Worked solution · Exam P
This Exam P sample reference tests Poisson Distribution. Writing the two Poisson cumulative probabilities cancels their common exponential factor. The stated ratio produces a quadratic equation with positive root 4. Because a Poisson mean equals its parameter, the answer is 4.0, choice B.
Independent solution
Setup
Write the cumulative probabilities through counts one and two.
Model
Insert the reported ratio and cancel the common factor.
Compute
Clear denominators and solve the resulting quadratic, rejecting the negative root.
Answer
The Poisson expected value is 4.0.
Test the method on unseen problems
30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.
Error diagnosis
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Original practice · fully worked
The number N of flaws found on a panel has a Poisson distribution. The expected number of unordered pairs that can be formed from the flaws on one panel is 6. Calculate the probability that a panel has no flaws.
For a Poisson count, the expected number of unordered pairs is one-half the squared mean. Setting this equal to 6 gives mean square root of 12, so the zero-flaw probability is approximately 0.0313 and choice B.
Setup
Write the pair count as a falling-factorial function of N.
Model
Use the second factorial moment of a Poisson variable.
Compute
Evaluate the zero-count mass at the recovered mean.
Answer
The zero-flaw probability is approximately 0.0313.
Keep practicing
These references share syllabus concepts or sit nearby in the official sequence. All 387 Exam P univariate random variables practice problems.
Drill the method: Exam P Poisson Distribution problems
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