This Exam P sample reference tests Conditional Probability. The joint probability is 0.70×0.15=0.105; complementing the resulting union gives 0.205, which rounds to choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.13 results if the overlap is incorrectly computed as 0.15(0.20)=0.03. The conditional probability must multiply Pr(M)=0.70, not Pr(S).
EThe value 0.30 is Pr(M complement). It includes outcomes where S occurs and therefore is larger than the probability that neither event occurs.
Original practice · fully worked
Original variant: probability of exactly one screening flag
A screening system marks 55% of transactions for manual review and 30% for a sanctions check. Among manually reviewed transactions, 20% also receive a sanctions check. Compute the share bearing one mark but not both.
A 0.63
B 0.55
C 0.44
D 0.30
E 0.19
Variant answer in brief
The overlap is 0.11. The two only-one cells are 0.44 and 0.19, totaling 0.63, choice A.
Setup
Setup
First compute the overlap by multiplying the manual-review marginal by the conditional sanctions-check rate.
Pr(M∩S)=0.55(0.20)=0.11
Model
Model
Subtract the overlap from each marginal to obtain the two disjoint cells with exactly one mark.
Pr(M∩Sc)=0.55−0.11=0.44
Pr(Mc∩S)=0.30−0.11=0.19
Compute
Compute
The manual-only and sanctions-only probabilities are 0.44 and 0.19; adding these disjoint cells gives 0.63.
Pr(exactly one)=0.44+0.19=0.63
Answer
Answer
Thus 63% of transactions bear exactly one mark, corresponding to choice A.
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