This Exam P sample reference tests Conditional Probability. List the three mutually exclusive ways that exactly one loss category can occur. The target category alone has probability 0.072 and all exactly-one cases total 0.306, so the conditional probability is 0.235294 and choice B is correct.
How to solve this Conditional Probability question
Setup
Setup
Write F, W, and T for occurrence of a loss in the first, second, and third categories. Their loss probabilities are the complements of the given no-loss probabilities.
Pr(F)=0.10,Pr(W)=0.10,Pr(T)=0.20
Model
Model
Under independence, multiply across categories for each of the three disjoint exactly-one cases.
Pr(F∩Wc∩Tc)=0.10(0.90)(0.80)=0.072
Pr(Fc∩W∩Tc)=0.90(0.10)(0.80)=0.072
Pr(Fc∩Wc∩T)=0.90(0.90)(0.20)=0.162
Compute
Compute
Add the exactly-one cases for the conditioning probability, then divide the target case by that total.
Pr(exactly one category)=0.072+0.072+0.162=0.306
Pr(F∣exactly one category)=0.3060.072=0.2352941176…
Answer
Answer
The conditional probability rounds to 0.235.
0.235(B)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is the joint probability that the target category is the sole loss category, before division by the exactly-one conditioning probability.
CNormalizing the three marginal loss probabilities 0.10, 0.10, and 0.20 gives 0.10/0.40=0.25, but the exactly-one cases require the accompanying no-loss factors.
DThis is the probability of exactly one loss category, which belongs in the denominator of the conditional probability.
ENormalizing the no-loss probabilities gives 0.90/(0.90+0.90+0.80)=0.346; the conditioning event concerns loss occurrences, not no-loss weights.
Original practice · fully worked
Original variant: greenhouse alert diagnosis
A greenhouse controller monitors four independent alert channels: moisture, frost, pests, and backup power. Their probabilities of triggering during a shift are 0.20, 0.10, 0.05, and 0.02, respectively. The shift log shows that exactly one channel triggered. Calculate the conditional probability that it was the moisture channel.
A 0.1676
B 0.2000
C 0.2910
D 0.5405
E 0.5758
Variant answer in brief
The moisture-only probability is 0.16758, and the four mutually exclusive one-alert probabilities total 0.29102. Their ratio is 0.575837, so the correct choice is E.
Setup
Setup
Denote the four trigger probabilities by 0.20, 0.10, 0.05, and 0.02 in the order listed.
(pM,pF,pP,pB)=(0.20,0.10,0.05,0.02)
Model
Model
For a named channel to be the sole trigger, that channel must trigger while all other independent channels remain quiet.
wi=pij=i∏(1−pj)
Compute
Compute
Evaluate each one-trigger cell and normalize the moisture cell by their sum.
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.
