Verified solution · Exam P
This Exam P sample reference tests Conditional Probability. Split the callers into attendant, same-day return, and next-day return groups. Their joint application probabilities are 0.48, 0.18, and 0.04, so Bayes' theorem gives 0.48/0.70=24/35=0.6857, choice E.
Independent solution
Setup
Let A denote initial attendant contact, S a same-day return call, N a next-day return call, and M a mortgage application.
Model
Weight each caller group by its conditional application probability.
Compute
First find the total application probability, then normalize the attendant branch.
Answer
The conditional probability rounds to 0.69.
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Original practice · fully worked
A laboratory receives sealed samples from sources A, B, and C in proportions 0.40, 0.35, and 0.25. Conditional on the source, each of two screening checks passes independently. The pass probabilities per check are 0.80 for A, 0.50 for B, and 0.20 for C. A selected sample passes exactly one of the two checks. Calculate the probability that it came from source B.
For a source with per-check pass probability p, exactly one pass has probability 2p(1-p). The prior-weighted likelihoods are 0.128, 0.175, and 0.080, so the posterior for source B is 0.175/0.383=0.4569, choice D.
Setup
Let E be the event that exactly one screening check passes.
Model
Compute the likelihood of the observed split result for each possible source.
Compute
Multiply by the source proportions and normalize the weight for source B.
Answer
Among samples with one passing check, about 45.7% came from source B.
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Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
