Worked solution · Exam P

SOA Exam P Sample Question #348 Solution

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.

Independent analysis by ActuaryProof · reviewed July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Independence question

Setup

For one policy, identify the event that the loss does not cross the deductible.

\Pr(\text{no benefit from one policy})=\frac{30}{100}=0.30

Model

The three losses are independent, so multiply the three no-benefit probabilities.

\Pr(\text{no benefit from all three})=(0.30)^3=0.027

Compute

Receiving a benefit from any policy is the complement of receiving none.

\Pr(\text{at least one benefit})=1-0.027=0.973

Answer

The policyholder receives at least one benefit with probability 0.973.

\boxed{0.973\quad\text{(E)}}

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Error diagnosis

Why the other letters tempt

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

Answer and calculation

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

Use complements within each weather state.

\Pr(A\mid S)=1-(0.60)^3=0.784

\Pr(A\mid O)=1-(0.90)^3=0.271

Model

Average the two conditional probabilities over the shared daily state.

\Pr(A)=\Pr(S)\Pr(A\mid S)+\Pr(O)\Pr(A\mid O)

Compute

Substitute the severe-day and ordinary-day weights.

\Pr(A)=0.20(0.784)+0.80(0.271)=0.3736

Answer

At least one sensor crosses its threshold with probability approximately 0.374.

\boxed{0.374\quad\text{(B)}}

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How to solve this Independence question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: shared weather state across three sensors

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability