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:

Given X follows binomial distribution n=5

Given $\mathrm{P}(\mathrm{x}=1)=0.4096$

$$\begin{gathered} \Rightarrow {}^5{\mathrm{c}}_1 p^1 {\left(\mathrm{q}\right)}^{5-1}=0.4096 \\ \Rightarrow 5p {(1-p)}^4=0.4096 -\left(1\right) \end{gathered}$$

and $\mathrm{P}(\mathrm{x}=2)=0.2048$

$$\begin{gathered} \Rightarrow {}^5{\mathrm{c}}_2 p^2 {\left(\mathrm{q}\right)}^{5-2}=0.2048 \\ \Rightarrow 10 p^2 {(1-\mathrm{P})}^3=0.2048 \\ \frac{\left(1\right)}{\left(2\right)} \Rightarrow \frac{(1-p)}{2p}=2 \\ \Rightarrow 1-p=4p \\ \Rightarrow 1=5p \\ \Rightarrow p=\frac{1}{5} \\ \Rightarrow \mathrm{q}=1-p=1-\frac{1}{5}=\frac{4}{5} \end{gathered}$$

Now $\mathrm{P}(X=3)={}^5{\mathrm{c}}_3 {\mathrm{p}}^3 {\mathrm{q}}^{5-3}$

                         $$\begin{gathered} =10\times {\left(\frac{1}{5}\right)}^3 {\left(\frac{4}{5}\right)}^2 \\ =10\times \frac{1}{125}\times \frac{16}{25} \\ =\frac{32}{625} \end{gathered}$$