This Exam P sample reference tests Independence. One loss remains below its deductible with probability 0.30. Independence makes the probability of no benefit from all three policies equal to the cube of 0.30, so the probability of at least one benefit is 0.973 and choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is the probability that all three losses remain below their deductibles, the complement of the requested event.
BThis is the probability that all three losses exceed their deductibles. It requires benefits from every policy rather than from at least one.
CThis is the complement of all three policies paying. It describes at least one policy not paying, which is a different union.
DThis is the benefit probability for one specified policy and ignores the other two independent opportunities.
Original practice · fully worked
Original variant: shared weather state across three sensors
A monitoring day is severe with probability 0.20 and ordinary otherwise. Conditional on the day's state, three pressure sensors act independently. Each sensor crosses its alert threshold with probability 0.40 on a severe day and 0.10 on an ordinary day. Calculate the probability that at least one sensor crosses its threshold.
A 0.271
B 0.374
C 0.400
D 0.407
E 0.784
Variant answer in brief
The at-least-one probabilities are 0.784 on a severe day and 0.271 on an ordinary day. Weighting them by the weather-state probabilities gives 0.3736, which rounds to 0.374 and choice B.
Setup
Setup
Use complements within each weather state.
Pr(A∣S)=1−(0.60)3=0.784
Pr(A∣O)=1−(0.90)3=0.271
Model
Model
Average the two conditional probabilities over the shared daily state.
Pr(A)=Pr(S)Pr(A∣S)+Pr(O)Pr(A∣O)
Compute
Compute
Substitute the severe-day and ordinary-day weights.
Pr(A)=0.20(0.784)+0.80(0.271)=0.3736
Answer
Answer
At least one sensor crosses its threshold with probability approximately 0.374.
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