Poisson Storms & Hypergeometric Cameras: 04-BS-2, May 2014, Q2

Question 2: Poisson Storms & Hypergeometric Cameras (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) Storms per year Y ∼ Poisson(λ=3). (B) A shipment of N=15 cameras (K=6 are 10 MP, 9 are 20 MP); n=5 drawn without replacement.

Find. (A)(a) P(Y<4) next year; (A)(b) P(4<Y<8) over 2 years; (B)(a) P(X>2); (B)(b) the full pmf of X.

Approach. Part (A) uses the Poisson pmf $P(Y=k)=e^{-\lambda}\lambda^k/k!$, re-scaling λ to 2×3=6 for the two-year window since the Poisson rate is additive over disjoint time intervals. Part (B) is sampling without replacement from a finite population of two types — the Hypergeometric distribution.

(A)(a) P(fewer than 4 storms next year). With $\lambda=3$, $$P(Y<4)=\sum_{k=0}^{3}\dfrac{e^{-3}3^k}{k!}=0.0498+0.1494+0.2240+0.2240=\boxed{0.6472}.$$
(A)(b) P(4<Y<8) over two years. Over a 2-year window the Poisson rate is $\lambda_2=2(3)=6$ (storms in disjoint years are independent Poisson counts, and the sum of independent Poissons is Poisson with summed rate). "More than four but fewer than eight" means $Y\in\{5,6,7\}$: $$P(5)+P(6)+P(7)=\dfrac{e^{-6}6^5}{5!}+\dfrac{e^{-6}6^6}{6!}+\dfrac{e^{-6}6^7}{7!}=0.1606+0.1606+0.1377=\boxed{0.4589}.$$
(B)(a) P(X > 2) for the Hypergeometric camera draw. With population N=15 (K=6 "successes" = 10 MP cameras) and a sample of n=5, $$P(X=k)=\dfrac{\binom{6}{k}\binom{9}{5-k}}{\binom{15}{5}},\qquad k=0,1,\dots,5.$$ $$P(X>2)=P(3)+P(4)+P(5)=0.2398+0.0450+0.0020=\boxed{0.2867}.$$
(B)(b) Full probability distribution of X. Evaluating the same Hypergeometric pmf at every k gives the table below (the five values sum to 1, confirming the enumeration).
pmf of X (number of 10 MP cameras among 5 drawn)
k 0 1 2 3 4 5

P(X=k) 0.0420 0.2517 0.4196 0.2398 0.0450 0.0020

pmf of X (number of 10 MP cameras among 5 drawn)
k	0	1	2	3	4	5
P(X=k)	0.0420	0.2517	0.4196	0.2398	0.0450	0.0020

Final results — Question 2
Part	Quantity	Result
(A)(a)	P(Y < 4), λ=3	0.6472
(A)(b)	P(4 < Y < 8), λ=6	0.4589
(B)(a)	P(X > 2)	0.2867
(B)(b)	pmf of X	0.0420, 0.2517, 0.4196, 0.2398, 0.0450, 0.0020

Topic: The Poisson distribution (rare independent events per unit time) and the Hypergeometric distribution (sampling without replacement from a finite dichotomous population).

Key relations: Poisson $P(Y=k)=e^{-\lambda}\lambda^k/k!$, additive over independent intervals; Hypergeometric $P(X=k)=\binom{K}{k}\binom{N-K}{n-k}/\binom{N}{n}$.

Why this works: the Poisson law models a count of rare independent occurrences in a fixed exposure, and rescaling the exposure (1 year → 2 years) simply rescales λ; the Hypergeometric law replaces the Binomial's fixed per-trial probability with an evolving without-replacement probability, which matters here because n=5 is a non-negligible fraction of N=15.

Common pitfall: using the Binomial with p=K/N=0.40 instead of the Hypergeometric — acceptable only when n/N is small (a rule of thumb is n/N<0.05); here n/N=1/3, so the Binomial approximation would materially understate the variance.

Source: Walpole, Myers, Myers & Ye, Ch. 5 (Discrete Probability Distributions).

Question 2 of 8: Poisson Storms & Hypergeometric Cameras

Question 2: Poisson Storms & Hypergeometric Cameras (20 marks)