Worked solution · Exam P
This Exam P sample reference tests Set Probability. Because B is contained in A while C is disjoint from both, the three union probabilities are 0.30, 1.00, and 0.80. Their sum is 2.10, so choice C is correct.
Independent solution
Setup
Use the event definitions to identify the relevant set relationships.
Model
A union with a subset equals the larger event. A union of disjoint events has probability equal to the sum of their probabilities.
Compute
Substitute the three supplied event probabilities.
Answer
The requested sum of union probabilities is 2.10.
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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.
Error diagnosis
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Original practice · fully worked
In a production audit, R is the event that an item receives a red flag, Q is the event that it receives any quality flag, and S is the event that it is diverted for a separate random inspection. Every red flag is a quality flag, and diversion is disjoint from either flag event. Suppose P(Q)=0.42, P(R)=0.18, and P(S)=0.31. Calculate P(Q union R)+P(Q union S)+P(R union S).
R is contained in Q, while S is disjoint from both. The three unions are 0.42, 0.73, and 0.49, which sum to 1.64. Choice D is correct.
Setup
Translate the stated relationships into set notation.
Model
Containment fixes the first union, while disjointness turns the other two into sums.
Compute
Evaluate and total the three terms.
Answer
The sum of the three union probabilities is 1.64.
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