Worked solution · Exam P
This Exam P sample reference tests Inclusion–Exclusion. Because event C cannot accompany either other event, exactly two events means the overlap of A and B, with probability 0.06. Adding that overlap to the 0.38 exactly-one probability gives 0.44 for the union of A and B. The remaining 0.46 belongs to C, choice B.
Independent solution
Setup
Use the two mutual-exclusivity conditions to identify the only possible two-event overlap.
Model
Relate the exactly-one probability for A and B to their union.
Compute
Recover the union of A and B, then use its disjoint union with C.
Answer
Event C has probability 0.46.
Test the method on unseen problems
30 original general probability 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
An integer is selected uniformly from 1 through 60. Calculate the probability that it is divisible by exactly one of 2, 3, and 5.
There are 16 integers divisible only by 2, eight divisible only by 3, and four divisible only by 5. Thus 28 of the 60 integers satisfy exactly one rule, giving 7/15, approximately 0.4667 and choice C.
Setup
Count the multiples and their intersections within the selected range.
Model
Remove the other two divisibility rules from each single-rule count and restore the triple intersection.
Compute
Add the disjoint exactly-one regions and divide by 60.
Answer
The selected integer satisfies exactly one rule with probability approximately 0.4667.
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