Verified solution · Exam P
This Exam P sample reference tests Hypergeometric Distribution. This is a two-case hypergeometric acceptance probability. An accepted sample contains either two good items and one defective item or three good items; the two cases each have 56 favorable samples, giving 112/120=14/15≈0.933 and choice E.
Independent solution
Setup
The finite box contains eight good items and two defective items, and three are sampled without replacement.
Model
Acceptance occurs for exactly two good items or exactly three good items.
Compute
Evaluate the two disjoint composition counts.
Answer
The acceptance probability rounds to 0.93.
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Original practice · fully worked
A project group contains two supervisors and seven analysts. Three people are selected uniformly without replacement for a review panel. Given that the panel contains at least one supervisor, calculate the probability that it contains both supervisors.
Seven of the 84 possible panels contain both supervisors. Forty-nine panels contain at least one supervisor. Conditioning gives 7/49, or 1/7, approximately 0.1429. This is choice B.
Setup
Count all three-person panels and panels with no supervisor.
Model
Use the complement to count panels represented by at least one supervisor.
Compute
Count panels with both supervisors and divide by the conditioning count.
Answer
Given supervisor representation, both supervisors serve with probability approximately 0.1429.
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ActuaryProof. “SOA Exam P Sample Question #335 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
