Worked solution · Exam P
This Exam P sample reference tests Conditional Probability. Costs exceed the deductible only for four, five, or six visits, whose total probability is 0.07. The exactly-five mass is 0.02, so the conditional probability is 0.02/0.07=0.2857, choice C.
Independent solution
Setup
Translate the strict cost threshold into the allowed visit counts.
Model
Form the conditional probability of exactly five visits within the qualifying tail.
Compute
Insert the three relevant table masses and normalize.
Answer
The conditional probability rounds to 0.286.
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Original practice · fully worked
A quality count N has probabilities 0.50, 0.25, 0.15, 0.07, and 0.03 at N=0,1,2,3,4, respectively. A review score is 0 for N at most 1, 10 for N=2, 30 for N=3, and 60 for N=4. Given that the score is at least 30, calculate its conditional mean.
A score of at least 30 restricts the count to three or four, with total probability 0.10. Weighting scores 30 and 60 by masses 0.07 and 0.03 gives conditional mean 39, choice C.
Setup
Identify the two count values admitted by the score condition.
Model
Form the conditional score expectation from the two retained masses.
Compute
Evaluate the weighted numerator and normalize.
Answer
The conditional mean score is 39.
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Drill the method: Exam P Conditional Probability problems
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