This Exam P sample reference tests Joint Distributions. Marginalizing the joint mass function over the theft coordinate gives fire-loss probabilities 17/60, 20/60, and 23/60, so E[X] = 2.1. The policyholder retains 44% of that loss, producing 0.924 and choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is consistent with weighting the marginal probabilities without the loss values, which understates the expected fire loss.
BThis results from an incomplete marginal total, typically omitting part of the highest-loss row or column.
DThis applies the 56% reimbursement rate as though it were the retained fraction.
EThis overweights the larger fire-loss values before applying the retained percentage.
Original practice · fully worked
Original variant: automated fault diagnosis
A repair center observes X electrical faults and Y cosmetic faults in a device, where each count can be 0, 1, or 2 and their joint probability mass is p(x,y) = (x + 2y + 1)/36. Each electrical fault normally requires 50 minutes of diagnosis, but an automated bench completes 60% of that work. Calculate the expected number of technician minutes spent diagnosing electrical faults.
A 10.000
B 20.000
C 23.333
D 35.000
E 58.333
Variant answer in brief
Summing the joint mass over cosmetic faults gives f_X(x) = (x + 3)/12 and E[X] = 7/6. Technicians perform 40% of 50 minutes per electrical fault, so the expected manual time is 20(7/6) = 23.333 minutes.
Setup
Setup
Marginalize over the cosmetic-fault count because the labor calculation depends only on X.
fX(x)=y=0∑236x+2y+1
Model
Model
Simplify the marginal and compute the expected electrical-fault count.
fX(x)=363x+9=12x+3
E[X]=0(123)+1(124)+2(125)=67
Compute
Compute
The technician supplies the 40% of diagnosis time not handled by the bench.
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.