Verified solution · Exam P
This Exam P sample reference tests Conditional Probability. The probability of having neither procedure is the complement of 0.35, namely 0.65. Dividing this joint no-procedure probability by the 0.70 probability of no fillings gives 13/14, approximately 0.93, and choice E.
Independent solution
Setup
Let F and R denote having at least one filling and at least one root canal. Complement the reported union.
Model
The requested event conditions the joint absence of both procedures on having no fillings.
Compute
Substitute the joint and marginal probabilities.
Answer
Given no fillings, the probability of no root canals is approximately 0.93.
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Original practice · fully worked
During a monitoring interval, a server has probability 0.30 of raising a capacity alert and probability 0.25 of raising a latency alert. The probability of raising neither alert is 0.55. Given that at least one alert is raised, calculate the probability that exactly one alert is raised.
At least one alert occurs with probability 0.45. Inclusion–exclusion gives overlap 0.10, so exactly one alert occurs with probability 0.35. Dividing by the conditioning probability gives approximately 0.778 and choice E.
Setup
Complement the no-alert probability.
Model
Use inclusion–exclusion to recover the overlap.
Compute
Remove the overlap twice from the marginal sum, then condition on the union.
Answer
Given at least one alert, exactly one is raised with probability approximately 0.778.
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ActuaryProof. “SOA Exam P Sample Question #365 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
