Question

The mean and variance of a random variable \( X \) having a Binomial distribution are 4 and 2 respectively, then \( P(X=1) \) is

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

Hint:

For binomial distribution, always use \(\text{Variance} / \text{Mean} = q\) to quickly find \( p \) or \( q \).

Answer 1

The Correct Option is A

For a binomial distribution: \[ \text{Mean} = n p = 4 \] \[ \text{Variance} = n p q = 2 \] where \( q = 1-p \) Now, \[ \frac{\text{Variance}}{\text{Mean}} = q \] \[ \frac{2}{4} = 0.5 = q \] So, \( p = 0.5 \) Then, find \( n \): \[ n \times 0.5 = 4 \Rightarrow n = 8 \] Now, \[ P(X=1) = \binom{8}{1} (0.5)^1 (0.5)^{7} \] \[ = 8 \times (0.5)^8 \] \[ = 8 \times \frac{1}{256} \] \[ = \frac{8}{256} = \frac{1}{32} \]