Difficulty: Medium · Puzzle 6 of 6 · Topic: Probability Puzzles
A silicon issue occurs in 1% of chips. A screening test has 99% sensitivity and 95% specificity. If a chip tests positive, what is P(issue is real)?
Use Bayes with prevalence, not just test accuracy.
1) Let D be real defect, + be positive test.
2) Given: P(D)=0.01, P(+|D)=0.99, P(+|~D)=0.05 (from 95% specificity).
3) Bayes: P(D|+) = P(+|D)P(D) / [P(+|D)P(D) + P(+|~D)P(~D)].
4) Numerator = 0.99*0.01 = 0.0099.
5) Denominator = 0.0099 + 0.05*0.99 = 0.0099 + 0.0495 = 0.0594.
6) P(D|+) = 0.0099/0.0594 = 1/6 approx 0.1667.Answer: Only about 16.67% of positive results are true defects in this setup.
Checked by running a model of the rule in a Python model. A computational check: exact Bayes arithmetic with fractions gives 1/6 at 1% prevalence and 11/16 at 10% prevalence.
Reading the 99% sensitivity as P(defect | positive). With 1% prevalence the 5% false positives from the 99% good chips (4.95 per 100) swamp the 0.99 true positives per 100 chips.
If prevalence rises to 10%, what does P(D|+) become? Model answer: 0.099 / (0.099 + 0.05*0.9) = 0.099/0.144 = 11/16 = 0.6875, about 69%. The same test, a much more trustworthy positive.
I use Bayes with the defect rate at one percent, sensitivity 99 percent, and a false positive rate of 5 percent from the 95 percent specificity. Out of 10,000 chips, 100 are defective and 99 of those test positive, while about 495 of the good chips also test positive. So the true positives are 99 out of 594, which is one sixth, about 16.7 percent. The trap is reading the 99 percent sensitivity as the chance that a positive is a real defect.