The random variable X follows binomial distribution B(n,p), for which the difference of th…

Question

The random variable $X$ follows binomial distribution $B(n,p)$, for which the difference of the mean and the variance is 1 . If $2P(X=2)=3P(X=1)$, then ${n}^{2}P(X>1)$ is equal to :

Accepted Answer

$\begin{gathered} np-npq=1 \ \ \Rightarrow np(1-q)=1 \ \ \Rightarrow n{p}^{2}=1 $

$ 2P(X=2)=3P(X=1) \ \ 2\cdot {}^{n}{C}_{2}{p}^{2}q{}^{n-2}=3\cdot {}^{n}{C}_{1}p\cdot {q}^{n-1} \ \ \Rightarrow 2\cdot \frac{n\cdot (n-1)}{2}\cdot p=3\cdot n\cdot q \ \ \Rightarrow (n-1)p=3(1-p) \ \ \Rightarrow (\frac{1}{{p}^{2}}-1)p=3(1-p) \ \ \Rightarrow \frac{(1-p)(1+p)}{p}=3(1-p) \ \ \Rightarrow 1+p=3p \ \ \Rightarrow p=\frac{1}{2} \ \ 	herefore n=4 $

$ {n}^{2}P(x>1) ={n}^{2}(1-P(x=1)-P(x=0)) \ \ =16(1-{}^{4}{C}_{1}\cdot {(\frac{1}{2})}^{4}-{(\frac{1}{2})}^{4})=11 \end{gathered}$

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