X B(n,p), mean - variance =1.

Question

$X\sim B(n,p)$, mean $-$ variance $=1$. If $2P(X=2)=3P(X=1)$, find $n^2P(X>1)$.

Accepted Answer

From $2P(X=2)=3P(X=1)$: $(n-1)P=3(1-P)$. From $np(1-(1-p))=1\Rightarrow nP^2=1\Rightarrow n=1/P^2$. Substituting and solving the quadratic $2P^2-3P+1=0$ gives $P=\dfrac12$ (rejecting $P=1$), so $n=4$.

$$n^2P(X>1) = 16\left[1-\left(\frac12\right)^4-4\left(\frac12\right)^4\right] = 16-1-4=11$$

Solutions · 1