Verified solution · Exam P
This Exam P sample reference tests Poisson Distribution. Each count has Poisson mean 0.25, and independence makes their aggregate Poisson with mean 0.50. Complementing the zero- and one-count masses gives 0.090204, so choice B is correct.
Independent solution
Setup
For a Poisson variable, the variance equals the mean. The supplied variance therefore identifies the rate for one item.
Model
Independent Poisson variables add to another Poisson variable whose mean is the sum of the component means.
Compute
The required upper tail is most directly evaluated by removing the masses at zero and one.
Answer
The result rounds to the second listed probability.
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Error diagnosis
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Original practice · fully worked
Three sensor clusters generate false alerts as independent Poisson processes with rates 0.15, 0.35, and 0.20 per hour. During a two-hour diagnostic window, calculate the probability that the monitoring service records at least three false alerts.
Pooling the three streams over two hours gives a Poisson mean of 1.4. Complementing the masses for zero, one, and two alerts produces 0.166502, so choice A is correct.
Setup
Add the three hourly rates before applying the two-hour exposure.
Model
Superposition and time scaling give the diagnostic-window count.
Compute
Remove all outcomes below three.
Answer
The pooled upper tail rounds to 0.1665.
Keep practicing
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 #529 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
