If the mean and the variance of a binomial distribution are 6 and 4 respectively, then the probability of no success is :
Answer & explanation
Correct answer: option 1
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, $P(X=0)=(1-p)^n=\left(\frac{2}{3}\right)^{18}$