Independent solution

How to solve this Poisson Superposition question

Setup

Setup

Let λ be the common mean for one location and translate the supplied relationship between its first two Poisson masses.

Pr⁡(X=0)=e−λ,Pr⁡(X=1)=λe−λ\Pr(X=0)=e^{-\lambda},\qquad \Pr(X=1)=\lambda e^{-\lambda}
λe−λ=2e−λ\lambda e^{-\lambda}=2e^{-\lambda}

Model

Model

Cancel the positive exponential factor to find the individual mean, then superpose the three independent Poisson counts.

λ=2\lambda=2
T=X1+X2+X3∼Poisson⁡(3λ)=Poisson⁡(6)T=X_1+X_2+X_3\sim\operatorname{Poisson}(3\lambda)=\operatorname{Poisson}(6)

Compute

Compute

Use the complement of total counts zero and one.

Pr⁡(T≥2)=1−Pr⁡(T=0)−Pr⁡(T=1)\Pr(T\ge 2)=1-\Pr(T=0)-\Pr(T=1)
Pr⁡(T≥2)=1−e−6(1+6)=0.9826487348…\Pr(T\ge 2)=1-e^{-6}(1+6)=0.9826487348\ldots

Answer

Answer

Rounding the computed probability to three decimals gives the final listed value.

Pr⁡(T≥2)≈0.983(E)\boxed{\Pr(T\ge 2)\approx 0.983\quad\text{(E)}}