Consider a Poisson distribution with μ=3. What is P(x≥2)?

Explanation:
The Poisson distribution is
${\displaystyle P\left(x\right)=\frac{e^{-\mu }{\mu }^x}{x!}}$
${\displaystyle \mu =3}$
3! = 3*2*1
${\displaystyle P(x\ge 2)=1-P\left(1\right)-P\left(0\right)}$
${\displaystyle P\left(1\right)=\frac{e^{-3}\cdot 3^1}{1!}=\frac{3}{e^3}}$
${\displaystyle P\left(0\right)=e^{-3}=\frac{1}{e^3}}$
Therefore,
${\displaystyle P(x\ge 2)=1-P\left(1\right)-P\left(0\right)=1-0.0498-0.1494=0.8009}$
Explanation:
In a Poisson probability distribution, if mean value of success is ${\displaystyle \mu }$,
the probability of getting x successes is given by
${\displaystyle P\left(x\right)=\frac{e^{-\mu }{\mu }^x}{x!}}$
Now ${\displaystyle P(x\ge 2)}$ means 1-P(x=0)-P(x=1)
Here ${\displaystyle \mu =3}$ and ${\displaystyle e^{-\mu }=e^{-3}=0.049787}$
and hence, desired probability is
${\displaystyle 1-\frac{e^{-3}{3}^0}{0!}-\frac{e^{-3}{3}^1}{1!}}$
${\displaystyle =1-e^{-3}\times (1+3)}$
${\displaystyle =1-0.049787\times 4}$
=1-0.199148
=0.800852

Compute ${\displaystyle P(X\ge 2)}$
The required probability is
${\displaystyle P(X\ge 2)=1-P\left(X<2\right)}$
=1-[P(X=0)+P(X=1)]
=1-[0.049787+0.149361] from statistical table
=0.8009
The required ${\displaystyle P(X\ge 2)=0.8009}$