Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let N be the annual count. Convert the supplied positive-count probability into the Poisson zero mass.

Pr⁡(N=0)=1−0.10=0.90\Pr(N=0)=1-0.10=0.90
Pr⁡(N=0)=e−λ\Pr(N=0)=e^{-\lambda}

Model

Model

Solve for the mean parameter, then express the desired upper tail by excluding counts zero and one.

λ=−ln⁡(0.90)=0.1053605157\lambda=-\ln(0.90)=0.1053605157
Pr⁡(N≥2)=1−Pr⁡(N=0)−Pr⁡(N=1)\Pr(N\ge2)=1-\Pr(N=0)-\Pr(N=1)

Compute

Compute

Use the recovered zero mass to evaluate the one-count mass without introducing additional rounding.

Pr⁡(N=1)=e−λλ=0.90[−ln⁡(0.90)]\Pr(N=1)=e^{-\lambda}\lambda=0.90[-\ln(0.90)]
Pr⁡(N≥2)=1−0.90−0.90[−ln⁡(0.90)]\Pr(N\ge2)=1-0.90-0.90[-\ln(0.90)]
Pr⁡(N≥2)=0.005175535908\Pr(N\ge2)=0.005175535908

Answer

Answer

The probability of two or more events rounds to 0.0052.

0.0052(A)\boxed{0.0052\quad\text{(A)}}