Tree Diagram and Bayes' Theorem for Customer Satisfaction: 04-BS-2, December 2013, Q4

Question 4: Tree Diagram and Bayes' Theorem for Customer Satisfaction (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. Education groups partition the customer base: $P(U)=0.25,\ P(C)=0.30,\ P(H)=0.25,\ P(R)=0.20$. Conditional dissatisfaction rates: $P(X'|U)=0.03,\ P(X'|C)=0.04,\ P(X'|H)=0.02,\ P(X'|R)=0.03$.

Find. (a) tree diagram; (b) $Pr(X)$, $Pr(H\cap X)$, $Pr(U\cap X')$; (c) $P(C\mid X')$; (d) $P(\text{fewer than 2 of 10 dissatisfied})$.

Approach. Use the law of total probability across the four education branches for (b)(i); multiply along a branch for the joint probabilities; invert with Bayes' theorem for (c); treat the ten files as Binomial trials with the overall dissatisfaction rate from (b) for (d).

Fig. 4(a) — tree diagram: first-stage branches U/C/H/R with their marginal probabilities, second-stage branches X/X' with the conditional satisfaction/dissatisfaction rates given each group.

(b)(i) $Pr(X)$ via total probability. First find $Pr(X')=\sum P(\text{group})P(X'|\text{group})$: $$Pr(X')=(0.25)(0.03)+(0.30)(0.04)+(0.25)(0.02)+(0.20)(0.03)=0.0075+0.0120+0.0050+0.0060=0.0305$$ so $$Pr(X)=1-Pr(X')=1-0.0305=\boxed{0.9695}$$
(b)(ii) $Pr(H\cap X)$. Multiply along the H branch to its satisfied leaf: $Pr(H\cap X)=P(H)\big(1-P(X'|H)\big)=(0.25)(0.98)=\boxed{0.245}$
(b)(iii) $Pr(U\cap X')$. Multiply along the U branch to its dissatisfied leaf: $Pr(U\cap X')=P(U)P(X'|U)=(0.25)(0.03)=\boxed{0.0075}$
(c) Bayes' theorem for $P(C|X')$. Given that the customer is dissatisfied, the probability they came from group C reweights $Pr(C\cap X')$ against the total dissatisfaction probability found in (b)(i): $$P(C\mid X')=\dfrac{P(C)P(X'|C)}{Pr(X')}=\dfrac{(0.30)(0.04)}{0.0305}=\dfrac{0.0120}{0.0305}=\boxed{0.3934}$$
(d) Ten independent files, Binomial. Treat each file as an independent Bernoulli trial with dissatisfaction probability $\pi=Pr(X')=0.0305$ from (b)(i), so the count of dissatisfied files among $n=10$ is $W\sim Binomial(10,0.0305)$. $$P(W\lt2)=P(W=0)+P(W=1)=(1-\pi)^{10}+10\pi(1-\pi)^{9}=0.7325+0.2319=\boxed{0.9644}$$

Final results — Question 4
Part	Result
(b)(i) $Pr(X)$	0.9695
(b)(ii) $Pr(H\cap X)$	0.245
(b)(iii) $Pr(U\cap X')$	0.0075
(c) $P(C\mid X')$	0.3934
(d) $P(W\lt2)$	0.9644

Topic: Total probability, joint probability via tree diagrams, and Bayes' theorem.

Key relations: $P(A)=\sum_i P(B_i)P(A|B_i)$; $P(B_i\cap A)=P(B_i)P(A|B_i)$; $P(B_i|A)=\dfrac{P(B_i)P(A|B_i)}{P(A)}$.

Why this works: The four education groups partition the whole customer population, so any event's probability can be reconstructed by summing its probability within each partition weighted by that partition's own probability; Bayes' theorem is simply this same weighted sum used as the common denominator when "reversing" a conditional.

Common pitfall: Computing $P(C|X')$ using $P(X')$ from a different (marginal, unconditioned) source instead of the value derived from the SAME tree in part (b)(i) — the two must be numerically identical for the branch probabilities to be internally consistent.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 2 (probability, Bayes' theorem).

Question 4 of 8: Tree Diagram and Bayes' Theorem for Customer Satisfaction

Question 4: Tree Diagram and Bayes' Theorem for Customer Satisfaction (20 marks)