Question

If the mean and the variance of a binomial variate are $2$ and $1$ respectively, the probability that $X$ takes a value greater than $1$ is equal to $1-k/16$.Find the value of $k$.

Answer 1

For Binomial distribution, Mean $=np$ and variance $=npq$
$\therefore np=2$ and $npq=1$
$\Rightarrow q=\frac{1}{2}$ and $p+q=1$
$\Rightarrow p=\frac{1}{2}$
$\therefore n=4,p=q=\frac{1}{2}$

Now, $P(X>1)=1-(P(X=0)+P(X=1))$
$=1{-}^4{C}_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^4{-}^4{C}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^3=1-\frac{1}{16}-\frac{4}{16}=\frac{11}{16}\therefore k=5$