Question

The mean and variance of a binomial variable $X$ are $2$ and $1$ respectively. The probability that $X$ takes values greater than $1$ is $1-\frac{k}{16}$.The value of $k$ is

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