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

Given X follows binomial distribution with Parameters n and P.

$$\begin{gathered} Given that \mathrm{P}(\mathrm{x}=2)=\mathrm{P}(\mathrm{x}=3) \\ \Rightarrow {}^{\mathrm{n}}{\mathrm{c}}_2 {\mathrm{p}}^2 {\mathrm{q}}^{\mathrm{n}-2}={}^{\mathrm{n}}{\mathrm{c}}_3 {\mathrm{p}}^3 {\mathrm{q}}^{\mathrm{n}-3} \\ \Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{2\times 1}\times p^2{(1-p)}^{\mathrm{n}-2}=\frac{\mathrm{n}(\mathrm{n}-1) (\mathrm{n}-2)}{3\times 2\times 1}\times p^3\times {(1-p)}^{\mathrm{n}-3} \\ \Rightarrow \frac{p^2{(1-p)}^{\mathrm{n}-2}}{p^3{(1-p)}^{\mathrm{n}-3}}=\frac{\mathrm{n}-2}{3} \\ \Rightarrow \frac{1-p}{p}=\frac{\mathrm{n}-2}{3} \\ \Rightarrow 3(1-p)=p(\mathrm{n}-2) \\ \Rightarrow 3-3p=np-2p \\ \Rightarrow np=3-p \end{gathered}$$

$\therefore$ mean of $X=\mathrm{E}\left(\mathrm{X}\right)=3-\mathrm{P}$