Solve this following
Question:

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 :

1. $\frac{32}{625}$

2. $\frac{80}{243}$

3. $\frac{40}{243}$

4. $\frac{128}{625}$

Correct Option: 1

Solution:

$\mathrm{P}(\mathrm{X}=1)={ }^{5} \mathrm{C}_{1} \cdot \mathrm{p} \cdot \mathrm{q}^{4}=0.4096$

$\mathrm{P}(\mathrm{X}=2)={ }^{5} \mathrm{C}_{2} \cdot \mathrm{p}^{2} \cdot \mathrm{q}^{3}=0.2048$

$\Rightarrow \frac{\mathrm{q}}{2 \mathrm{p}}=2$

$\Rightarrow \mathrm{q}=4 \mathrm{p}$ and $\mathrm{p}+\mathrm{q}=1$

$\Rightarrow \mathrm{p}=\frac{1}{5}$ and $\mathrm{q}=\frac{4}{5}$

Now

$\mathrm{P}(\mathrm{X}=3)={ }^{5} \mathrm{C}_{3} \cdot\left(\frac{1}{5}\right)^{3} \cdot\left(\frac{4}{5}\right)^{2}=\frac{10 \times 16}{125 \times 25}=\frac{32}{625}$