Question

The mean and variance of a random variable \( X \) having binomial distribution are 4 and 2, respectively. Find \( P(X = 1) \).

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{32} \)
• C. \( \frac{1}{16} \)
• D. \( \frac{1}{8} \)

Hint:

For binomial distributions, use \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \).

Answer 1

The Correct Option is B

Step 1: Understanding binomial distribution properties. 
For a binomial distribution \( B(n, p) \), - Mean \( \mu = np \) - Variance \( \sigma^2 = np(1 - p) \) 
Step 2: Finding \( n \) and \( p \). 
Given: \[ np = 4 \] \[ np(1 - p) = 2 \] Substituting \( np = 4 \) into variance equation: \[ 4(1 - p) = 2 \] \[ 1 - p = \frac{1}{2} \Rightarrow p = \frac{1}{2} \] Using \( np = 4 \): \[ n \times \frac{1}{2} = 4 \Rightarrow n = 8 \] 
Step 3: Computing \( P(X = 1) \). 
\[ P(X = 1) = \binom{8}{1} p^1 (1 - p)^{8 - 1} \] \[ = \binom{8}{1} \times \left( \frac{1}{2} \right)^1 \times \left( \frac{1}{2} \right)^7 \] \[ = 8 \times \frac{1}{2} \times \frac{1}{128} = \frac{8}{256} = \frac{1}{32} \] Thus, the correct answer is (B).