Verified solution · Exam P

SOA Exam P Sample Question #532 Solution

This Exam P sample reference tests Conditional Probability. The insured-and-failing joint probability is 0.80(0.03)=0.024. Dividing this by the overall failure probability 0.05 gives 0.48, so choice D is correct.

Independent analysis by ActuaryProof · reviewed August 4, 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

Let I identify an insured institution and F identify failure during the stated horizon.

\Pr(I)=0.80,\qquad \Pr(F\mid I)=0.03,\qquad \Pr(F)=0.05

Model

First form the joint mass associated with insured failures.

\Pr(I\cap F)=\Pr(I)\Pr(F\mid I)=0.80(0.03)

Compute

Condition the joint mass on the full failure population.

\Pr(I\mid F)=\frac{\Pr(I\cap F)}{\Pr(F)}

\Pr(I\mid F)=\frac{0.024}{0.05}=0.48

Answer

Forty-eight percent of the failing institutions belong to the insured group.

\boxed{0.48\quad\text{(D)}}

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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 joint mass 0.80(0.03)=0.024 rounded to 0.02; it does not condition on the fact that failure occurred.
BThis comes from dividing the conditional failure rate by the insured prior, 0.03/0.80=0.0375, which reverses the multiplication needed to form the joint event.
CThis doubles the 0.03 conditional failure rate. Bayes' theorem requires weighting by the 0.80 prior and dividing by 0.05, not an arbitrary factor of two.
EThis is 0.03/0.05=0.60. It omits the insured prior 0.80 from the Bayes numerator.

Original practice · fully worked

Original variant: route identification after an alert

A data center routes 40% of packets through channel A, 35% through channel B, and 25% through channel C. The corresponding checksum-alert probabilities are 0.02, 0.05, and 0.01. Given that a packet generated a checksum alert, calculate the probability that it used channel B.

A
0.0175
B
0.0500
C
0.3500
D
0.6863
E
0.6250

Variant answer in brief

The total alert probability is 0.40(0.02)+0.35(0.05)+0.25(0.01)=0.028. Channel B contributes 0.0175, so its posterior probability is 0.625 and choice E is correct.

Setup

Weight each channel's alert rate by its traffic share.

w_A=0.40(0.02)=0.008

w_B=0.35(0.05)=0.0175

w_C=0.25(0.01)=0.0025

Model

The three routes partition packet traffic, so their alert masses add to the denominator.

\Pr(L)=w_A+w_B+w_C=0.028

Compute

Divide channel B's joint alert mass by the total alert mass.

\Pr(B\mid L)=\frac{w_B}{\Pr(L)}

\Pr(B\mid L)=\frac{0.0175}{0.028}=0.625

Answer

The posterior channel-B share is 62.5%.

\boxed{0.6250\quad\text{(E)}}

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ActuaryProof. “SOA Exam P Sample Question #532 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Conditional Probability question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: route identification after an alert

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution