If the mean and the variance of a binomial distribution are 6 and 3 respectively, then the probability of five successes is :

${^{12}C}_5 \left(\frac{1}{2}\right)^{12}$

The correct answer is Option (1) → ${^{12}C}_5 \left(\frac{1}{2}\right)^{12}$

In a Binomial Distribution,

Mean, $μ=np$

Variance, $σ^2=np(1-p)$

and,

$μ=6,σ^2=3$

$⇒np=6$

and,

$np(1-p)=3$

$6(1-p)=3$

$p=\frac{1}{2}$ and $n=12$

the probability mass function (PMF) is,

$P(X=k)={^nC}_kp^k(1-p)^{n-k}$

$P(X=5)={^{12}C}_5(\frac{1}{2})^5(\frac{1}{2})^{12-5}$

$={^{12}C}_5 \left(\frac{1}{2}\right)^{12}$