This Exam P sample reference tests Conditional Probability. Costs exceed the deductible only for four, five, or six visits, whose total probability is 0.07. The exactly-five mass is 0.02, so the conditional probability is 0.02/0.07=0.2857, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.050 is 0.02/0.40, the probability of five visits conditional on at least one visit. The actual condition is cost above the deductible.
BThe value 0.133 is 0.02/0.15, obtained by admitting three visits. Three visits cost 300 and do not exceed the 350 deductible.
DThe value 0.333 is 0.02/(0.04+0.02), which omits the six-visit outcomes from the conditioning event.
EThe value 0.429 is (0.02+0.01)/0.07, the conditional probability of at least five visits rather than exactly five.
Original practice · fully worked
Original variant: conditional mean of a tiered review score
A quality count N has probabilities 0.50, 0.25, 0.15, 0.07, and 0.03 at N=0,1,2,3,4, respectively. A review score is 0 for N at most 1, 10 for N=2, 30 for N=3, and 60 for N=4. Given that the score is at least 30, calculate its conditional mean.
A 5.4
B 30.0
C 39.0
D 45.0
E 60.0
Variant answer in brief
A score of at least 30 restricts the count to three or four, with total probability 0.10. Weighting scores 30 and 60 by masses 0.07 and 0.03 gives conditional mean 39, choice C.
Setup
Setup
Identify the two count values admitted by the score condition.
{S≥30}={N=3}∪{N=4}
Pr(S≥30)=0.07+0.03=0.10
Model
Model
Form the conditional score expectation from the two retained masses.
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