Worked solution · Exam P
This Exam P sample reference tests Conditional Probability. The insured-and-failing joint probability is 0.80(0.03)=0.024. Dividing this by the overall failure probability 0.05 gives 0.48, so choice D is correct.
Independent solution
Setup
Let I identify an insured institution and F identify failure during the stated horizon.
Model
First form the joint mass associated with insured failures.
Compute
Condition the joint mass on the full failure population.
Answer
Forty-eight percent of the failing institutions belong to the insured group.
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Original practice · fully worked
A data center routes 40% of packets through channel A, 35% through channel B, and 25% through channel C. The corresponding checksum-alert probabilities are 0.02, 0.05, and 0.01. Given that a packet generated a checksum alert, calculate the probability that it used channel B.
The total alert probability is 0.40(0.02)+0.35(0.05)+0.25(0.01)=0.028. Channel B contributes 0.0175, so its posterior probability is 0.625 and choice E is correct.
Setup
Weight each channel's alert rate by its traffic share.
Model
The three routes partition packet traffic, so their alert masses add to the denominator.
Compute
Divide channel B's joint alert mass by the total alert mass.
Answer
The posterior channel-B share is 62.5%.
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
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