Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let N be the annual count and use its zero probability to recover the Poisson parameter.

Pr⁡(N=0)=e−λ=e−1.5\Pr(N=0)=e^{-\lambda}=e^{-1.5}
λ=1.5\lambda=1.5

Model

Model

Because the upper-tail event is contained in the conditioning event, divide its probability by the probability of at least one occurrence.

Pr⁡(N≥4∣N≥1)=Pr⁡(N≥4)Pr⁡(N≥1)\Pr(N\ge4\mid N\ge1)=\frac{\Pr(N\ge4)}{\Pr(N\ge1)}

Compute

Compute

Evaluate the numerator by complementing the first four Poisson masses and then normalize.

Pr⁡(N≥4)=1−e−1.5(1+1.5+1.522+1.536)\Pr(N\ge4)=1-e^{-1.5}\left(1+1.5+\frac{1.5^2}{2}+\frac{1.5^3}{6}\right)
Pr⁡(N≥4)=0.06564245438\Pr(N\ge4)=0.06564245438
Pr⁡(N≥1)=1−e−1.5=0.7768698399\Pr(N\ge1)=1-e^{-1.5}=0.7768698399
0.065642454380.7768698399=0.08449607774\frac{0.06564245438}{0.7768698399}=0.08449607774

Answer

Answer

The conditional upper-tail probability rounds to 0.084.

0.084(B)\boxed{0.084\quad\text{(B)}}