Answer:
Option C
Explanation:
Here, mean =4 and variance =2
$\Rightarrow$ np=4 and npq=2
so, $\frac{npq}{np}=\frac{2}{4}\Rightarrow q=\frac{1}{2}$
Then $p=1-q=1-\frac{1}{2}=\frac{1}{2}$
Mean= np=4
$\Rightarrow $ $n\times\frac{1}{2}=4\Rightarrow n=8$
$\therefore$ $P(X=r)=^{n}C_{r} p^{r}q^{n-r}$
$=^{8}C_{r} \left(\frac{1}{2}\right)^{8}$ $ \left[\because p=q=\frac{1}{2}\right]$
The required probabilty of atleast 7 successes is
$P(X\geq 7)=P(X=7)+P(X=8)$
= $\left(^{8}C_{7}+^{8}C_{8}\right)\left(\frac{1}{2}\right)^{8}$
= $\left(\frac{8!}{7!1!}+\frac{8!}{8!0!}\right)\left(\frac{1}{2}\right)^{8}$
= $\left(8+1\right)\left(\frac{1}{2}\right)^{8}=\frac{9}{256}$
