04-BS-2, May 2017, 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. Market shares $P(M_1)=0.35,\ P(M_2)=0.20,\ P(M_3)=0.30,\ P(M_4)=0.15$; conditional satisfaction rates $P(B|M_1)=0.80,\ P(B|M_2)=0.90,\ P(B|M_3)=0.85,\ P(B|M_4)=0.75$.

Find. (a) tree diagram; (b) $P(B\cap M_2)$, $P(M_3\cap B^C)$, $P(B)$; (c) $P(M_3|B)$; (d) $P(\text{fewer than 3 satisfied of 6})$.

Approach. Multiply along each branch for a joint (intersection) probability, sum all branches ending in $B$ for the law of total probability, then apply Bayes' theorem for the reverse conditional; part (d) treats "satisfied" as a fixed-probability trial and applies the binomial.

(a) Tree diagram. The first stage branches on the provider $M_1,\dots,M_4$ with the given market-share probabilities; each provider branch then splits into $B$ (satisfied) and $B^C$ (not satisfied) with the given conditional rates (Fig. 4).
(b) Joint probabilities and P(B). $P(B\cap M_2)=P(M_2)P(B|M_2)=0.20(0.90)=0.180$. $P(M_3\cap B^C)=P(M_3)\bigl(1-P(B|M_3)\bigr)=0.30(0.15)=0.045$. By the law of total probability, $P(B)=\sum_i P(M_i)P(B|M_i)=0.35(0.80)+0.20(0.90)+0.30(0.85)+0.15(0.75)=0.280+0.180+0.255+0.1125=0.8275$. So $\boxed{P(B\cap M_2)=0.180,\ P(M_3\cap B^C)=0.045,\ P(B)=0.8275}$.
(c) Bayes' theorem. $P(M_3|B)=\dfrac{P(M_3\cap B)}{P(B)}=\dfrac{P(M_3)P(B|M_3)}{P(B)}=\dfrac{0.30(0.85)}{0.8275}=\dfrac{0.255}{0.8275}=0.3082$, so $\boxed{P(M_3|B)=0.3082}$.
(d) Binomial, fewer than three satisfied of six. Each of the 6 sampled customers is satisfied independently with probability $P(B)=0.8275$ (a fixed population-level rate), so the count $S\sim\text{Bin}(6,0.8275)$. $P(S<3)=\sum_{k=0}^{2}\binom{6}{k}(0.8275)^k(0.1725)^{6-k}=0.00988$, giving $\boxed{P(S<3)=0.00988}$ — a rare event since satisfaction is high overall.

Fig. 4 — tree diagram of provider market share (first stage) and satisfaction outcome (second stage).

Question 4 — final results
Part	Quantity	Value
(b)(i)	P(B ∩ M2)	0.1800
(b)(ii)	P(M3 ∩ B^c)	0.0450
(b)(iii)	P(B)	0.8275
(c)	P(M3 \| B)	0.3082
(d)	P(fewer than 3 of 6 satisfied)	0.00988

Topic: Law of total probability, Bayes' theorem, and the binomial applied to a Bernoulli outcome derived from a mixture.

Key relations: $P(B)=\sum_i P(M_i)P(B|M_i)$ (law of total probability); $P(M_i|B)=\dfrac{P(M_i)P(B|M_i)}{P(B)}$ (Bayes' theorem); once $P(B)$ is known, repeated independent customers satisfied with probability $P(B)$ form a $\text{Bin}(n,P(B))$ count.

Why this works: the four provider events partition the population ($\sum P(M_i)=1$), so any event's total probability decomposes additively across the partition; Bayes' theorem is just the definition of conditional probability applied in both directions using that same decomposition in the denominator.

Common pitfall: computing $P(M_3|B)$ as $P(B|M_3)$ itself (confusing the two conditional directions), or forgetting that part (d) needs the population-level $P(B)$ (not a provider-specific rate) since the sampled customers are not restricted to one provider.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 2 (Probability) and Ch. 5 (Discrete Distributions).

Question 4 of 8

Question 4 (20 marks)