Verified solution · Exam P
This Exam P sample reference tests Mixture Distributions. This is a mixture probability over four latent count states. Weighting each conditional zero-visit probability gives 0.389948, so the probability of one or more visits is 0.610052 and choice C.
Independent solution
Setup
Let X be the latent illness count and V the resulting visit count. Work through the complement event V=0.
Model
The zero state produces no visits with certainty; each positive state produces a Poisson zero probability exp(-k).
Compute
Evaluate the weighted mixture and take its complement.
Answer
The probability of at least one visit rounds to 0.61.
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Original practice · fully worked
A control panel activates 1, 2, or 4 scanners with probabilities 0.25, 0.50, and 0.25, respectively. Conditional on the number activated, scanners pass a diagnostic independently with probability 0.80. Calculate the probability that at least two scanners pass.
The one-scanner state contributes zero. The two- and four-scanner states contribute 0.50(0.64)=0.32 and 0.25(0.9728)=0.2432, totaling 0.5632 and choice C.
Setup
Let N be the active-scanner count and K the number that pass. Conditional on N, K is binomial.
Model
Evaluate the target under each possible panel size.
Compute
Weight the three conditional probabilities by the panel-size probabilities.
Answer
The probability that at least two scanners pass is 0.5632.
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These references share syllabus concepts or sit nearby in the official sequence. All 388 Exam P univariate random variables practice problems.
Drill the method: Exam P Poisson Distribution problems
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ActuaryProof. “SOA Exam P Sample Question #457 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
