Question

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

Answer 1

The answer is `=0.8009`

Explanation:

The Poisson distribution is

`P(x)=(e^-mu mu^x)/(x!)`

`mu=3`

`3! =3*2*1`

`P(x>=2)=1-P(1)-P(0)`

`P(1)=(e^-3 *3^1)/(1!)=3/e^3`

`P(0)=e^-3=1/e^3`

Therefore,

`P(x>=2)=1-P(1)-P(0)=1-0.0498-0.1494=0.8009`

Answer 2

`P(x>=2)=0.800852`

Explanation:

In a Poisson probability distribution, if mean value of success is `mu`,

the probability of getting `x` successes is given by

`P(x)=(e^(-mu)mu^x)/(x!)`

Now `P(x>=2)` means `1-P(x=0)-P(x=1)`

Here `mu=3` and `e^(-mu)=e^(-3)=0.049787`

and hence, desired probability is

`1-(e^(-3)3^0)/(0!)-(e^(-3)3^1)/(1!)`

= `1-e^(-3)xx(1+3)`

= `1-0.049787xx4`

= `1-0.199148`

= `0.800852`