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(C) Given that in $3$ trials of a binomial distribution,the probability of $2$ successes is $9$ times the probability of $3$ successes. Let $p$ be the

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) Given that in $3$ trials of a binomial distribution,the probability of $2$ successes is $9$ times the probability of $3$ successes. Let $p$ be the probability of success and $q$ be the probability of failure,where $p + q = 1$. The probability of $r$ successes in $n$ trials is given by $P(X = r) = {}^nC_r p^r q^{n-r}$. For $n = 3$: $P(X = 2) = 9 	imes P(X = 3)$ ${}^3C_2 p^2 q^1 = 9 	imes {}^3C_3 p^3 q^0$ $3 p^2 q = 9 p^3$ Since $p 
eq 0$,we can divide by $3p^2$: $q = 3p$ Substitute $q = 1 - p$: $1 - p = 3p$ $1 = 4p$ $p = \frac{1}{4}$ Thus,the probability of success in each trial is $\frac{1}{4}$. Hence,option $C$ is correct.

In $3$ trials of a binomial distribution,the probability of $2$ successes is $9$ times the probability of $3$ successes. Then the probability of success in each trial is

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