Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to

Let p be the probability of success.
\(\begin{aligned}
& \mathrm{P}(\mathrm{X}=1)=0.4096 \text { and } \mathrm{P}(\mathrm{X}=2)=0.2048 \\
& \Rightarrow{ }^5 \mathrm{C}_1 \mathrm{p}^1 \mathrm{q}^4=0.4096 \text { and }{ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3=0.2048 \\
& \Rightarrow 5 \mathrm{pq}^4=0.4096 \text { and } 10 \mathrm{p}^2 \mathrm{q}^3=0.2048 \\
& \Rightarrow \frac{10 \mathrm{p}^2 \mathrm{q}^3}{5 \mathrm{pq}^4}=\frac{0.2048}{0.4096}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{2 \mathrm{p}}{\mathrm{q}}=\frac{1}{2} \\ & \Rightarrow 4 \mathrm{p}=\mathrm{q} \\ & \Rightarrow 4 \mathrm{p}=1-\mathrm{p} \\ & \Rightarrow 5 \mathrm{p}=1 \\ & \Rightarrow \mathrm{p}=\frac{1}{5}\end{aligned}\)
\(\begin{aligned} & \Rightarrow q=1-\frac{1}{5}=\frac{4}{5} \\ & \begin{aligned} P(X=3) & ={ }^5 C_3 p^3 q^2 \\ & =10\left(\frac{1}{5}\right)^3\left(\frac{4}{5}\right)^2 \\ & =10 \times \frac{1}{125} \times \frac{16}{25}=\frac{32}{625}\end{aligned}\end{aligned}\)