If a random variable X follows binomial distribution with mean 5 and variance $\frac{5}{2},$ then $P(X≤9)$ is :

$\frac{1023}{1024}$

The correct answer is Option (1) → $\frac{1023}{1024}$

Given that X follow a binomial distribution X ∼ B in (n, p)

Mean, $np=5$   ....(1)

Variance = $np(1-p)=\frac{5}{2}$    ....(2)

Using (1) and (2),

$⇒5(1-p)=\frac{5}{2}$

$⇒p=1-\frac{1}{2}=\frac{1}{2}$

From eq. (1)

$n×\frac{1}{2}$

$n=10$

$P(X≤9)=1-P(X=10)$

$=1-{^{10}C}_{10}(\frac{1}{2})^{10}(1-\frac{1}{2})^0$

$=1-\frac{1}{2^{10}}=\frac{1023}{1024}$