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(B) Given that $X \sim B(n, p)$,the probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$. Given $P(X=2) = P(X=3)$,we have:

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$3-p$

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(B) Given that $X \sim B(n, p)$,the probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$. Given $P(X=2) = P(X=3)$,we have: $\binom{n}{2} p^2 q^{n-2} = \binom{n}{3} p^3 q^{n-3}$ $\frac{n!}{2!(n-2)!} p^2 q^{n-2} = \frac{n!}{3!(n-3)!} p^3 q^{n-3}$ Dividing both sides by $n! p^2 q^{n-3}$: $\frac{1}{2(n-2)!} q = \frac{1}{6(n-3)!} p$ $\frac{q}{2} = \frac{p}{6(n-2)}$ $3q = p(n-2)$ Since $q = 1-p$,we substitute: $3(1-p) = np - 2p$ $3 - 3p = np - 2p$ $np = 3 - p$ The mean of a binomial distribution is given by $E(X) = np$. Therefore,the mean is $3-p$.

It is given that the discrete random variable is $X \sim B(n, p)$ and $P(X=2)=P(X=3)$. Find the mean of the distribution.

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