Question 2 of 8: Poisson Process — Road Cracks

Question 2: Poisson Process — Road Cracks (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Road cracks follow a Poisson process with rate 1.5 cracks per 10 km; road network length 1,500 km; per-car accident probability 0.00001 on a sample of 100,000 cars.

Find. (a) P(fewer than 3 cracks in 20 km); (b) P(2 cracks in one 10 km AND 1 crack in a different non-overlapping 10 km); (c) P(fewer than 200 cracks in 1,500 km); (d) P(more than 2 accidents in 100,000 cars).

Approach. Rescale the Poisson rate λ to match each interval by the additive property of independent Poisson counts over disjoint intervals; use the Normal approximation to the Poisson (with continuity correction) for the large-λ case in (c), and the Poisson approximation to the Binomial for the rare-event case in (d).

1. (a) Poisson count over a rescaled interval. A 20 km stretch has twice the length, so λ₂₀=2(1.5)=3.0 cracks. With $X\sim\text{Poisson}(3.0)$, $$P(X<3)=P(X\le 2)=\sum_{k=0}^{2}\dfrac{e^{-3}3^k}{k!}=e^{-3}(1+3+4.5)=\boxed{0.4232}.$$
2. (b) Independent non-overlapping intervals. Because the Poisson process has independent increments, the crack counts in two non-overlapping 10 km stretches are independent Poisson(1.5) variables; the joint probability is therefore the PRODUCT of the two marginal probabilities (never their sum, since "and" here means a joint, not a mutually-exclusive, event). With $Y_1,Y_2\stackrel{iid}\sim\text{Poisson}(1.5)$, $$P(Y_1=2)\,P(Y_2=1)=\left[\dfrac{e^{-1.5}1.5^2}{2!}\right]\left[\dfrac{e^{-1.5}1.5^1}{1!}\right]=(0.2510)(0.3347)=\boxed{0.0840}.$$
3. (c) Normal approximation for a large-λ Poisson count. 1,500 km is 150 stretches of 10 km, so $\lambda_{1500}=150(1.5)=225$ cracks, giving $\mu=\sigma^2=225$, $\sigma=15$. Since λ is large, $X\approx N(225,15^2)$ is an accurate approximation; applying the continuity correction for the discrete-to-continuous switch, "fewer than 200" becomes $X\le 199$, $$P(X<200)\approx\Phi\!\left(\dfrac{199.5-225}{15}\right)=\Phi(-1.70)=\boxed{0.0446}.$$
4. (d) Poisson approximation to the Binomial. With n=100,000 trials (cars) and a small per-trial probability p=0.00001, the Poisson approximation to the Binomial applies because n is large and p is small, so that $\lambda=np$ stays moderate (here $\lambda=100{,}000(0.00001)=1.0$) — this is the classical "rare events" regime where the Binomial's discreteness in n is well approximated by a Poisson count, unlike the large-λ Normal approximation used in part (c). With $X\sim\text{Poisson}(1.0)$, $$P(X>2)=1-P(X\le 2)=1-e^{-1}(1+1+0.5)=1-0.9197=\boxed{0.0803}.$$