Verified solution · Exam P
This Exam P sample reference tests Hypergeometric Distribution. This is a hypergeometric probability for selecting exactly two members of a six-person target group in a four-person sample. The count is C(6,2)C(4,2)/C(10,4)=3/7=0.4286, so choice E.
Independent solution
Setup
Partition the population into six target members and four other members. Every four-person subset is equally likely.
Model
A favorable subset chooses two members from each of the two groups.
Compute
Divide the favorable subset count by the total subset count.
Answer
The probability rounds to 0.429.
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Original practice · fully worked
A locker contains 5 red, 4 blue, and 3 green tokens. Five tokens are drawn uniformly without replacement. Calculate the probability that the sample contains at least one token of every color.
There are C(12,5)=792 samples. Inclusion-exclusion removes 21, 56, and 126 samples missing each color and restores one all-red sample counted twice, leaving 590; thus 590/792=0.74495, choice D.
Setup
Count unordered five-token samples because all such subsets are equally likely.
Model
Use inclusion-exclusion on samples missing red, blue, or green. Only the all-red sample survives a pairwise intersection.
Compute
Subtract the three missing-color counts and add back the one nonempty intersection.
Answer
The probability of covering all three colors is approximately 0.745.
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ActuaryProof. “SOA Exam P Sample Question #456 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
