If the sum and product of the mean and variance of a binomial distrib…

The correct answer is Option (1) → $25\left(\frac{1}{2}\right)^{24}$

In Binomial distribution,

Mean = $np$

Variance = $np(1-p)$

Let,

$x=np$ and $y=np(1-p)$

given,

$x+y=18$   ...(1)

$xy=72$   ...(2)

$⇒x^2-18t+72=0$

$⇒(x-12)(x-6)=0$

$x=12$ or 6

$y=6$ or 12

Case 1: $np=12,np(1-p)=6$

$12(1-p)=6$

$⇒p=\frac{1}{2}⇒n=24$

$∴P(X≤1)=P(0)+P(1)$

$={^nC}_kP^k(1-P)^{n-k}$

$={^{24}C}_0(\frac{1}{2})^0(\frac{1}{2})^{24}+{^{24}C}_1(\frac{1}{2})^1(\frac{1}{2})^{23}$

$=(\frac{1}{2})^{24}+24×(\frac{1}{2})^{24}$

$=\frac{25}{2^{24}}$