Question 4 of 8

Question 4

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. $P(A)=0.25,\ P(B)=0.30,\ P(C)=0.25,\ P(D)=0.20$; conditional claim rates $P(X\mid A)=0.04,\ P(X\mid B)=0.04,\ P(X\mid C)=0.02,\ P(X\mid D)=0.05$.

Find. (a) tree diagram; (b) $P(X),\ P(B\cap X),\ P(D\cap X^c)$; (c) $P(B\mid X)$; (d) for $n=10$ independent households, $P(\text{fewer than 2 claims})$.

Approach. Total probability across the 4 mutually-exclusive cities gives $P(X)$; intersection and complement probabilities follow directly from the multiplication rule; Bayes' theorem inverts the conditioning for (c); (d) treats the 10 households as Binomial trials with success probability $P(X)$ from part (b).

1. a) Tree diagram. Four first-stage branches (A, B, C, D) each split into a second-stage $X$/$X^c$ pair, shown below.
Probability tree: city selection (A/B/C/D) then claim event X / X⁶ (complement).
2. b) Pr(X), Pr(B∩X), Pr(D∩X^c). (i) By the law of total probability, $$P(X)=\sum_{\text{city}}P(\text{city})P(X\mid\text{city})=.25(.04)+.30(.04)+.25(.02)+.20(.05)=\boxed{0.0370}.$$ (ii) By the multiplication rule, $P(B\cap X)=P(B)P(X\mid B)=0.30(0.04)=\boxed{0.0120}$. (iii) $$P(D\cap X^c)=P(D)\,[1-P(X\mid D)]=0.20(0.95)=\boxed{0.1900}.$$
3. c) P(B|X) via Bayes' theorem. $$P(B\mid X)=\frac{P(B\cap X)}{P(X)}=\frac{0.0120}{0.0370}=\boxed{0.3243}.$$
4. d) Fewer than two of ten claims. Each of 10 independently-selected households makes a claim with probability $p=P(X)=0.0370$, so the claim count $K\sim\text{Bin}(10,0.0370)$: $$P(K<2)=P(K=0)+P(K=1)=(0.963)^{10}+10(0.0370)(0.963)^{9}=0.6866+0.2628=\boxed{0.9494}.$$