If the mean and the variance of a binomial distribution are 6 and 4 respectively, then the probability of no success is :

$\left(\frac{2}{3}\right)^{18}$

The correct answer is Option (1) → $\left(\frac{2}{3}\right)^{18}$

Mean, $μ=np=6$ [given]

Variance, $σ^2=np(1-p)=4$ [given]

from (1) and (2)

$6(1-p)=4$

$1-p=\frac{4}{6}⇒p=1-\frac{2}{3}=\frac{1}{3}$

$∴n=\frac{6}{p}=6×3=18$

Probability, $P(X=0)=(1-p)^n=\left(\frac{2}{3}\right)^{18}$