Verified solution · Exam P
This Exam P sample reference tests Conditional Probability. Summing the masses from one through four and dividing by the mass through four gives 20/50=2/5, choice B.
Independent solution
Setup
Use the supplied probability mass function to list the masses for N=1 through N=4.
Model
The desired conditional numerator is the sum of the four positive masses allowed by N≤4.
Compute
The numerator is 1/3. Adding the N=0 mass 1/2 gives denominator 5/6, and their ratio is 2/5.
Answer
Thus Pr(N≥1 given N≤4)=2/5, corresponding to choice B.
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Original practice · fully worked
A count N takes values 0 through 4 with probability proportional to n+1. Given that N is even, find the probability that N equals 4.
The even outcomes 0, 2, and 4 have weights 1, 3, and 5. The largest outcome therefore has conditional probability 5/9, choice E.
Setup
Under the proportional mass rule, assign weights n+1 to the possible values n=0 through 4.
Model
After conditioning on an even count, only the weights for 0, 2, and 4 remain in the normalizing sum.
Compute
The target weight is 5 and the even-count weights total 1+3+5=9, so the conditional probability is 5/9.
Answer
Therefore Pr(N=4 given N even)=5/9, which is choice E.
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Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
