Question

The mean and variance of a random variable $X$ having binomial distribution are $4$ and $2$ respectively. Then $P(X>6)$ =

• A. $\frac{9}{128}$
• B. $\frac{81}{128}$
• C. $\frac{9}{256}$
• D. $\frac{3}{11}$

Answer 1

Answer: (C)

$\frac{9}{256}$

$Mean=np$
$=4$
$Variance=np(1-p)$
$=2$
Hence $4(1-p)=2$
$1-p=\frac{1}{2}$

$p=\frac{1}{2}$

No.of trials will be
$n$
$=2\left(4\right)$
$=8$.
Therefore for $P(X>6)$ we get
${}^8{C}_7\frac{1}{2^8}+{}^8{C}_8\frac{1}{2^8}$

$=\frac{1}{2^8}[8+1]$

$=\frac{9}{2^8}$

$=\frac{9}{256}$