04-BS-2, May 2015, 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 $N=10$ treadmills, $K=5$ need adjustment; a sample of $n=7$ is drawn without replacement. (B) Subway delays follow a Poisson process with rate $\lambda=0.6$ delays/day.

Find. (A)(i) $P(X\ge 3)$; (ii) $E[X]$, $\mathrm{Var}(X)$ for the number needing adjustment in the sample of 7. (B)(a) $P(X<3)$ in one day; (b) $P(2<X<6)$ over 5 days.

Approach. Sampling 7 of 10 known items without replacement is Hypergeometric, not Binomial (the trials are not independent); the delay counts are Poisson, with the rate scaled by the time window ($\lambda t$) for the 5-day part.

(A)(i) Hypergeometric upper tail. $X\sim\text{Hypergeometric}(N=10,\,K=5,\,n=7)$. $$P(X\ge3)=1-P(X=0)-P(X=1)-P(X=2)=1-\sum_{x=0}^{2}\frac{\binom{5}{x}\binom{5}{7-x}}{\binom{10}{7}}=\boxed{0.9167}$$ (Note: since only 5 of the 10 treadmills are *good*, at least $7-5=2$ of any 7 selected must need adjustment — the distribution is concentrated at high $x$, so a high probability is expected.)
(A)(ii) Hypergeometric mean and variance. $$E[X]=n\frac{K}{N}=7\cdot\frac{5}{10}=\boxed{3.5}$$ $$\mathrm{Var}(X)=n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}=7(0.5)(0.5)\frac{3}{9}=\boxed{0.5833}$$
(B)(a) Poisson, one day ($\lambda=0.6$). $$P(X<3)=P(0)+P(1)+P(2)=e^{-0.6}\left(1+0.6+\frac{0.6^2}{2}\right)=\boxed{0.9769}$$
(B)(b) Poisson, five-day window. Over 5 days the rate scales to $\lambda_5=5(0.6)=3.0$ delays. "More than two but fewer than six" is $X=3,4,5$. $$P(3\le X\le5)=\sum_{x=3}^{5}\frac{e^{-3}3^x}{x!}=\boxed{0.4929}$$

Question 3 — final results
Part	Value
(A)(i) $P(X\ge3)$	0.9167
(A)(ii) mean, variance	3.5, 0.5833
(B)(a) $P(X<3)$, 1 day	0.9769
(B)(b) $P(3\le X\le5)$, 5 days	0.4929

Topic: Hypergeometric sampling and the Poisson process.

Key relations: Hypergeometric: $P(X=x)=\binom{K}{x}\binom{N-K}{n-x}/\binom{N}{n}$, $E[X]=nK/N$, $\mathrm{Var}(X)=n\frac{K}{N}\frac{N-K}{N}\frac{N-n}{N-1}$. Poisson over a window of length $t$: $X\sim\text{Poisson}(\lambda t)$, $P(X=x)=e^{-\lambda t}(\lambda t)^x/x!$.

Why this works: Sampling without replacement from a finite, already-fixed lot changes the success probability trial-to-trial, which is exactly what the Hypergeometric model captures (unlike the Binomial, which assumes independent trials with constant $p$); the Poisson process's defining property is that counts over disjoint/scaled time windows simply rescale the rate parameter.

Common pitfall: Treating the treadmill sample as Binomial($n=7,p=0.5$) instead of Hypergeometric — the population is small (10) and fixed, so sampling without replacement matters; also forgetting to rescale $\lambda$ by the window length in part (B)(b).

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 5 (Hypergeometric & Poisson Distributions).

Question 3 of 8

Question 3 (20 marks)