Verified solution · Exam P
This Exam P sample reference tests Independence. The probability of no accidents is 0.2916. Exactly one accident contributes 0.3888 from a high-risk employee and 0.0648 from a low-risk employee. Adding these disjoint cases gives 0.7452 and choice E.
Independent solution
Setup
Convert the no-accident probabilities into accident probabilities for the two employee classes.
Model
Calculate the probability that no employee has an accident.
Compute
Separate the exactly-one case according to whether the accident belongs to a high-risk or low-risk employee.
Answer
At most one employee has an accident with probability 0.7452.
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Original practice · fully worked
Four independent sensors have outage probabilities 0.15, 0.25, 0.35, and 0.45 during a test window. Calculate the probability that an even number of sensors are offline.
For independent Bernoulli indicators, the even-parity probability is one-half of one plus the product of one minus twice each outage probability. The product is 0.0105, giving 0.50525 and choice D.
Setup
Encode outage-count parity through the expectation of minus one raised to the count.
Model
Independence makes the parity expectation a product of four Bernoulli factors.
Compute
Evaluate the product and combine it with the probabilities summing to one.
Answer
An even number of sensors are offline with probability approximately 0.5053.
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ActuaryProof. “SOA Exam P Sample Question #351 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
