23-Ind-B1 Reliability and Maintainability · December 2014

Question 3 of 9: Binomial and Poisson Probabilities

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2014 — 98-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: four sections — Section A: do 2 of 3 questions (20 marks); Section B: do 1 of 2 (25 marks); Section C: do 1 of 2 (25 marks); Section D: do 1 of 2 (30 marks) — a 5-question, 100-mark paper as printed. All nine questions across the four sections are solved below for completeness. Page 1's own NOTES list is mis-numbered (two items both labelled "4."), transcribed as printed.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — point/interval estimation (ch. 8–9), hypothesis testing (ch. 9–10), simple/multiple linear regression (ch. 11–12), single- and two-factor ANOVA (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).

Question 3 (Section A.3): Binomial and Poisson Probabilities (10 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)–(b) each birth is modelled as an independent Bernoulli trial with $P(\text{girl})=P(\text{boy})=0.5$; (c) family size $X\sim\text{Poisson}(\lambda=1.5)$.

Find. (a) $P(3\text{ girls, }1\text{ boy in }4\text{ children})$; (b) $P(\text{5th child is a girl})$; (c) $P(X\ge 4)$.

Approach. (a) is a Binomial$(n=4,p=0.5)$ count of girls; (b) uses the independence of successive births; (c) uses the Poisson complement rule $P(X\ge 4)=1-P(X\le 3)$.

Part (a) — binomial probability. Let $G\sim\text{Binomial}(4,0.5)$ be the number of girls among 4 children. There are $\binom{4}{3}=4$ orderings of exactly 3 girls and 1 boy, so $$P(G=3) = \binom{4}{3}(0.5)^3(0.5)^1 = 4\times 0.5^4 = \boxed{0.2500}.$$
Part (b) — next child. Each birth is an independent trial — the sexes of the first four children carry no information about the fifth (this is not a "gambler's fallacy" situation; births are not drawn without replacement from a fixed pool). Therefore $$P(\text{5th child is female}) = \boxed{0.50}.$$
Part (c) — Poisson tail probability. With $\lambda=1.5$, $$P(X\ge 4) = 1-\sum_{k=0}^{3}\frac{e^{-1.5}1.5^k}{k!} = 1-\big(0.2231+0.3347+0.2510+0.1255\big) = 1-0.9344 = \boxed{0.0656}.$$

Part	Result
(a) $P(3$ girls, 1 boy$)$	0.2500
(b) $P(\text{next child female})$	0.50
(c) $P(X\ge 4)$, $\lambda=1.5$	0.0656

Concept & Theory

Topic. Binomial and Poisson probability models for counts.

Key relations. Binomial pmf $P(X{=}k)=\binom{n}{k}p^k(1-p)^{n-k}$; Poisson pmf $P(X{=}k)=e^{-\lambda}\lambda^k/k!$; independent-trials rule $P(\text{next trial}\mid\text{past trials})=p$.

Why this works. Births in a family are well modelled as i.i.d. Bernoulli trials (absent specific genetic factors), so part (a) is a direct Binomial pmf evaluation and part (b) is simply the marginal probability restated — the four prior outcomes do not update it. Family-size counts across a population are a classic Poisson application: a large number of families, each with a small individual probability of any specific count, summing to a stable population rate $\lambda$.

Common pitfall. Answering part (b) as anything other than 0.5 by (incorrectly) conditioning on the observed 3-girls-1-boy history — a textbook instance of the gambler's fallacy; and forgetting the $k=0$ term when computing the Poisson complement in part (c).

Source. Montgomery & Runger, Applied Statistics and Probability for Engineers, ch. 3 (discrete random variables — Binomial and Poisson distributions).

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