Verified solution · Exam P

SOA Exam P Sample Question #370 Solution

This Exam P sample reference tests Law of Total Probability. Weighting each bankruptcy rate by its category share gives an overall bankruptcy probability of 0.098. Service businesses contribute 0.020 of that total, so their conditional share is 0.020/0.098, approximately 0.204. This is choice C.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Law of Total Probability question

Setup

Calculate the joint probability of belonging to each category and going bankrupt.

\Pr(B\cap R)=(0.60)(0.12)=0.072

\Pr(B\cap S)=(0.25)(0.08)=0.020

\Pr(B\cap T)=(0.10)(0.06)=0.006

\Pr(B\cap O)=(0.05)(0)=0

Model

Sum the mutually exclusive category contributions to obtain the total bankruptcy probability.

\Pr(B)=0.072+0.020+0.006+0=0.098

Compute

Apply Bayes' theorem using the service contribution as the numerator.

\Pr(S\mid B)=\frac{\Pr(B\cap S)}{\Pr(B)}=\frac{0.020}{0.098}=0.2040816\ldots

Answer

A bankrupt business has probability approximately 0.204 of being a service business.

\boxed{0.204\quad\text{(C)}}

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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 value 0.020 is the joint probability that a business is both a service business and bankrupt. It is the Bayes numerator, not the posterior probability.
BThe value 0.080 is the bankruptcy rate within the service category. The question reverses the conditioning and asks for the category given bankruptcy.
DThe value 0.250 is the unconditional service-business share. Selection from bankrupt businesses changes the category mix.
EA posterior of 0.308 would require a total bankruptcy probability near 0.065. The weighted contributions from all four categories instead total 0.098.

Original practice · fully worked

Original variant: calibration mode from an interval reading

A controller selects broad calibration mode with probability 0.40 and focused calibration mode with probability 0.60. In broad mode, a reading X is uniform from 0 to 5; in focused mode, X is uniform from 2 to 4. A report shows that X lies between 2.5 and 3.5. Calculate the conditional probability that broad mode was selected.

A
0.080
B
0.200
C
0.211
D
0.300
E
0.500

Variant answer in brief

The reported unit-length interval has likelihood 1/5 in broad mode and 1/2 in focused mode. The prior-weighted contributions are therefore 0.08 and 0.30, so the posterior broad-mode probability is 0.08/0.38=4/19, approximately 0.211 and choice C.

Setup

Let I denote the reported interval and calculate its likelihood under each uniform mode.

\Pr(I\mid B)=\frac{3.5-2.5}{5-0}=0.20

\Pr(I\mid F)=\frac{3.5-2.5}{4-2}=0.50

Model

Weight the interval likelihoods by the two mode priors.

\Pr(B\cap I)=0.40(0.20)=0.08

\Pr(F\cap I)=0.60(0.50)=0.30

Compute

Normalize the broad-mode contribution by the total probability of the reported interval.

\Pr(B\mid I)=\frac{0.08}{0.08+0.30}=\frac4{19}=0.2105263\ldots

Answer

Given the interval reading, the probability of broad mode is approximately 0.211.

\boxed{0.211\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #370 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Law of Total Probability question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: calibration mode from an interval reading

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution