Question

If a random variable X satisfies the Poisson's distribution with a mean value of 2, then the probability that X > 2 is

• A. $2e^{-2}$
• B. $1-2e^{-2}$
• C. $3e^{-2}$
• D. $1-3e^{-2}$

Answer 1

The correct option is D $1-3e^{-2}$

$\text{Mean}\left(\lambda \right)=2\text{(given)}$ & $P(x=r)=\frac{e^{-\lambda }{\lambda }^r}{r!}$

Required probability is

$P(x\ge 2)=1-P(x\le 1)$

$=1-\left[P\right(X=0)+P(X=1\left)\right]$

$=1[\frac{{\lambda }^0.e^{-2}}{0!}+e^{-2}\frac{\lambda }{1!}]$

$=1-\frac{1}{e^2}(1+2)=1-3e^{-2}$