Q. The mean and the variance of a binomial distribution are 4 and 2 respectively. Then, the probability of 2 successes is
A) $\frac{37}{256}$
B) $\frac{219}{256}$
C) $\frac{128}{256}$
D) $\frac{28}{256}$
Explanation
(d)
Given, $np = 4,npq = 2$
$\Rightarrow \ p = q = \frac{1}{2}$
Also, $n = 8$
$\therefore$ Probability of 2 successes,
$P(X = 2) = \frac{8!}{2! \times 6!} \times \left( \frac{1}{2} \right)^{2} \times \left( \frac{1}{2} \right)^{6} = \frac{28}{256}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.