Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let N be Poisson with rate λ and recover that rate from its stated second moment.

E⁡[N2]=Var⁡(N)+E⁡[N]2=λ+λ2\operatorname{E}[N^2]=\operatorname{Var}(N)+\operatorname{E}[N]^2=\lambda+\lambda^2
λ2+λ=12\lambda^2+\lambda=12

Model

Model

The nonnegative solution is λ equal to 3. The requested conditional event uses the probability of two or more counts divided by the probability of one or more.

λ=3\lambda=3
Pr⁡(N≥2∣N≥1)=Pr⁡(N≥2)Pr⁡(N≥1)\Pr(N\ge2\mid N\ge1)=\frac{\Pr(N\ge2)}{\Pr(N\ge1)}

Compute

Compute

Use the Poisson probabilities at zero and one to evaluate both the numerator and denominator.

Pr⁡(N≥2)=1−e−3−3e−3=1−4e−3\Pr(N\ge2)=1-e^{-3}-3e^{-3}=1-4e^{-3}
Pr⁡(N≥1)=1−e−3\Pr(N\ge1)=1-e^{-3}
1−4e−31−e−3=0.8428129105…\frac{1-4e^{-3}}{1-e^{-3}}=0.8428129105\ldots

Answer

Answer

The conditional probability rounds to 0.84.

Pr⁡(N≥2∣N≥1)≈0.84(C)\boxed{\Pr(N\ge2\mid N\ge1)\approx0.84\quad\text{(C)}}