This Exam P sample reference tests Bayes' Theorem. Zone A contributes 0.40(0.015)=0.006 to the fire-loss probability. The three zones contribute 0.01185 in total, so Bayes' theorem gives 0.006/0.01185=0.506329 and choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.349 is essentially the 0.35 prior share of policies in Zone B. It uses the wrong zone and does not condition on a fire loss.
BThe value 0.400 is Zone A's prior policy share. It ignores the differing zone-specific fire rates.
CThe value 0.441 is 0.015/(0.015+0.011+0.008). It normalizes the three fire rates as though the zones had equal policy shares.
DWith the verified numerator 0.006, a posterior of 0.465 would require total fire probability about 0.012903. The prior-weighted zone contributions instead sum to 0.01185, so this choice reflects incompatible weights.
Original practice · fully worked
Original variant: infer whether paired outputs are linked
A test unit operates in linked mode with probability 0.30 and independent mode with probability 0.70. It emits two binary outputs. In linked mode the outputs are always identical, with 0 and 1 equally likely. In independent mode the outputs are independent fair bits. The two observed outputs match. Calculate the conditional probability that the unit was in linked mode.
A 0.3000
B 0.3500
C 0.4615
D 0.5000
E 0.5385
Variant answer in brief
A match is certain in linked mode and has probability 1/2 in independent mode. The prior-weighted match probabilities are 0.30 and 0.35, so the linked-mode posterior is 0.30/0.65=6/13, approximately 0.4615 and choice C.
Setup
Setup
Let M denote the event that the two outputs match.
Pr(M∣L)=1
Model
Model
In independent mode, the matching patterns 00 and 11 are two of four equally likely patterns.
Pr(M∣I)=42=21
Pr(L∩M)=0.30,Pr(I∩M)=0.70(21)=0.35
Compute
Compute
Normalize the linked-mode contribution over all matching outputs.
Pr(L∣M)=0.30+0.350.30=136=0.46153846…
Answer
Answer
The conditional linked-mode probability rounds to 0.4615.
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