Question

If the mean and variance of binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than one is equal to

• A. $\frac{5}{16}$
• B. $\frac{7}{16}$
• C. $\frac{11}{16}$
• D. $\frac{9}{16}$

Answer 1

The correct option is C $\frac{11}{16}$

Mean $=np$ where p is the probability of success.
Variance $=np(1-p)$
Variance $=(S.D{)}^2$
$1=2(1-p)$
$\frac{1}{2}=1-p$
$p=\frac{1}{2}$
$np=2$

$\frac{n}{2}=2$
$n=4$
Hence the required probability is $1-\frac{1}{2^4}-{}^4{C}_1\frac{1}{2^4}$
$=1-\left[\frac{1+4}{16}\right]$
$=\frac{11}{16}$