Question
At a repair desk, an electronic device is either sound or faulty. Before testing, two out of every three devices are expected to be sound. The desk first runs a quick check, which shows green with probability \(3/4\) for a sound device and \(1/4\) for a faulty one.
A green result sends the device to scanner A; a red result sends it to scanner B. Scanner A reports clear with probability \(4/5\) if the device is sound and \(1/5\) if it is faulty. Scanner B reports clear with probability \(1/2\) if sound and \(1/10\) if faulty. Conditional on the device's actual condition and which scanner is used, the follow-up result is independent of the quick check.
A supervisor sees only the final label clear, not the earlier colour or which scanner was used. What is the probability that this device is sound? Explain why using the accuracy figures for scanner A alone is not sufficient.
Get hint
Hint 1
A final clear label can come from two distinct test routes. Account for both green→A and red→B.
Check your answer
Solution
The final label is a mixture of two tests selected by the earlier check, so we must first compute how often the complete procedure reports clear in each true condition.
For a sound device, the check is green with probability \(3/4\) and red with probability \(1/4\). The selected scanner's clear probabilities are therefore weighted as
\[
P(C\mid S)=(3/4)(4/5)+(1/4)(1/2)=3/5+1/8=\frac{29}{40}.
\]
For a faulty device, the quick check is green with probability \(1/4\) and red with probability \(3/4\), so
\[
P(C\mid F)=(1/4)(1/5)+(3/4)(1/10)=1/20+3/40=\frac18.
\]
With \(P(S)=2/3\) and \(P(F)=1/3\), the probability of a final clear label is
\[
P(C)=(2/3)(29/40)+(1/3)(1/8)=\frac{29}{60}+\frac1{24}=\frac{21}{40}.
\]
Bayes' rule now gives
\[
P(S\mid C)=\frac{P(S)P(C\mid S)}{P(C)}
=\frac{29/60}{21/40}=\boxed{\frac{58}{63}}.
\]
The two routes are mutually exclusive, and all four combinations of true condition and route are included. Scanner A's individual clear rate cannot replace \(P(C\mid S)\) or \(P(C\mid F)\), because devices sent to B also contribute to the merged final report.
Assumptions
Prior sound probability is 2/3, and the device’s true condition does not change during inspection.
The green/red check determines exactly one of the two scanners.
The scanner’s stated conditional clear rates apply on its assigned route, independent of the quick check given quality and route.
Only the merged clear label is observed by the supervisor.
Final Answer
\(P(\text{sound}\mid\text{final clear})=\frac{58}{63}\approx92.06\%\). The full routing procedure gives \(P(\text{clear}\mid\text{sound})=\frac{29}{40}\) and \(P(\text{clear}\mid\text{faulty})=\frac18\).