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(B) Given that the random variable $X$ follows a binomial distribution with parameters $n=5$ and $p$. The probability mass function is given by $P(X=k

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$1/10$

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(B) Given that the random variable $X$ follows a binomial distribution with parameters $n=5$ and $p$. The probability mass function is given by $P(X=k) = { }^n C_k p^k (1-p)^{n-k}$.Given $P(X=2) = 9 P(X=3)$.Substituting the values: ${ }^5 C_2 p^2 (1-p)^{5-2} = 9 	imes { }^5 C_3 p^3 (1-p)^{5-3}$.Since ${ }^5 C_2 = \frac{5 	imes 4}{2 	imes 1} = 10$ and ${ }^5 C_3 = \frac{5 	imes 4 	imes 3}{3 	imes 2 	imes 1} = 10$,we have:$10 p^2 (1-p)^3 = 9 	imes 10 p^3 (1-p)^2$.Dividing both sides by $10 p^2 (1-p)^2$ (assuming $p 
eq 0, 1$):$(1-p) = 9p$.$1 = 10p$.$p = 1/10$.

If a random variable $X$ follows the binomial distribution with parameters $n=5, p$ and $P(X=2)=9 P(X=3)$,then $p$ is equal to

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