Verified solution · Exam P
This Exam P sample reference tests Law of Total Probability. Class A contributes 0.008 to the hospitalization probability and Class B contributes 0.003. Thus Class A accounts for eight elevenths of all hospitalizations, approximately 0.727, and choice E.
Independent solution
Setup
Convert the class counts to proportions and complement the two no-hospitalization rates.
Model
Weight each class-specific hospitalization rate by its class proportion.
Compute
Condition the Class A contribution on the total hospitalization probability.
Answer
A hospitalized policyholder has probability approximately 0.727 of belonging to Class A.
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Original practice · fully worked
A simulation selects a fair six-sided die with probability two-thirds and a fair eight-sided die otherwise. The selected die is rolled twice. The first roll is greater than four. Given this result, calculate the probability that the second roll is also greater than four.
The first result updates the die probabilities to four-sevenths for the six-sided die and three-sevenths for the eight-sided die. Averaging their chances of another result above four gives 17/42, approximately 0.405, and choice C.
Setup
Let E denote a roll greater than four. Calculate its likelihood under each possible die.
Model
Use the first roll to update the probability assigned to each die.
Compute
Conditional on the selected die, the two rolls are independent. Average the second-roll likelihoods over the updated die probabilities.
Answer
The second roll exceeds four with conditional probability approximately 0.405.
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ActuaryProof. “SOA Exam P Sample Question #363 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
