Answer:
Option B
Explanation:
Given , mean E(X)=5, and variance , Var(X)=2.5
$\therefore$ np=5 and npq=2.5
$\Rightarrow$ 5q=2.5 $\Rightarrow$ q=$\frac{1}{2}$
$\Rightarrow$ p+q=1
$\therefore$ $p=1-\frac{1}{2}=\frac{1}{2}$
$\therefore$ np=5
$\Rightarrow$ $n\times\frac{1}{2}=5$ $\left[\because p=\frac{1}{2}\right]$
$\Rightarrow$ n=10
$p(X<1)=p(X=0)=^{n}C_{r}p^{r}q^{n-r}$
= $p(X<1)=p(X=0)=^{10}C_{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{10-0}$
$=1\times1\times\left(\frac{1}{2}\right)^{10}=\left(\frac{1}{2}\right)^{10}$
