Worked solution · Exam P

SOA Exam P Sample Question #312 Solution

This Exam P sample reference tests Conditional Probability. Inclusion-exclusion gives overlap 0.30+0.42-0.60=0.12. Exactly one type has probability 0.48 and the conditioning event has probability 0.88, so the ratio is 6/11=0.545455 and choice D.

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 Conditional Probability question

Setup

Recover the overlap from the two marginal probabilities and their union.

\Pr(M\cap P)=0.30+0.42-0.60=0.12

Model

Compute the numerator by removing the overlap from both marginal events, and compute the conditioning probability by complementing the overlap.

\Pr(\text{exactly one})=0.30+0.42-2(0.12)=0.48

\Pr(\text{not both})=1-0.12=0.88

Compute

Divide the target region by the conditioning region.

\Pr(\text{exactly one}\mid\text{not both})=\frac{0.48}{0.88}

=\frac6{11}=0.5454545\ldots

Answer

The conditional probability rounds to 0.55.

\boxed{0.55\quad\text{(D)}}

Test the method on unseen problems

General Probability Practice Workbook

30 original general probability problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

AThe verified exactly-one numerator is 0.48. A conditional value of 0.45 would require a denominator greater than one, so this choice results from mixing incompatible Venn regions.
BThe value 0.48 is the unconditional probability of exactly one claim type. It is the numerator before conditioning on not filing both types.
CThe value 0.52 is 1-0.48, the probability of either both or neither. It complements the numerator rather than applying the stated condition.
EThe value 0.60 is the probability of at least one claim type. The conditioning event also includes no claim, but excludes the overlap.

Original practice · fully worked

Original variant: rail use among two-mode riders

A city survey records whether each rider uses rail, bus, and ferry service. The respective use probabilities are 0.55, 0.45, and 0.30. The rail-bus, rail-ferry, and bus-ferry intersection probabilities are 0.25, 0.15, and 0.10, and 0.05 use all three. A rider is known to use exactly two of the modes. Calculate the conditional probability that rail is one of them.

A
0.3000
B
0.3500
C
0.5500
D
0.7143
E
0.8571

Answer and calculation

The exclusive rail-bus and rail-ferry cells have probabilities 0.20 and 0.10. All exclusive two-mode cells total 0.20+0.10+0.05=0.35, so the conditional rail share is 0.30/0.35=6/7=0.8571, choice E.

Setup

Remove the all-three cell from each inclusive pairwise intersection.

\Pr(RB\text{ only})=0.25-0.05=0.20

\Pr(RF\text{ only})=0.15-0.05=0.10

\Pr(BF\text{ only})=0.10-0.05=0.05

Model

Exactly two modes is the disjoint union of the three pair-only cells.

\Pr(\text{exactly two})=0.20+0.10+0.05=0.35

Compute

Rail appears in the rail-bus-only and rail-ferry-only cells.

\Pr(R\cap\text{exactly two})=0.20+0.10=0.30

\Pr(R\mid\text{exactly two})=\frac{0.30}{0.35}=\frac67=0.8571428571

Answer

Among exactly-two-mode riders, the rail share rounds to 0.8571.

\boxed{0.8571\quad\text{(E)}}

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How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: rail use among two-mode riders

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability