Here, mean $=4$ and variance $=2$
$\Rightarrow \quad n p=4$ and $n p q=2$
So, $\frac{\mathrm{npq}}{\mathrm{np}}=\frac{2}{4} \Rightarrow \mathrm{q}=\frac{1}{2}$
Then, $\mathrm{p}=1-\mathrm{q}=1-\frac{1}{2}=\frac{1}{2}$
$\begin{aligned}
& \text { Mean }=\mathrm{np}=4 \\
\Rightarrow & \mathrm{n} \times \frac{1}{2}=4 \Rightarrow \mathrm{n}=8 \\
\therefore & \mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{p}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \\
&={ }^{8} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{8} \quad\left[\because \mathrm{p}=\mathrm{q}=\frac{1}{2}\right]
\end{aligned}$
The required probability of atleast 7 successes is
$\begin{array}{l}
P(X \geq 7)=P(X=7)+P(X=8) \\
=\left({ }^{8} C_{7}+{ }^{8} C_{8}\right)\left(\frac{1}{2}\right)^{8} \\
=\left(\frac{8 !}{7 ! ! !}+\frac{8 !}{8 ! 0 !}\right)\left(\frac{1}{2}\right)^{8}
\end{array}$
$=(8+1)\left(\frac{1}{2}\right)^{8}=\frac{9}{256}$