Question 5 of 8: Decision Tree for a Medical/Travel Decision

Question 5: Decision Tree for a Medical/Travel Decision (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. Trip value $10,000 (fully enjoyed only if well); $P(\text{problem})=0.20$, $P(\text{fine})=0.80$; Malo Clinic screening $2,000 (perfectly accurate per the source: "sure they can screen her"), plus $1,000 to treat/cure if needed; medical exam $500 (90% true-positive rate, 40% false-positive rate); second opinion $400 more, same 0.9/0.4 accuracy, treated as an independent repeat test.

Find. The optimal course of action (do nothing / go to Malo now / take the exam then decide / take the exam plus second opinion then decide) and its expected monetary value (EMV).

Approach. Compute the EMV of each of the four top-level strategies by rolling the tree back from its right-hand payoffs (trip value minus costs incurred, minus nothing if the trip cannot be enjoyed), using Bayes' rule to update $P(\text{problem})$ after each test result before deciding on Malo Clinic.

1. Alternative 1 — do nothing. No cost is spent and Delma simply finds out in 6 months: $$EMV_1 = 0.80(10{,}000)+0.20(0) = \boxed{8{,}000}.$$
2. Alternative 2 — go to Malo Clinic directly. The $2,000 screening is always paid; the extra $1,000 treatment is paid only if she truly has the problem (Malo's screening is stated as certain, so it always catches and cures a real problem in time): $$EMV_2 = 0.80(10{,}000-2{,}000)+0.20(10{,}000-2{,}000-1{,}000) = \boxed{7{,}800}.$$
3. Alternative 3 — take the exam ($500), then decide. By the total-probability rule, $P(+)=0.20(0.9)+0.80(0.4)=0.50$ and $P(-)=0.50$. Bayes' rule gives the posteriors $$P(\text{problem}\mid +)=\dfrac{0.20(0.9)}{0.50}=0.36,\qquad P(\text{problem}\mid -)=\dfrac{0.20(0.1)}{0.50}=0.04.$$ At each result, compare "go to Malo" against "do nothing," both now net of the $500 already spent: after a positive result, $EMV(\text{Malo})=0.36(10{,}000{-}500{-}3{,}000)+0.64(10{,}000{-}500{-}2{,}000)=7{,}140$ beats $EMV(\text{nothing})=0.36(-500)+0.64(9{,}500)=5{,}900$, so Malo is chosen; after a negative result, $EMV(\text{Malo})=0.04(6{,}500)+0.96(7{,}500)=7{,}460$ is beaten by $EMV(\text{nothing})=0.04(-500)+0.96(9{,}500)=9{,}100$, so nothing is chosen. Rolling back to the exam decision: $$EMV_3 = 0.50(7{,}140)+0.50(9{,}100) = \boxed{8{,}120}.$$
4. Alternative 4 — exam + second opinion ($900 total), then decide. Treating the two exams as independent 0.9/0.4 tests, the four result combinations occur with probability 0.29 (++), 0.21 (+−), 0.21 (−+) and 0.29 (−−), giving posteriors $P(\text{problem})=$ 0.559, 0.086, 0.086 and 0.007 respectively. Comparing $EMV(\text{Malo}\mid\pi)=7{,}100-1{,}000\pi$ against $EMV(\text{nothing}\mid\pi)=9{,}100-10{,}000\pi$ (both net of the $900 sunk cost) shows the crossover posterior is $\pi^*=2/9\approx0.222$ — only the (++) branch ($\pi=0.559$) clears it, so Malo is chosen there (EMV 6,541) and "do nothing" is chosen on the other three branches (EMV 8,243, 8,243, 9,031). Weighting by branch probability: $$EMV_4 = 0.29(6{,}541)+0.21(8{,}243)+0.21(8{,}243)+0.29(9{,}031) = \boxed{7{,}978}.$$
5. Compare and decide. $EMV_3=8{,}120 > EMV_1=8{,}000 > EMV_4=7{,}978 > EMV_2=7{,}800$: the exam is worth its $500 cost (it beats doing nothing), but the second opinion is not worth its extra $400 (it makes things worse, not better, because the first exam's evidence is already strong enough that a second, equally noisy test rarely changes the decision).