This Exam P sample reference tests Expected Value. Multiplying each of the four positive benefits by its first-failure probability gives expected benefit 2,694.4, choice E.
The first-failure year has a geometric distribution with annual failure probability 0.4. The benefit is positive only for first failure in years one through four.
Pr(T=k)=0.6k−1(0.4),k=1,2,3,4
Model
Model
For each eligible year, multiply its benefit by the probability of surviving all earlier years and then failing in that year.
Original variant: expected odd-year inspection benefit
At the start of each year a functioning unit has probability 0.25 of failing during that year. An inspection benefit pays 1000 only when first failure occurs in year 1, 3, or 5, and pays zero otherwise. Find the expected benefit.
A 250.00
B 390.63
C 437.50
D 469.73
E 578.13
Variant answer in brief
The eligible first-failure probabilities are 0.25, 0.75²(0.25), and 0.75⁴(0.25). Their 1000-weighted sum is 469.73, choice D.
Setup
Setup
The first-failure year is geometric with annual failure probability 0.25.
Pr(T=k)=0.75k−1(0.25)
Model
Model
Only years one, three, and five qualify. Weight the 1,000 benefit by the first-failure probability at each of those three years.
E[B]=1000(0.25){1+0.752+0.754}
Compute
Compute
The three contributions sum to approximately 469.7266.
E[B]=469.7265625
Answer
Answer
The expected inspection benefit is approximately 469.73, selecting choice D.
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