If a random variable (x) follows binomial distribution with mean 5...

$\begin{gathered}\text{ Mean }=5 \text{\text{ and } variance }=4 \\ 5^{23} \mathrm{P}(\mathrm{X}=3)=\lambda 4^\lambda \qquad ...(i) \end{gathered}$$\begin{aligned} \because & \text{ Mean } & =5 \\ \Rightarrow & n p & =5 \\ \Rightarrow & \text{ Variance } & =4 \\ \Rightarrow & n p q & =4 \\ \Rightarrow & 5 q & =4 \\ \Rightarrow & q & =\frac{4}{5} \end{aligned}$Now, $\quad p=1-q=1-\frac{4}{5}=\frac{1}{5}\begin{aligned} \because & n p & =5 \\ \Rightarrow & n & =25 \end{aligned}$ Now, $5^{23} \mathrm{P}(\mathrm{X}=3)=5^{23}\left[n c^3 p^3 q^{n-3}\right]\begin{aligned} & =5^{23}\left[{ }^{25} \mathrm{C}_3\left(\frac{1}{5}\right)^3\left(\frac{4}{5}\right)^{22}\right] \\ & =5^{23}\left[\frac{25 \times 24 \times 23}{6} \times \frac{1}{5^3} \times \frac{4^{22}}{5^{22}}\right] \\ & =4 \times 23 \times 4^{22} \end{aligned}$$\Rightarrow \quad 5^{23} \mathrm{P}(\mathrm{X}=3)=23 \times 4^{22}\qquad ....(ii)$ From (i) and (ii) $\lambda=23$