \binom{n}{2} p^2 (1-p)^{n-2} = 3 \binom{n}{1} p (1-p)^{n-1} \] \[ 2 \frac{n(n-1)}{2} p^2 (1-p)^{n-2} = 3 n p (1-p)^{n-1} \] \[ n(n-1) p^2 = 3 n p (1-p) \] \[ (n-1) p = 3 (1-p) \] Step 4: Solve for \(p\)
\[ (n-1) p = 3 - 3p \implies p(n-1+3) = 3 \implies p(n+2) = 3 \implies p = \frac{3}{n+2} \] Step 5: Use \(np^2 = 1\)
\[ n \left(\frac{3}{n+2}\right)^2 = 1 \implies n \frac{9}{(n+2)^2} = 1 \implies 9n = (n+2)^2 \] Step 6: Solve quadratic for \(n\)
\[ 9n = n^2 + 4n + 4 \implies n^2 - 5n + 4 = 0 \] \[ (n-4)(n-1) = 0 \implies n = 4 \text{ or } 1 \] Step 7: Calculate \(P(x>1)\)
\[ P(x>1) = 1 - P(x=0) - P(x=1) = 1 - (1-p)^n - n p (1-p)^{n-1} \] For \(n=4\), \[ p = \frac{3}{6} = \frac{1}{2} \] \[ P(x>1) = 1 - (1/2)^4 - 4 \times \frac{1}{2} \times (1/2)^3 = 1 - \frac{1}{16} - \frac{4}{16} = 1 - \frac{5}{16} = \frac{11}{16} \] Step 8: Calculate \(n^2 P(x>1)\)
\[ 4^2 \times \frac{11}{16} = 16 \times \frac{11}{16} = 11 \]