Worked solution · Exam P
This Exam P sample reference tests Poisson Aggregation. Three independent weekly Poisson counts with mean 2 combine to Poisson mean 6. Summing masses zero through three gives 61 exp(−6)=0.1512, choice B.
Independent solution
Setup
The three independent weekly Poisson counts combine into one Poisson count with mean equal to the sum of the weekly means.
Model
The aggregate mean is six. Fewer than four events includes counts zero through three.
Compute
Adding the first four Poisson masses gives 61 times the zero-count mass, or approximately 0.151204.
Answer
The probability of fewer than four events is about 0.15.
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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.
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Original practice · fully worked
Weekly support-call counts are independent Poisson variables with mean 1.5. Find the probability that fewer than three calls occur over four weeks.
The four-week total is Poisson with mean 6. Summing the masses at zero, one, and two gives 25 exp(−6)=0.06197, choice B.
Setup
The four independent weekly Poisson counts combine into a Poisson total with mean six.
Model
Fewer than three calls includes counts zero, one, and two.
Compute
Adding those three masses gives 25 times the zero-count mass, or approximately 0.061969.
Answer
The probability is about 0.0620.
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