Worked solution · Exam P
This Exam P sample reference tests Independence. Exactly one loss type is the disjoint union of fire without theft and theft without fire. Independence gives 0.20(0.70)+0.80(0.30)=0.38, so choice B.
Independent solution
Setup
Let F and T denote the two annual loss events.
Model
Exactly one event occurs in either of two mutually exclusive patterns.
Compute
Use independence within each pattern and add the disjoint probabilities.
Answer
The exactly-one probability is 0.38.
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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.
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Original practice · fully worked
Two verification modules return independent pass indicators I and J, with pass probabilities 0.25 and 0.40. A status lamp has indicator Z=1 exactly when one module passes and the other fails. Calculate Cov(I,Z).
The lamp probability is 0.25(0.60)+0.75(0.40)=0.45, while IZ=1 only when I=1 and J=0, with probability 0.15. Thus Cov(I,Z)=0.15-(0.25)(0.45)=0.0375, choice C.
Setup
Compute the probability that the exclusive-status lamp turns on.
Model
The product IZ equals one only when module I passes and module J fails.
Compute
Subtract the product of the two means from the product moment.
Answer
The covariance is positive 0.0375.
Keep practicing
These references share syllabus concepts or sit nearby in the official sequence. All 192 Exam P general probability practice problems.
Drill the method: Exam P Conditional Probability problems · Exam P Independence problems
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