Verified solution · Exam P

SOA Exam P Sample Question #498 Solution

This Exam P sample reference tests Conditional Probability. The three prior-by-suburban weights are 0.1100, 0.1800, and 0.0525. Normalizing the under-25 weight gives 0.1100/0.3425=44/137=0.32117, so choice C is correct.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Conditional Probability question

Setup

The middle age group has the remaining prior probability after subtracting the two supplied extreme-group shares.

\Pr(M)=1-0.55-0.15=0.30

\Pr(U)=0.55,\qquad \Pr(O)=0.15

Model

Weight each group's suburban likelihood by its prior, then normalize the target group's joint weight.

\Pr(S)=\sum_{g\in\{U,M,O\}}\Pr(g)\Pr(S\mid g)

\Pr(U\mid S)=\frac{\Pr(U)\Pr(S\mid U)}{\Pr(S)}

Compute

Evaluate the three compatible joint weights and their Bayes ratio.

\Pr(S)=0.55(0.20)+0.30(0.60)+0.15(0.35)=0.3425

\Pr(U\cap S)=0.55(0.20)=0.1100

\Pr(U\mid S)=\frac{0.1100}{0.3425}=\frac{44}{137}=0.321167\ldots

Answer

The requested conditional probability rounds to 0.32.

\boxed{0.32\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.

AThis stops at the joint numerator P(U and S)=0.55(0.20)=0.11 and omits division by the total suburban probability.
BThis assumes the three age groups have equal priors and normalizes only the suburban likelihoods: 0.20/(0.20+0.60+0.35)=0.1739.
DThis repeats the prior P(U)=0.55 and ignores the information that the observed location was suburban.
EThis reverses the Bayes likelihood ratio. Using prior odds 0.55/0.45 times P(S|not U)/P(S|U) gives posterior probability 0.7595; the likelihood ratio must be P(S|U)/P(S|not U).

Original practice · fully worked

Original variant: recover a supplier share from defect evidence

A component comes from supplier A or supplier B. Components from A are defective with probability 0.08, while those from B are defective with probability 0.02. Among all defective components, half came from supplier A. Calculate the overall proportion of components supplied by A.

A
0.02
B
0.08
C
0.20
D
0.50
E
0.80

Variant answer in brief

If p is A's prior share, the two defective joint weights are 0.08p and 0.02(1-p). A posterior share of one half makes them equal, so p=0.20 and choice C is correct.

Setup

Let p denote the unknown proportion supplied by A and let D denote a defective component.

\Pr(A)=p,\qquad \Pr(B)=1-p

\Pr(D\mid A)=0.08,\qquad \Pr(D\mid B)=0.02

Model

Express the supplied posterior probability with Bayes' formula.

\Pr(A\mid D)=\frac{0.08p}{0.08p+0.02(1-p)}=\frac{1}{2}

Compute

A one-half posterior means the defective weights from the two suppliers are equal.

0.08p=0.02(1-p)

0.10p=0.02

p=0.20

Answer

Supplier A provides 20% of all components.

\boxed{0.20\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #498 — 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: recover a supplier share from defect evidence

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution