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(B) Given that $X$ follows a binomial distribution $B(n, p)$,the probability mass function is given by $P(X = k) = ^{n}C_{k} p^{k} (1-p)^{n-k}$.We are

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$3 - p$

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(B) Given that $X$ follows a binomial distribution $B(n, p)$,the probability mass function is given by $P(X = k) = ^{n}C_{k} p^{k} (1-p)^{n-k}$.We are given $P(X = 2) = P(X = 3)$.Substituting the formula,we get: $^{n}C_{2} p^{2} (1-p)^{n-2} = ^{n}C_{3} p^{3} (1-p)^{n-3}$.Dividing both sides by $p^{2} (1-p)^{n-3}$,we get: $^{n}C_{2} (1-p) = ^{n}C_{3} p$.Expanding the combinations: $\frac{n!}{2!(n-2)!} (1-p) = \frac{n!}{3!(n-3)!} p$.Simplifying the factorials: $\frac{1}{2} (1-p) = \frac{1}{3(n-2)} (n-2)! \cdot \frac{1}{(n-3)!} p = \frac{1}{6} \cdot (n-2) \cdot \frac{1}{(n-2)!} \dots$ actually,$\frac{1}{2} (1-p) = \frac{1}{3 \cdot 2 \cdot 1} \cdot \frac{n!}{(n-3)!} \cdot \frac{(n-2)!}{n!} p = \frac{1}{6} (n-2) p$.Thus,$3(1-p) = (n-2)p$.$3 - 3p = np - 2p$.$np = 3 - p$.Since the mean $E(X) = np$,we have $E(X) = 3 - p$.

If $X$ has a binomial distribution,$B(n, p)$ with parameters $n$ and $p$ such that $P(X = 2) = P(X = 3)$,then $E(X)$,the mean of variable $X$,is

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