The mean and variance of a binomial distributors are 4 and 2 respectively. Then the probability of atmost 2 successes is :

$\frac{37}{256}$

The correct answer is Option (4) → $\frac{37}{256}$

For a binomial distribution B(n, p), the mean (μ) and variance $(σ^2)$ are given by,

$μ=n.p$   ...(1)

$σ^2=n×p×(1-p)$   ...(2)

$np=4⇒p=\frac{4}{n}$

Simplify eqn (2)

$n×\frac{4}{n}\left(1-\frac{4}{n}\right)=2$

$4\left(1-\frac{4}{n}\right)=2$

$1-\frac{4}{n}=\frac{1}{2}$

$⇒\frac{4}{n}=\frac{1}{2}$

$⇒n=8$

$∴P=\frac{4}{8}=\frac{1}{2}$

$P(X=0)={^8C}_0(0.5)^0(0.5)^8=\frac{1}{2^8}=\frac{1}{256}$

$P(X=1)={^8C}_1(\frac{1}{2})^1(\frac{1}{2})^7=\frac{8}{256}$

$P(X=2)={^8C}_2(\frac{1}{2})^2(\frac{1}{2})^6=\frac{28}{256}$

$∴P(X≤2)=\frac{37}{256}$