Verified solution · Exam P
This Exam P sample reference tests Combinatorial Probability. This is an at-least probability under sampling without replacement. Adding the hypergeometric terms for exactly three and exactly four target members gives 25/42=0.595238, which rounds to choice E.
Independent solution
Setup
Partition the nine members into six target members and three others. Every four-member subset is equally likely.
Model
At least three target members means either exactly three or exactly four, and these two events are disjoint.
Compute
Evaluate the two favorable counts and reduce the result.
Answer
The probability rounds to 0.60.
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Original practice · fully worked
Seven capsules are placed in a uniformly random order. Four are marked priority and three are regular. Calculate the probability that the second priority capsule appears in position four.
Exactly one of the first three positions must be priority, position four must be priority, and two of the final three positions must be priority. Counting priority-position sets gives 9/35, so choice B.
Setup
Describe an ordering by the four positions occupied by priority capsules. All four-position subsets are equally likely.
Model
For the second priority capsule to occupy position four, choose one priority position among the first three, fix position four, and choose two priority positions among the last three.
Compute
Divide favorable position sets by all possible position sets.
Answer
The required probability is 9/35.
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ActuaryProof. “SOA Exam P Sample Question #414 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
