This Exam P sample reference tests Compound Uniform Distribution. Zero, one, and two claims contribute 0.7, 0.2(0.8), and 0.1(0.8²/2) to the event. Their sum is 0.892, choice D.
How to solve this Compound Uniform Distribution question
Setup
Setup
Condition on whether the claim count is zero, one, or two. These are the only count values with positive probability.
Pr(S≤48∣N=0)=1,Pr(S≤48∣N=1)=6048=0.8
Model
Model
With no claim the total is automatically within the limit. With one uniform claim the conditional probability is 0.8. With two claims, the qualifying region is a right triangle occupying 0.32 of the square support.
Pr(S≤48∣N=2)=21(6048)2=0.32
Compute
Compute
Weight the three conditional probabilities by the count probabilities. The contributions are 0.70, 0.16, and 0.032, totaling 0.892.
Pr(S≤48)=0.7+0.2(0.8)+0.1(0.32)=0.892
Answer
Answer
The total benefit is at most 48 with probability 0.892.
0.892(D)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.320 is only the conditional probability for two claims. It ignores both the claim-count probabilities and the zero- and one-claim cases.
CThe value 0.800 is only the conditional probability for one claim and is not averaged over the claim count.
EThe value 0.924 treats the two-claim qualifying region as a square of area 0.64 instead of the triangular region of area 0.32.
Original practice · fully worked
Original variant: total service credit under a mixed count
A customer receives no credit with probability 0.40, one independent Uniform(0,100) credit with probability 0.40, and two such credits with probability 0.20. Find the probability that the total credit is at most 50.
A 0.500
B 0.550
C 0.600
D 0.625
E 0.650
Variant answer in brief
The three count cases contribute 0.40, 0.40(0.50), and 0.20(0.50²/2). The sum is 0.625, choice D.
Setup
Setup
Condition on the number of credits, which may be zero, one, or two.
Pr(S≤50∣N=0)=1,Pr(S≤50∣N=1)=0.5
Model
Model
Zero credits always satisfy the limit, one uniform credit does so with probability one half, and two credits do so on a triangular region occupying one eighth of their square support.
Pr(S≤50∣N=2)=20.52=0.125
Compute
Compute
Weighting by the count probabilities gives contributions 0.40, 0.20, and 0.025, whose sum is 0.625.
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