Poisson and Hypergeometric Distributions: 04-BS-2, December 2015, 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. (A) buses needing maintenance $\sim$ Poisson, mean 4/day. (B) lot of 12 units, 5 substandard, sample of 8 drawn without replacement.

Find. (A)(a) P(more than 3 in a day); (A)(b) P(5,6,7 in two days); (B)(a) P(≤3 substandard of 8); (B)(b) full pmf of X and E(X).

Approach. A count of independent rare events over a fixed time/day is Poisson (scale $\lambda$ by the length of the period); a count drawn without replacement from a small finite lot with two categories is hypergeometric.

3(A)(a) Poisson, λ=4. $$P(X>3)=1-P(X\le3)=1-e^{-4}\left(1+4+\dfrac{4^2}{2}+\dfrac{4^3}{6}\right)=1-e^{-4}(21.333)=\boxed{0.5665}.$$
3(A)(b) two-day period, λ=8. Over two days the Poisson parameter scales to $\lambda=2\times4=8$. "More than four but fewer than eight" means $X=5,6,7$: $$P(5\le X\le7)=\sum_{k=5}^{7}\dfrac{e^{-8}8^k}{k!}=0.0916+0.1221+0.1396=\boxed{0.3533}.$$
3(B)(a) hypergeometric, N=12, K=5, n=8. $X$ (substandard among the 8 sold) has support $\max(0,n-(N-K))=1$ through $\min(n,K)=5$ (the 8 sold must include at least $8-7=1$ substandard unit, since only 7 good units exist). $$P(X\le3)=\sum_{k=1}^{3}\dfrac{\binom{5}{k}\binom{7}{8-k}}{\binom{12}{8}}=0.1414+0.3535+0.3535=\boxed{0.5758}.$$
3(B)(b) full pmf and E(X). $$P(X=k)=\dfrac{\binom{5}{k}\binom{7}{8-k}}{\binom{12}{8}},\qquad k=1,2,3,4,5.$$
pmf of X (substandard units among 8 sold)
k 1 2 3 4 5

P(X=k) 0.1414 0.3535 0.3535 0.1414 0.0101

By the hypergeometric mean formula, $$E(X)=\dfrac{nK}{N}=\dfrac{8\times5}{12}=\boxed{3.333}.$$

pmf of X (substandard units among 8 sold)
k	1	2	3	4	5
P(X=k)	0.1414	0.3535	0.3535	0.1414	0.0101

Final results — Question 3
Part	Result
3(A)(a)	0.5665
3(A)(b)	0.3533
3(B)(a)	0.5758
3(B)(b) E(X)	3.333

Topic: Poisson counting process scaling; hypergeometric sampling without replacement.

Key relations: Poisson over $t$ periods has $\lambda_t=t\lambda_1$; hypergeometric $E(X)=nK/N$, support runs from $\max(0,n-(N-K))$ to $\min(n,K)$.

Why this works: the Poisson process has independent increments, so combining two independent days simply adds their rates; the hypergeometric pmf directly counts favourable vs. total combinations for sampling without replacement from a finite lot.

Common pitfall: using the binomial instead of the hypergeometric for a small finite lot (no replacement matters here, since 8 of only 12 units are drawn); forgetting the hypergeometric's restricted support (X cannot be 0 here, since more items are wanted than "good" units available).

Source: Walpole et al., Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 5 (Poisson, Hypergeometric distributions).

Question 3 of 8: Poisson and Hypergeometric Distributions

Question 3: Poisson and Hypergeometric Distributions (20 marks)