04-BS-2, May 2017, Q3 | Nivaar Exam Prep

Question 3 (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) Lot of $N=14$ fans, $5$ substandard (so $9$ good); a sample of $n=6$ is drawn without replacement. (B) Service calls follow a Poisson process at rate $4$/hour.

Find. (A)(a) $P(\text{more than 3 good of 6})$; (b) the pmf, mean and variance of $X=$ number of substandard fans in the sample of 6. (B)(a) $P(\text{more than 3 calls in a 30-min window})$; (b) $P(2<\text{calls}<5\text{ in a 1-hr window})$.

Approach. Sampling without replacement from a finite lot of two types is hypergeometric (A); counts in disjoint time windows of a Poisson process scale the rate by the window length (B).

(A)(a) Hypergeometric, more than three good of six. Let $Y$ = number of good (non-substandard) fans in the sample, $Y\sim\text{Hypergeom}(N=14,\,K=9\text{ good},\,n=6)$. $P(Y>3)=1-P(Y\le3)=1-\sum_{y=0}^{3}\dfrac{\binom{9}{y}\binom{5}{6-y}}{\binom{14}{6}}$, giving $\boxed{P(Y>3)=0.6573}$.
(A)(b) pmf, mean, variance of X = substandard count. $X\sim\text{Hypergeom}(14,5,6)$: $P(X=x)=\dfrac{\binom{5}{x}\binom{9}{6-x}}{\binom{14}{6}}$ for $x=0,1,\dots,5$, with values $(0.0280,0.2098,0.4196,0.2797,0.0599,0.0030)$ summing to 1. Mean $E(X)=n\dfrac{K}{N}=6\left(\dfrac{5}{14}\right)=2.1429$. Variance $\text{Var}(X)=n\dfrac{K}{N}\left(1-\dfrac{K}{N}\right)\left(\dfrac{N-n}{N-1}\right)=6\left(\dfrac{5}{14}\right)\left(\dfrac{9}{14}\right)\left(\dfrac{8}{13}\right)=0.8477$. So $\boxed{E(X)=2.1429,\ \text{Var}(X)=0.8477}$.
(B)(a) Poisson, 30-minute window. Half an hour at 4/hr gives $\lambda=2.0$. $P(\text{calls}>3)=1-\sum_{k=0}^{3}\dfrac{e^{-2}2^k}{k!}=1-0.8571=0.1429$, so $\boxed{P(\text{calls}>3)=0.1429}$.
(B)(b) Poisson, one-hour window. One hour at 4/hr gives $\lambda=4.0$. $P(2<\text{calls}<5)=P(\text{calls}=3)+P(\text{calls}=4)=\dfrac{e^{-4}4^3}{3!}+\dfrac{e^{-4}4^4}{4!}=0.1954+0.1954=0.3907$, so $\boxed{P(2<\text{calls}<5)=0.3907}$.

Question 3 — final results
Part	Quantity	Value
(A)(a)	P(more than 3 of 6 good)	0.6573
(A)(b)	E(X); Var(X), X=substandard count	2.1429; 0.8477
(B)(a)	P(>3 calls, 30 min)	0.1429
(B)(b)	P(2<calls<5, 1 hr)	0.3907

Topic: Hypergeometric sampling from a finite lot; Poisson process over disjoint time windows.

Key relations: $\text{Hypergeom}(N,K,n)$: $P(X=x)=\binom{K}{x}\binom{N-K}{n-x}/\binom{N}{n}$, $E(X)=nK/N$, $\text{Var}(X)=n\frac{K}{N}\frac{N-K}{N}\frac{N-n}{N-1}$; Poisson over a window of length $t$ at rate $\lambda_0$ per unit time: $\lambda=\lambda_0 t$, $P(X=k)=e^{-\lambda}\lambda^k/k!$.

Why this works: sampling without replacement from a fixed 2-class lot is exactly the hypergeometric setting (no independence between draws); a Poisson process's counts in non-overlapping intervals are themselves Poisson with mean proportional to interval length, a consequence of the process's stationary, independent-increments definition.

Common pitfall: using the binomial (with replacement) formula for the fan-sampling question instead of the hypergeometric — the lot is small (14) relative to the sample (6), so the without-replacement correction matters; and forgetting to rescale $\lambda$ to the actual window length in the Poisson parts.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 5 (Discrete Distributions).

Question 3 of 8

Question 3 (20 marks)