If the mean and standard deviation of a binomial variate X are 8 and 2 respectively, then P(X≥1) is :

$1-\frac{1}{2^{16}}$

The correct answer is Option (4) → $1-\frac{1}{2^{16}}$

In Binomial Distribution,

Mean, $μ=np=8$  ....(1)

Standard Deviation, $σ=\sqrt{np(1-p)}=2$   ....(2)

from (1) and (2) we get,

$np(1-p)=2$

$⇒8(1-p)=2$

$⇒p=\frac{1}{2}$

$⇒n=\frac{8}{p}=16$

$∴P(X≥1)=1-P(X=0)$

$=1-{^{16}C}_0(\frac{1}{2})^0(\frac{1}{2})^{16}$

$=1-\frac{1}{2^{16}}$