Q44 / 90 · JEE Main 2021 (18 Mar Shift 2) · Mathematics

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 :

Solution

$P (X = 1) = {}^{5}C_{1} \cdot p . q^{4} = 0.4096$

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

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

$\Rightarrow q = 4 p$ and $p + q = 1$

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

Now

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

Hence, the correct answer is option: 'A'.