Verified solution · Exam P

SOA Exam P Sample Question #313 Solution

This Exam P sample reference tests Independence. Eight annual updates separate Year 1 from Year 9. Independence lets the two marginal multipliers combine, so the Year 1 joint probability is multiplied by (1.25 × 0.75) to the eighth power, giving 0.005967 and choice B.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Independence question

Setup

Count the changes between the reference year and the requested year.

9-1=8\quad\text{annual updates}

Model

Under independence, the joint probability is the product of the two marginal probabilities. Therefore their annual relative changes multiply.

m=(1+0.25)(1-0.25)=1.25(0.75)=0.9375

Compute

Compound the joint multiplier for eight updates from the supplied Year 1 joint probability.

p_9=0.01(0.9375)^8

=0.005967194738\ldots

Answer

The requested joint probability rounds to 0.006.

\boxed{0.006\quad\text{(B)}}

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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.

AUsing ten annual updates instead of eight gives 0.01(0.9375)¹⁰=0.005245, which rounds to this choice.
CThis leaves the Year 1 joint probability unchanged, as if a 25% rise and a 25% fall canceled additively. Their multiplicative net factor is 0.9375, not 1.
DTreating the decline factor as 1/1.20 instead of 0.75 gives 0.01(1.25/1.20)⁸=0.01387, approximately this choice.
EThis results from increasing 0.01 once by both 25% magnitudes, 0.01(1+0.25+0.25)=0.015. The liability change has the opposite direction and all eight updates must be compounded.

Original practice · fully worked

Original variant: acceptance after odds adjustments

Two screening programs make independent approval decisions. Initially, program R approves with probability 0.72 and program S approves with probability 0.65. An update doubles R's approval odds and halves S's approval odds. A record is accepted when exactly one program approves it. Calculate the acceptance probability after the update.

A
0.403
B
0.481
C
0.512
D
0.837
E
0.916

Variant answer in brief

The updated approval probabilities are 36/43 and 13/27. Independence gives an exactly-one probability of 595/1161, or approximately 0.512, so choice C.

Setup

Convert each initial probability to odds and apply the stated odds multiplier.

O_R=\frac{0.72}{0.28}=\frac{18}{7},\qquad O_R'=\frac{36}{7}

O_S=\frac{0.65}{0.35}=\frac{13}{7},\qquad O_S'=\frac{13}{14}

Model

Convert the adjusted odds back to probabilities.

p_R=\frac{36/7}{1+36/7}=\frac{36}{43}

p_S=\frac{13/14}{1+13/14}=\frac{13}{27}

Compute

The two disjoint acceptance cases are R-only and S-only.

\Pr(\text{exactly one})=p_R(1-p_S)+(1-p_R)p_S

=\frac{36}{43}\frac{14}{27}+\frac{7}{43}\frac{13}{27}=\frac{595}{1161}=0.512489\ldots

Answer

The acceptance probability rounds to 0.512.

\boxed{0.512\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #313 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Independence question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: acceptance after odds adjustments

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution