Poisson Traffic Jams and Hypergeometric Snow-blowers: 04-BS-2, Undated paper, Q3

Question 3: Poisson Traffic Jams and Hypergeometric Snow-blowers (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. (A) traffic jams $\sim$ Poisson, rate $\lambda=2.4$/week. (B) lot of $N=16$ snow-blowers, $K=6$ substandard, $N-K=10$ standard.

Find. (A)(a) $P(X\lt 4)$ in one week. (b) $P(2\lt X\lt 7)$ in two weeks. (B)(a) $P(X\le 2)$ substandard in a sample of 8 sold. (b) distribution, mean, variance of X = number standard in a sample of 3.

Approach. Poisson pmf/cdf for (A), rescaling λ by the exposure period; Hypergeometric pmf for (B), since the lot is finite and sampling is without replacement.

(A)(a) Fewer than four jams in one week. $$P(X\lt 4)=P(X\le 3)=\sum_{k=0}^{3}\dfrac{e^{-2.4}(2.4)^k}{k!}=0.7787$$ $$\boxed{P(X\lt 4)=0.7787}$$
(A)(b) More than two but fewer than seven jams in two weeks. Over a 2-week window, $\lambda'=2(2.4)=4.8$. "More than two but fewer than seven" means $X\in\{3,4,5,6\}$: $$P(3\le X\le 6)=\sum_{k=3}^{6}\dfrac{e^{-4.8}(4.8)^k}{k!}=0.6483$$ $$\boxed{P(2\lt X\lt 7)=0.6483}$$
(B)(a) At most two of eight sold are substandard. With $N=16$, $K=6$ substandard, sample $n=8$: $$P(X\le 2)=\sum_{k=0}^{2}\dfrac{\binom{6}{k}\binom{10}{8-k}}{\binom{16}{8}}=0.3042$$ $$\boxed{P(X\le 2)=0.3042}$$
(B)(b) Distribution of X = number standard in a sample of three. Here the "successes" are the 10 standard units, $n=3$: $$P(X=x)=\dfrac{\binom{10}{x}\binom{6}{3-x}}{\binom{16}{3}},\qquad x=0,1,2,3$$ giving $P(0)=0.0357$, $P(1)=0.2679$, $P(2)=0.4821$, $P(3)=0.2143$ (sums to 1.0000). The mean and variance follow the hypergeometric formulas: $$\mu_X=n\dfrac{K}{N}=3\left(\dfrac{10}{16}\right)=1.875$$ $$\sigma_X^2=n\dfrac{K}{N}\left(1-\dfrac{K}{N}\right)\dfrac{N-n}{N-1}=3(0.625)(0.375)\left(\dfrac{13}{15}\right)=0.6094$$ $$\boxed{\mu_X=1.875,\ \ \sigma_X^2=0.6094}$$

Question 3 — final results
Part	Quantity	Result
(A)(a)	P(X<4), 1 week	0.7787
(A)(b)	P(2<X<7), 2 weeks	0.6483
(B)(a)	P(X≤2 substandard), n=8	0.3042
(B)(b)	P(0),P(1),P(2),P(3); μ, σ²	0.0357, 0.2679, 0.4821, 0.2143; 1.875, 0.6094

Topic: Poisson process rescaling and the Hypergeometric distribution.

Key relations: Poisson over $t$ periods: $\lambda_t=t\lambda$. Hypergeometric: $P(X=x)=\dfrac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}$, $\mu=n\dfrac{K}{N}$, $\sigma^2=n\dfrac{K}{N}\left(1-\dfrac{K}{N}\right)\dfrac{N-n}{N-1}$.

Why this works: The Poisson count over a longer window is itself Poisson with a proportionally larger rate (independent increments). Sampling without replacement from a finite lot of two types is exactly Hypergeometric, not Binomial — the finite-population correction factor $\frac{N-n}{N-1}$ shrinks the variance versus an equivalent Binomial.

Common pitfall: Using Binomial(n,K/N) instead of Hypergeometric when the lot is small relative to the sample (here $n/N=8/16=0.5$, a large fraction, so the correction matters); forgetting to switch the "success" category between (a) substandard and (b) standard.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 5.

Question 3 of 8: Poisson Traffic Jams and Hypergeometric Snow-blowers

Question 3: Poisson Traffic Jams and Hypergeometric Snow-blowers (20 marks)