This Exam P sample reference tests Conditional Probability. The amount threshold is met with probabilities 3/4 and 1/2 in the one- and two-claim branches. Their weighted contributions are 0.06 and 0.01, so the posterior one-claim probability is 0.06/(0.06+0.01)=6/7, choice E.
How to solve this Conditional Probability question
Setup
Setup
Let N denote the claim count and let H denote the event that the aggregate amount does not exceed the stated threshold. Only the two positive-count branches can enter the condition.
Pr(N=1)=0.08,Pr(N=2)=0.02
Model
Model
Evaluate the conditional amount distribution at the threshold separately for each branch.
Pr(H∣N=1)=1−2000500=43
Pr(H∣N=2)=1−20001000=21
Compute
Compute
Weight each likelihood by its claim-count probability and normalize the one-claim contribution.
Pr(N=1,H)=0.08(43)=0.06
Pr(N=2,H)=0.02(21)=0.01
Pr(N=1∣N≥1,H)=0.06+0.010.06=76
Answer
Answer
The conditional probability is six-sevenths.
76(E)
Continue without hunting through PDFs
All 718 Exam P sample solutions in syllabus order
The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis reports Pr(H given N = 2) = 1/2, the amount-threshold likelihood in the two-claim branch, rather than the posterior probability of one claim.
BThis divides the filtered one-claim contribution by the unfiltered positive-count probability: 0.06/(0.08 + 0.02) = 3/5. The denominator must impose the amount condition too.
CThis weights the two-claim prior twice and ignores both amount likelihoods, giving 0.08/[0.08 + 2(0.02)] = 2/3. Claim count is not a multiplicity weight in Bayes normalization.
DThis reports Pr(H given N = 1) = 3/4. The observed threshold changes the relative branch weights and must be combined with both prior probabilities.
Original practice · fully worked
Original variant: identify a high-drift instrument line
A laboratory receives 60% of its instruments from line A and 40% from line B. Conditional on line A, absolute calibration drift is uniform from 0 to 10 units. Conditional on line B, it is uniform from 0 to 20 units. An instrument is observed to have drift greater than 8 units. Calculate the probability that it came from line B.
A 0.200
B 0.400
C 0.600
D 0.667
E 0.750
Variant answer in brief
The upper-tail likelihoods are 0.20 for line A and 0.60 for line B. Their prior-weighted contributions are 0.12 and 0.24, so the line-B posterior is 0.24/0.36=2/3, choice D.
Setup
Setup
Let H denote the observed high-drift event. Compute its probability under each uniform source.
Pr(H∣A)=1010−8=0.20
Pr(H∣B)=2020−8=0.60
Model
Model
Combine each likelihood with its source proportion.
Pr(A∩H)=0.60(0.20)=0.12
Pr(B∩H)=0.40(0.60)=0.24
Compute
Compute
Normalize the line-B contribution over all instruments satisfying the observation.
Pr(B∣H)=0.12+0.240.24=32=0.666666…
Answer
Answer
The observed instrument came from line B with probability two-thirds.
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.
