Q. The mean and variance of a binomial distribution are 4 and 3 respectively, then the probability of getting exactly six successes in this distribution is
A) ${\ ^{16}C}_{6}\left( \frac{1}{4} \right)^{10}\left( \frac{3}{4} \right)^{6}$
B) ${\ ^{16}C}_{6}\left( \frac{1}{4} \right)^{6}\left( \frac{3}{4} \right)^{10}$
C) ${\ ^{12}C}_{6}\left( \frac{1}{4} \right)^{10}\left( \frac{3}{4} \right)^{6}$
D) ${\ ^{12}C}_{6}\left( \frac{1}{4} \right)^{6}\left( \frac{3}{4} \right)^{6}$
Explanation
(b)
In binomial distribution, variance $= npq$ and mean $= np$.
From the given condition
$npq = 3\ and\ np = 4$
$\therefore\frac{npq}{np} = \frac{3}{4}$
$\Rightarrow \ q = \frac{3}{4},\ p = \frac{1}{4}\ and\ n = 16$
Probability of exactly six success $= {\ ^{16}C}_{6}\left( \frac{1}{4} \right)^{6}\left( \frac{3}{4} \right)^{10}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.