Question 6 of 10: Decision Analysis — Machine-Screening Policy Under an Unstated Prior

Question 6: Decision Analysis — Machine-Screening Policy Under an Unstated Prior (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. Sale price $10,000 if the machine works, $0 if returned; scan cost $500 (once) or $900 (twice), sensitivity 90% (P(flags fault | faulty)), false-positive rate 40% (P(flags fault | not faulty)); overhaul cost $2,000, plus $1,000 to fix a part found not flat during overhaul (then 100% works). No prior probability that the part is not sufficiently flat is stated anywhere on the source page.

Find. The expected-profit-maximizing course of action — expressed as a policy over the (unstated) prior probability $p$ that the part is not sufficiently flat, with the exact break-even values of $p$ at which the optimal action switches.

Approach. Compute the EMV of each of the four candidate strategies (ship as-is, overhaul unconditionally, scan once then decide, scan twice then decide) as a function of $p$, using Bayes' rule to update $p$ on a scan result, then compare the four EMV functions on $p\in[0,1]$.

1. Ship as-is / overhaul unconditionally. With $p=P(\text{not flat})$: $\text{EMV}_A(p)=10000(1-p)$ (ship as-is); $\text{EMV}_B(p)=10000-2000-1000p=8000-1000p$ (overhaul, fixing the fraction $p$ that needs it, then guaranteed sale). These cross at $$10000(1-p)=8000-1000p\ \Rightarrow\ \boxed{p=2/9\approx0.222.}$$
2. Scan once, then decide. $P(\text{pos})=0.4+0.5p$, $P(\text{neg})=0.6-0.5p$; posteriors by Bayes' rule: $P(\text{not flat}\mid\text{pos})=\dfrac{0.9p}{0.4+0.5p}$, $P(\text{not flat}\mid\text{neg})=\dfrac{0.1p}{0.6-0.5p}$. After either result, the better follow-up is $\max\{10000(1-\text{post}),\,8000-1000\,\text{post}\}$ — ship if the posterior is below $2/9$, else overhaul-and-fix (same $2/9$ threshold as Step 1). Solving where the posterior crosses $2/9$: the "pos" branch crosses at $p=8/71$ and the "neg" branch at $p=12/19$ — over the interval $8/71
3. Compare all three (scan twice is checked and dominated — Step 5). Solve $\text{EMV}_A(p)=\text{EMV}_1(p)$ and $\text{EMV}_B(p)=\text{EMV}_1(p)$: $$10000(1-p)=8700-2900p\ \Rightarrow\ \boxed{p=13/71\approx0.183}\qquad\qquad 8000-1000p=8700-2900p\ \Rightarrow\ \boxed{p=7/19\approx0.368.}$$ Both lie inside the $8/71
4. Resulting policy. Comparing $\text{EMV}_A,\text{EMV}_B,\text{EMV}_1$ on $[0,1]$: $$\boxed{0\le p<\tfrac{13}{71}:\ \text{ship as-is}\qquad \tfrac{13}{71}\le p<\tfrac{7}{19}:\ \text{scan once, then ship/overhaul on the result}\qquad p\ge\tfrac{7}{19}:\ \text{overhaul unconditionally}}$$
5. Scanning twice is never optimal. Repeating the Step 2–3 construction with two (conditionally independent) scans and the same $500 vs.\ 900$ cost gives $\text{EMV}_2(p)\le\max(\text{EMV}_A,\text{EMV}_B,\text{EMV}_1)$ for every $p\in[0,1]$: the extra $400 for a second scan never buys enough additional certainty to beat scanning once, because the 40% false-positive rate on each individual scan limits how much a second reading can sharpen the posterior.