This Exam P sample reference tests Bayes’ Theorem. A positive test comes from disease with weight 0.0095 and without disease with weight 0.00495. Normalization gives 0.65744, choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
DThe value 0.950 is the test sensitivity Pr(positive given disease), which reverses the conditioning direction.
EThe value 0.995 is the specificity, the probability of a negative result when disease is absent, not the posterior after a positive result.
Original practice · fully worked
Original variant: posterior defect probability after two positive screens
A manufacturing defect occurs in 4% of items. Each of two conditionally independent screens is positive for 90% of defective items and for 10% of nondefective items. Given that both screens are positive, find the probability that the item is defective.
A 0.04000
B 0.77143
C 0.27273
D 0.90000
E 0.96000
Variant answer in brief
The two-positive weights are 0.04(0.9²)=0.0324 and 0.96(0.1²)=0.0096. Their normalization gives 0.77143, choice B.
Setup
Setup
For each defect state, multiply its prior probability by the likelihood that both conditionally independent screens are positive.
wD=0.04(0.90)2=0.0324
wDc=0.96(0.10)2=0.0096
Model
Model
Normalize the defective two-positive weight over the defective and nondefective two-positive weights.
Pr(D∣++)=0.0324+0.00960.0324
Compute
Compute
The weights are 0.0324 and 0.0096, so the posterior defect probability is 0.0324/0.0420=27/35.
Pr(D∣++)=27/35=0.771429
Answer
Answer
Given two positive screens, the defect probability is 0.77143, which is choice B.
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