In 3 trials of a binomial distribution, the probability of 2 successe…

The correct answer is Option (3) → $\frac{1}{4}$

Given, in 3 trials of a binomial distribution

$P(2) = 9P(3)$   ...(i)

Let the probability of success be p,

then probability of failure = $q = 1-p$.

Here, the number of Bernoullian trials is 3 i.e. $n = 3$.

From (i), we get

${^3C}_2 p^2q = 9.{^3C}_3 p^3$   $(∵p≠0)$

$⇒ 3p^2q=9.1.p^3⇒q=3p$  $(∵q=1-p)$

$⇒ 1-p=3p$

$⇒ 4p = 1⇒ p =\frac{1}{4}$

Hence, the probability of success in each trial = $\frac{1}{4}$