If the mean and variance of a binomial distribution are $\frac{5}{6}$ and $\frac{25}{36}$ respectively. They value of $P(X=2)$ is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → ${^5C}_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3$
Mean, $E(X)=np=\frac{5}{6}$ ...(1)
Variance, $Var(X)=np(1-p)=\frac{25}{36}$ ...(2)
from (1) and (2),
$\frac{5}{6}(1-p)=\frac{25}{36}$
$⇒P=1-\frac{5}{6}=\frac{1}{6}$
$P(X=2)={^5C}_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3$