If the mean and variance of a binomial variable $X$ are 2 and 1 respectively, then $P(X \geq 1)$ is equal to
If the mean and variance of a binomial variable $X$ are 2 and 1 respectively, then $P(X \geq 1)$ is equal to
$\frac{2}{3}$
$\frac{15}{16}$
$\frac{7}{8}$
$\frac{4}{5}$
Solution
Mean $\Rightarrow n p=2$ and variance $\Rightarrow n p q=1$
then $q=\frac{1}{2}, p=\frac{1}{2}$ and $n=4 \quad(\because p=1-q)$
$\therefore$ Required probability
$=P(X \geq 1)=1-P(X < 1)$
$=1-\left(\frac{1}{16}\right)$
$=\frac{15}{16}$