This Exam P sample reference tests Conditional Expectation. The hospitalization count has probabilities 0.49, 0.42, and 0.09. The loss condition always holds for counts 0 or 1 and with probability 1/2 for count 2, yielding conditional mean 0.534, choice B.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.510 is the count-weighted numerator before division by the conditioning probability 0.955.
CThe value 0.600 is the unconditional hospitalization mean. It ignores the observed loss condition.
Original practice · fully worked
Original variant: expected incident count after an audit pass
Half of service-desk shifts have zero incidents, three tenths have one, and the remainder have three. An audit passes in 90%, 60%, and 20% of those respective states. Find the expected incident count given that the audit passes.
C 0.2687
B 0.3750
A 0.4478
D 0.6000
E 0.7463
Variant answer in brief
The pass-state weights are 0.45, 0.18, and 0.04. Their total is 0.67 and their count-weighted sum is 0.30, giving conditional mean 0.4478.
Setup
Setup
Combine each incident-count probability with its audit-pass probability to form the three joint pass weights.
P(N=0,1,3)=(0.50,0.30,0.20)
Model
Model
The joint weights for counts 0, 1, and 3 are 0.45, 0.18, and 0.04.
P(A∣N=0,1,3)=(0.90,0.60,0.20)
Compute
Compute
The pass probability is 0.67 and the count-weighted pass total is 0.30, giving conditional mean 0.447761.
P(A)=0.45+0.18+0.04=0.67
E[N1A]=1(0.18)+3(0.04)=0.30
E[N∣A]=0.30/0.67=0.447761
Answer
Answer
The expected incident count given a pass is 0.4478, corresponding to choice A.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
