Q. The mean and variance of binomial distribution are 4 and 3 respectively. Then, the probability of getting exactly six success in this distribution is
A) $\ ^{16}C_{6}\left( \frac{1}{4} \right)^{6}\left( \frac{3}{4} \right)^{10}$
B) $\ ^{16}C_{6}\left( \frac{1}{4} \right)^{6}\left( \frac{3}{4} \right)^{20}$
C) $\ ^{16}C_{6}\left( \frac{1}{4} \right)^{8}\left( \frac{3}{4} \right)^{12}$
D) $\ ^{16}C_{9}\left( \frac{1}{4} \right)^{16}\left( \frac{3}{4} \right)^{20}$
Explanation
(a)
Given, $np = 4,\ \ npq = 3$
$\Rightarrow \ P = \frac{1}{4},\ q = \frac{3}{4}$
$\therefore\ P(X = 6) = {\ ^{16}C}_{6}\left( \frac{1}{4} \right)^{6}.\left( \frac{3}{4} \right)^{10}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.