Independent solution

How to solve this Poisson Distribution question

Answer in brief

For a Poisson count, the second moment is lambda squared plus lambda, so the rate is 3. Conditioning the upper count event on at least one count gives 0.842813 and choice C.

Setup

Setup

Let N be Poisson with rate lambda 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 lambda 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)}}