The variance of a Poisson variate $X$ is $2$. Then $P\left(X\ge 3\right)=$

We know that formula of poisson distribution is given by,

$P\left(X=n\right)=\frac{{\lambda }^n{e}^{-\lambda }}{n!}$ where $\lambda =$ mean or variance and given $\lambda =2$,

Now using the above formula in $P\left(X\ge 3\right)=1-\left\{P\left(0\right)+P\left(1\right)+P\left(2\right)\right\}$ we get,

$P\left(X\ge 3\right)=1-\left\{P\left(0\right)+P\left(1\right)+P\left(2\right)\right\}$

$\Rightarrow P\left(X\ge 3\right)=1-\left\{\frac{2^0{e}^{-2}}{0!}+\frac{2^1{e}^{-2}}{1!}+\frac{2^2{e}^{-2}}{2!}\right\}$

$\Rightarrow P\left(X\ge 3\right)=1-e^{-2}\left\{\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}\right\}$

$\Rightarrow P\left(X\ge 3\right)=1-\frac{1}{e^2}\left\{1+2+2\right\}$

$\Rightarrow P\left(X\ge 3\right)=\frac{e^2-5}{e^2}$