This Exam P sample reference tests Conditional Distributions. First average the conditional low-cost probabilities over the market states to obtain 0.32. The complementary high-cost probability is 0.68, so the expected cost is 5(0.32)+10(0.68)=8.4 million, choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AA value of 6.4 million would require a high-cost probability of only 0.28, contradicting the weighted total of 0.68.
BThis is 10(0.68), so it counts the high-cost contribution but omits the low-cost contribution 5(0.32).
CThis is the unweighted midpoint of the two possible costs and ignores all conditional and market-state probabilities.
EThis would require a high-cost probability of 0.84, overstating the high-cost branch relative to the weighted table.
Original practice · fully worked
Original variant: field-station repair cost
A field station faces either a 2-thousand-dollar routine repair or an 8-thousand-dollar major repair. Weather is clear, rainy, or snowy with probabilities 0.50, 0.30, and 0.20. Conditional on those states, the probabilities of a major repair are 0.10, 0.40, and 0.80, respectively. Calculate the expected repair cost.
A 2.64 thousand dollars
B 3.02 thousand dollars
C 3.98 thousand dollars
D 4.60 thousand dollars
E 9.80 thousand dollars
Variant answer in brief
The total probability of a major repair is 0.50(0.10)+0.30(0.40)+0.20(0.80)=0.33. The expected cost is therefore 2(0.67)+8(0.33)=3.98 thousand dollars, choice C.
Setup
Setup
Average the major-repair probabilities over the weather distribution.
Pr(M)=w∑Pr(M∣W=w)Pr(W=w)
Model
Model
Insert the three weather-state probabilities.
Pr(M)=0.50(0.10)+0.30(0.40)+0.20(0.80)
Compute
Compute
Use the resulting two-point distribution for repair cost, measured in thousands.
Pr(M)=0.33,Pr(Mc)=0.67
E[C]=8(0.33)+2(0.67)=3.98
Answer
Answer
The expected repair cost is 3.98 thousand dollars.
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