Verified solution · Exam P
This Exam P sample reference tests Conditional Probability. Inclusion-exclusion gives overlap 0.30+0.42-0.60=0.12. Exactly one type has probability 0.48 and the conditioning event has probability 0.88, so the ratio is 6/11=0.545455 and choice D.
Independent solution
Setup
Recover the overlap from the two marginal probabilities and their union.
Model
Compute the numerator by removing the overlap from both marginal events, and compute the conditioning probability by complementing the overlap.
Compute
Divide the target region by the conditioning region.
Answer
The conditional probability rounds to 0.55.
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Original practice · fully worked
A city survey records whether each rider uses rail, bus, and ferry service. The respective use probabilities are 0.55, 0.45, and 0.30. The rail-bus, rail-ferry, and bus-ferry intersection probabilities are 0.25, 0.15, and 0.10, and 0.05 use all three. A rider is known to use exactly two of the modes. Calculate the conditional probability that rail is one of them.
The exclusive rail-bus and rail-ferry cells have probabilities 0.20 and 0.10. All exclusive two-mode cells total 0.20+0.10+0.05=0.35, so the conditional rail share is 0.30/0.35=6/7=0.8571, choice E.
Setup
Remove the all-three cell from each inclusive pairwise intersection.
Model
Exactly two modes is the disjoint union of the three pair-only cells.
Compute
Rail appears in the rail-bus-only and rail-ferry-only cells.
Answer
Among exactly-two-mode riders, the rail share rounds to 0.8571.
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ActuaryProof. “SOA Exam P Sample Question #312 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
