Question

If the mean and variance of a binomial variate $X$ are $8$ and $4$ respectively, then $P(X<3)=$

• A. $\frac{265}{2^{15}}$
• B. $\frac{137}{2^{16}}$
• C. $\frac{267}{2^{16}}$
• D. $\frac{265}{2^{16}}$

Answer 1

The correct option is B $\frac{137}{2^{16}}$

Let '$p$' denote the probability of success and '$n$' denote the number of trials.
$E\left(x\right)=np$$=8$.
And $Var\left(x\right)=np(1-p)$ $=4$
Therefore $np(1-p)=4$
$8(1-p)=4$
$1-p=\frac{1}{2}$
$p=\frac{1}{2}$ ...(i)
Hence $n\left(\frac{1}{2}\right)=8$
$n=16$
Therefore $P(X<3)$
$=P\left(0\right)+P\left(1\right)+P\left(2\right)$
$=\frac{1}{2^{16}}({}^{16}{C}_0+{}^{16}{C}_1+{}^{16}{C}_2)$ $=\frac{1+16+120}{2^{16}}$ $=\frac{137}{2^{16}}$