This Exam P sample reference tests Bayes' Theorem. A red insured contributes qualifying claims at rate 0.10(0.90)=0.09 and a green insured at rate 0.05(0.80)=0.04. Weighting by 300 and 700 insureds gives 27 red and 28 green qualifying claims, so the red share is 27/55=0.491, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.300 is the red share of all insured cars. Red and green cars have different qualifying-claim rates, so the prior share is not the posterior share.
BThe value 0.462 results from conditioning on accidents alone: 0.3(0.10)/[0.3(0.10)+0.7(0.05)]. It omits the deductible-crossing probabilities.
DThe value 0.667 is 0.10/(0.10+0.05), which compares accident rates without weighting by group sizes or later claim qualification.
EThe value 0.692 is 0.09/(0.09+0.04), which compares per-car qualifying rates but fails to weight them by the 300-to-700 portfolio mix.
Original practice · fully worked
Original variant: production line given either alarm
Twenty percent of components come from a high-heat line and the rest from a low-heat line. Conditional on the line, pressure and vibration alarms are independent. Their trigger probabilities are 0.50 and 0.40 on the high-heat line, and 0.10 and 0.20 on the low-heat line. A component triggered at least one alarm. Determine the probability it came from the high-heat line.
A 0.2000
B 0.2800
C 0.3500
D 0.3846
E 0.7000
Variant answer in brief
At least one alarm occurs with probability 0.70 on the high-heat line and 0.28 on the low-heat line. Bayes' rule gives 0.20(0.70)/[0.20(0.70)+0.80(0.28)]=0.3846, choice D.
Setup
Setup
Let H denote the high-heat line and A denote at least one alarm.
Pr(H)=0.20,Pr(Hc)=0.80
Model
Model
Within each line, use conditional independence to complement the probability that neither alarm triggers.
Pr(A∣H)=1−(0.50)(0.60)=0.70
Pr(A∣Hc)=1−(0.90)(0.80)=0.28
Compute
Compute
Weight the alarm probabilities by the line shares and normalize.
Pr(H∣A)=0.20(0.70)+0.80(0.28)0.20(0.70)
Pr(H∣A)=0.3640.14=0.3846154
Answer
Answer
The posterior high-heat probability is about 0.3846.
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