Verified solution · Exam P
This Exam P sample reference tests Geometric Distribution. This is a geometric stopping-time calculation. A round continues with probability (1/4)³+(3/4)³=7/16, so exactly five rounds requires four continuations followed by a stop: (7/16)⁴(9/16)=0.020608, making choice C correct.
Independent solution
Setup
In one round, continuation occurs only when all three outcomes match. Add the all-first-outcome and all-second-outcome probabilities.
Model
Exactly five rounds means that each of the first four rounds continues and the fifth round stops.
Compute
Substitute the one-round continuation and stopping probabilities.
Answer
The probability rounds to 0.0206.
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Error diagnosis
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Original practice · fully worked
An incident starts at level 0. During the first response cycle it is resolved with probability 0.3 or advances to level 1 with probability 0.7. Once at level 1, each later cycle resolves it with probability 0.6 or leaves it at level 1 with probability 0.4. Calculate the probability that the incident is resolved exactly on cycle 4.
Resolution on cycle 4 requires advancement on cycle 1, two further unresolved level-1 cycles, and resolution on cycle 4. The probability is 0.7(0.4)²(0.6)=0.0672, so choice B is correct.
Setup
Identify the unique state path that permits resolution on the fourth cycle.
Model
The first transition has probability 0.7, the next two cycles must remain unresolved, and the final cycle must resolve.
Compute
Multiply the transition probabilities along the required path.
Answer
The incident resolves on cycle 4 with probability 0.0672.
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ActuaryProof. “SOA Exam P Sample Question #590 — Verified Solution.” Verified 2026-07-30.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.
