X B(n,p), sum and product of mean & variance are 24 and 128.

Question

$X\sim B(n,p)$, sum and product of mean \& variance are 24 and 128. If $P(X>n-3)=\dfrac{k}{2^n}$, find $k$.

Accepted Answer

As before $p=q=\dfrac12,n=32$.

$$P(X>n-3) = ({}^{32}C_2+{}^{32}C_1+{}^{32}C_0)\left(\frac12\right)^{32} = \frac{529}{2^{32}} \implies k=529$$

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