Independent solution

How to solve this Poisson Distribution question

Setup

Setup

Let λ denote the common Poisson mean for one hour and recover it from the probability of a zero count.

e−λ=0.60653e^{-\lambda}=0.60653
λ=−log⁡(0.60653)=0.5000010877…\lambda=-\log(0.60653)=0.5000010877\ldots

Model

Model

The sum across two independent hours is Poisson with mean equal to the sum of the hourly means.

T∼Poisson⁡(2λ)T\sim\operatorname{Poisson}(2\lambda)
μT=2λ=1.0000021754…\mu_T=2\lambda=1.0000021754\ldots

Compute

Compute

Fewer than two arrivals means that the two-hour total is either zero or one.

Pr⁡(T<2)=Pr⁡(T=0)+Pr⁡(T=1)\Pr(T<2)=\Pr(T=0)+\Pr(T=1)
Pr⁡(T<2)=e−μT(1+μT)=0.7357580821…\Pr(T<2)=e^{-\mu_T}(1+\mu_T)=0.7357580821\ldots

Answer

Answer

The computed probability rounds to the first listed value.

Pr⁡(T<2)≈0.736(A)\boxed{\Pr(T<2)\approx 0.736\quad\text{(A)}}