In 5 trials of binomial distribution, the probability of 3 successes is 4 times the probability of 2 successes. The probability of success in each trial is:

$\frac{4}{5}$

The correct answer is Option (2) → $\frac{4}{5}$

Given: n = 5, P(3 \text{ successes}) = 4 × P(2 \text{ successes})

Binomial probability formula: $P(k) = \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}$

For 3 successes:

$P(3) = \frac{5!}{3!2!} p^3 (1-p)^2 = 10 p^3 (1-p)^2$

For 2 successes:

$P(2) = \frac{5!}{2!3!} p^2 (1-p)^3 = 10 p^2 (1-p)^3$

Given: $P(3) = 4 P(2)$

$10 p^3 (1-p)^2 = 4 * 10 p^2 (1-p)^3$

$p^3 (1-p)^2 = 4 p^2 (1-p)^3$

$p = 4 (1-p)$

$p = 4 - 4p \Rightarrow 5p = 4 \Rightarrow p = \frac{4}{5}$

Probability of success in each trial = 4/5