Verified solution · Exam P
This Exam P sample reference tests Hypergeometric Distribution. A qualifying inspection selects three of the seven intact boxes and two of the three damaged boxes. Dividing this favorable count by all five-box selections gives 105/252, or 5/12. The probability is approximately 0.417, choice D.
Independent solution
Setup
Treat the inspection as a sample without replacement from seven intact and three damaged boxes.
Model
Exactly three intact boxes requires choosing the remaining two boxes from the damaged group.
Compute
Evaluate the hypergeometric ratio.
Answer
The probability of finding exactly three intact boxes is approximately 0.417.
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Original practice · fully worked
A rack holds 10 removable drives in a uniformly random order. Four drives are encrypted and the other six are not. Calculate the probability that no two encrypted drives occupy adjacent positions.
The six unencrypted drives create seven gaps in which encrypted drives can be placed without adjacency. Choosing four of those gaps gives 35 valid encrypted-position sets among 210 unrestricted sets, so the probability is 1/6 and choice C.
Setup
Choose the four encrypted positions; all such position sets are equally likely.
Model
Place the six unencrypted drives first. Their seven surrounding gaps can each hold at most one encrypted drive.
Compute
Divide the valid encrypted-position sets by all possible sets.
Answer
The probability that all encrypted drives are separated is approximately 0.1667.
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ActuaryProof. “SOA Exam P Sample Question #366 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
