This Exam P sample reference tests Uniform Distribution. The first payment probability determines the deductible as 0.20. A payment of at least 1.44 then requires a loss of at least 1.64, whose uniform upper-tail probability is 0.18, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.06 subtracts the 0.24 increase in payment threshold directly from 0.30. Under a uniform distribution on an interval of length 2, the tail falls by 0.24/2=0.12.
BThe value 0.16 uses 0.24, the increase between the two payment thresholds, as though it were the deductible. That produces the incorrect loss threshold 1.68.
DThe value 0.20 reports the deductible rather than the requested tail probability.
EThe value 0.28 treats the tail reduction as only 0.02. The payment threshold rises by 0.24, reducing a uniform tail by 0.12.
Original practice · fully worked
Original variant: capped reimbursement among eligible invoices
A maintenance invoice X is uniformly distributed from 0 to 15 thousand dollars. A service fund reimburses 60% of the portion above 3 thousand dollars, subject to a maximum reimbursement of 6 thousand dollars. Given that an invoice produces a positive reimbursement, find the probability that the reimbursement reaches its maximum.
A 0.1333
B 0.1667
C 0.2500
D 0.5000
E 0.8333
Variant answer in brief
A positive reimbursement requires X>3, while the cap is reached once 0.6(X-3) is at least 6, or X at least 13. The conditional uniform length ratio is 2/12=1/6, choice B.
Setup
Setup
Translate positivity and the capped-payment event into invoice thresholds.
{Y>0}={X>3}
{Y=6}={0.6(X−3)≥6}
Model
Model
Solve the cap inequality within the uniform support.
0.6(X−3)≥6⟺X≥13
Compute
Compute
Condition by dividing the cap interval length by the positive-payment interval length.
Pr(Y=6∣Y>0)=15−315−13
Pr(Y=6∣Y>0)=61=0.1666667
Answer
Answer
One sixth of eligible invoices reach the reimbursement cap.
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