If the probability of two successes is 9 times the probability of 3 successes in 3 trials of a binomial distribution, then the probability of success in each trial is:

1/4

The correct answer is Option (1) → 1/4

Let probability of success in each trial be $p$ and failure be $q=1-p$

Given number of trials $n=3$

Probability of exactly $2$ successes

$P(2)=\frac{3!}{2!1!}p^2q=3p^2q$

Probability of exactly $3$ successes

$P(3)=p^3$

Given $P(2)=9P(3)$

$3p^2q=9p^3$

Divide by $3p^2$

$q=3p$

$1-p=3p$

$1=4p$

$p=\frac{1}{4}$

The probability of success in each trial is $\frac{1}{4}$.