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(B) For a binomial distribution $X \sim B(n, p)$,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.Given $\mu - \sigma^2 = 1

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$11$

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(B) For a binomial distribution $X \sim B(n, p)$,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.Given $\mu - \sigma^2 = 1$,we have $np - npq = 1$,which simplifies to $np(1-q) = 1$,so $np^2 = 1$.Given $2 P(X=2) = 3 P(X=1)$,we use the formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$.$2 \binom{n}{2} p^2 q^{n-2} = 3 \binom{n}{1} p^1 q^{n-1}$.$2 \cdot \frac{n(n-1)}{2} p^2 q^{n-2} = 3n p q^{n-1}$.$(n-1) p = 3q = 3(1-p)$.$np - p = 3 - 3p \implies np + 2p = 3$.Since $np^2 = 1$,we have $n = \frac{1}{p^2}$.Substituting $n$: $\frac{1}{p^2} \cdot p + 2p = 3 \implies \frac{1}{p} + 2p = 3$.$1 + 2p^2 = 3p \implies 2p^2 - 3p + 1 = 0$.$(2p-1)(p-1) = 0$. Since $p Then $n = \frac{1}{(1/2)^2} = 4$.We need $n^2 P(X>1) = 16(1 - P(X=0) - P(X=1))$.$P(X=0) = (1/2)^4 = 1/16$.$P(X=1) = \binom{4}{1} (1/2)^1 (1/2)^3 = 4 \cdot (1/16) = 4/16$.$P(X>1) = 1 - (1/16 + 4/16) = 11/16$.$n^2 P(X>1) = 16 \cdot (11/16) = 11$.

$A$ random variable $X$ follows a binomial distribution in which the difference between its mean and variance is $1$. If $2 P(X=2)=3 P(X=1)$,then $n^2 P(X>1)=$

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