In a Binomial distribution consisting of 5 independent trials, probabilities of exactly 1 and 2 successes are 0.4096 and 0.2048 respectively, then the probability, of getting exactly 4 successes, is

Let \(\mathrm{P}(\mathrm{X}=1)\) be probability of one success and \(\mathrm{P}(\mathrm{X}=2)\) be probability of two success
\(\begin{array}{ll}
\therefore \quad & P(X=1)={ }^5 \mathrm{C}_1 \mathrm{p}^1 \mathrm{q}^4=0.4096 ...(i)\\
& \mathrm{P}(\mathrm{X}=2)={ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3=0.2048...(ii)
\end{array}\)
Where \(p=\) probability of success
\(\mathrm{q}=\) probability of failure
\(\therefore \quad\) Dividing (i) by (ii), we get
\(\begin{aligned}
& \frac{{ }^5 \mathrm{C}_1 \mathrm{pq}^4}{{ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3}=\frac{0.4096}{0.2048} \\
& \frac{5 \mathrm{q}}{10 \mathrm{p}}=2 \\
& \Rightarrow \mathrm{q}=4 \mathrm{p} \\
\therefore \quad & \text { Also, } \mathrm{p}+\mathrm{q}=1 \\
\therefore \quad & p+4 p=1
\end{aligned}\)
\(\begin{array}{ll}\therefore & p=\frac{1}{5} \\ & \therefore \\ & q=\frac{4}{5}\end{array}\)
Now, Probability of getting 4 successes
\(\begin{aligned}
& =\mathrm{P}(\mathrm{X}=4) \\
& ={ }^5 \mathrm{C}_4 \mathrm{p}^4 \mathrm{q} \\
& =5 \times\left(\frac{1}{5}\right)^4 \times \frac{4}{5} \\
& =\frac{4}{625}
\end{aligned}\)