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It is given that the discrete random variable is $X \sim B(n, p)$ and $P(X=2)=P(X=3)$.
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Verified Answer
The correct answer is:
$3-p$
$\mathrm{P}(x=2)=\mathrm{P}(x=3)$
${ }^n C_2 p^2 q^{\mathrm{n}-2}={ }^n C_3 p^3 q^{\mathrm{n}-3}$
$\begin{aligned} & \frac{n !}{2 !(n-2) !} p^2 \frac{q^n}{q^2}=\frac{n !}{3 !(n-3) !} p^3 \frac{q^n}{q^3} \\ & \frac{1}{n-2}=\frac{p}{3 q} \\ & 3 q=p(n-2) \\ & q=1-p \\ & 3(1-p)=n p-2 p \\ & n p=2 p+3(1-p)\end{aligned}$
$\begin{aligned} & =2 p+3-3 p \\ & =3-p\end{aligned}$
mean $=3-p$
${ }^n C_2 p^2 q^{\mathrm{n}-2}={ }^n C_3 p^3 q^{\mathrm{n}-3}$
$\begin{aligned} & \frac{n !}{2 !(n-2) !} p^2 \frac{q^n}{q^2}=\frac{n !}{3 !(n-3) !} p^3 \frac{q^n}{q^3} \\ & \frac{1}{n-2}=\frac{p}{3 q} \\ & 3 q=p(n-2) \\ & q=1-p \\ & 3(1-p)=n p-2 p \\ & n p=2 p+3(1-p)\end{aligned}$
$\begin{aligned} & =2 p+3-3 p \\ & =3-p\end{aligned}$
mean $=3-p$
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