Poisson and Hypergeometric Distributions: 04-BS-2, May 2013, Q3

Question 3: Poisson and Hypergeometric Distributions (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. Part A: traffic jams ∼ Poisson, rate 3.2/week. Part B: lot of N=18 snow-blowers, 8 substandard, 10 standard.

Find. (A)(a) P(X<3) in one week; (A)(b) P(2<X<6) in two weeks; (B)(a) P(at most 2 substandard) among 8 sold; (B)(b) the probability distribution and E(X) of the number of standard units in a sample of 3.

Approach. Poisson counts scale linearly with the observation window (λ for 2 weeks = 2×λ for 1 week). The snow-blower lot is finite and sampled without replacement, so use the Hypergeometric distribution, not the Binomial.

3A(a) One-week Poisson, λ=3.2. $$P(X<3)=P(X\le2)=e^{-3.2}\left(1+3.2+\dfrac{3.2^2}{2!}\right)=\boxed{0.3799}.$$
3A(b) Two-week Poisson, λ=2(3.2)=6.4. $$P(2<X<6)=P(X=3)+P(X=4)+P(X=5)=e^{-6.4}\left(\dfrac{6.4^3}{3!}+\dfrac{6.4^4}{4!}+\dfrac{6.4^5}{5!}\right)=\boxed{0.3374}.$$
3B(a) Hypergeometric, N=18, defective K=8, n=8 sold. Let X = number substandard among the 8 sold. $$P(X\le2)=\sum_{k=0}^{2}\dfrac{\binom{8}{k}\binom{10}{8-k}}{\binom{18}{8}}=\boxed{0.1573}.$$
3B(b) Hypergeometric, N=18, standard K=10, n=3. Let X = number standard among a sample of 3. $$P(X=x)=\dfrac{\binom{10}{x}\binom{8}{3-x}}{\binom{18}{3}},\qquad x=0,1,2,3.$$ Evaluating: $P(0)=0.0686,\ P(1)=0.3431,\ P(2)=0.4412,\ P(3)=0.1471$ (these sum to 1.0000, confirming the pmf). The mean follows the shortcut $E(X)=n\dfrac{K}{N}$: $$E(X)=3\left(\dfrac{10}{18}\right)=\boxed{1.667}.$$

Question 3 — final results
Part	Quantity	Value
A(a)	P(X<3), λ=3.2	0.3799
A(b)	P(2<X<6), λ=6.4	0.3374
B(a)	P(at most 2 substandard)	0.1573
B(b)	P(X=0,1,2,3)	0.0686, 0.3431, 0.4412, 0.1471
B(b)	E(X)	1.667

Topic: Poisson process scaling and the Hypergeometric distribution for sampling without replacement.

Key relations: Poisson pmf $P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}$ with $\lambda$ proportional to the observation interval; Hypergeometric pmf $P(X=x)=\binom{K}{x}\binom{N-K}{n-x}/\binom{N}{n}$, mean $nK/N$, variance $n\dfrac{K}{N}\dfrac{N-K}{N}\dfrac{N-n}{N-1}$.

Why this works: a Poisson process has independent increments, so the count over a 2-week window is the sum of two independent Poisson(3.2) counts, itself Poisson(6.4); the snow-blower lot is small and finite with no replacement, so each draw changes the composition of what remains — exactly the Hypergeometric setting, unlike the Binomial which assumes independent trials with constant p.

Common pitfall: applying the Binomial to a small finite lot sampled without replacement (ignoring the changing composition), or forgetting to double λ when the observation window doubles.

Source: Walpole, Myers, Myers & Ye, Ch. 5 (Poisson Distribution and Poisson Process, Hypergeometric Distribution).

Question 3 of 8: Poisson and Hypergeometric Distributions

Question 3: Poisson and Hypergeometric Distributions (20 marks)