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 :

${^5C}_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3$

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),

$\frac{5}{6}(1-p)=\frac{25}{36}$

$⇒P=1-\frac{5}{6}=\frac{1}{6}$

$P(X=2)={^5C}_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3$