Q. The mean and variance of a random variable $X$ having a binomial distribution are 4 and 2 respectively. Then, $P(X > 6)$ is equal to
A) $\frac{1}{256}$
B) $\frac{3}{256}$
C) $\frac{9}{256}$
D) $\frac{7}{256}$
Explanation
(c)
Given, $np = 4,npq = 2$
$\Rightarrow \ p = q = \frac{1}{2}$
$\therefore\ n = 4 \times 2 = 8$
$\therefore\ P(X > 6) = {\ ^{8}C}_{7}\left( \frac{1}{2} \right)^{7}\left( \frac{1}{2} \right) +^{8}C_{8}\left( \frac{1}{2} \right)^{8}$
$= \frac{8}{256} + \frac{1}{256} = \frac{9}{256}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.