04-BS-2, May 2016, Q4 | Nivaar Exam Prep

Question 4 (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. First-stage (residency) probabilities $P(M_1)=0.40$, $P(M_2)=0.25$, $P(M_3)=0.15$, $P(M_4)=0.20$; second-stage (favour) conditionals $P(B|M_1)=0.75$, $P(B|M_2)=0.80$, $P(B|M_3)=0.90$, $P(B|M_4)=0.80$.

Find. (a) the tree; (b) $P(B\cap M_2)$, $P(M_3\cap B^c)$, $P(B)$; (c) $P(M_3|B)$; (d) $P(\text{fewer than 2 of 5 in favour})$.

Approach. Multiply along each branch of the two-stage tree for joint probabilities, sum all four "in favour" branches (law of total probability) for $P(B)$, apply Bayes' theorem for the reverse conditional in (c), and treat "favour" as a Bernoulli success probability across an independent sample of 5 citizens for (d).

(b)(i) Joint probability along the M2→B branch. $\Pr(B\cap M_2)=P(M_2)\,P(B|M_2)=0.25(0.80)=\boxed{0.20}$.
(b)(ii) Joint probability along the M3→Bc branch. $\Pr(M_3\cap B^c)=P(M_3)\big[1-P(B|M_3)\big]=0.15(0.10)=\boxed{0.015}$.
(b)(iii) Total probability of B. $\Pr(B)=\sum_i P(M_i)P(B|M_i)=0.40(0.75)+0.25(0.80)+0.15(0.90)+0.20(0.80)=0.300+0.200+0.135+0.160=\boxed{0.795}$.
(c) Bayes' theorem for P(M3 | B). $P(M_3|B)=\dfrac{P(M_3)P(B|M_3)}{P(B)}=\dfrac{0.15(0.90)}{0.795}=\dfrac{0.135}{0.795}=\boxed{0.1698}$.
(d) Binomial with success probability P(B)=0.795, n=5. $P(\text{fewer than 2})=P(Y\le1)=\sum_{k=0}^{1}\binom{5}{k}(0.795)^k(0.205)^{5-k}=0.000363+0.00702=\boxed{0.00738}$.

Fig. 4 — two-stage tree: residency (M1–M4) then favour/against (B, Bc).

Question 4 — final results
Part	Quantity	Value
(b)(i)	Pr(B ∩ M2)	0.2000
(b)(ii)	Pr(M3 ∩ Bc)	0.0150
(b)(iii)	Pr(B)	0.7950
(c)	P(M3 \| B)	0.1698
(d)	P(fewer than 2 of 5 in favour)	0.0074

Topic: Two-stage tree diagrams, total probability, and Bayes' theorem.

Key relations: total probability $P(B)=\sum_i P(M_i)P(B|M_i)$; Bayes' theorem $P(M_i|B)=\dfrac{P(M_i)P(B|M_i)}{P(B)}$; each tree branch's joint probability is the product of the probabilities along it.

Why this works: the municipalities partition the whole population (residency probabilities sum to 1), so summing the "favour" branch of every partition recovers the overall $P(B)$ exactly; Bayes' theorem then reverses the conditioning by comparing one partition's joint probability against that same total.

Common pitfall: confusing $P(M_3|B)$ with $P(B|M_3)$ (the given conditional), and forgetting that (d) treats "favour" as a fixed Bernoulli probability across a fresh sample of people (not the same tree again).

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 2 (Total Probability and Bayes' Theorem).

Question 4 of 8

Question 4 (20 marks)