Worked solution · Exam P

SOA Exam P Sample Question #132 Solution

This Exam P sample reference tests Conditional Probability. The under-25 group has size 505. Its female portion is 0.57(414)=235.98, leaving 269.02 males, so the requested conditional probability is 269.02/505=0.5327, choice B.

Independent analysis by ActuaryProof · reviewed October 6, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

Independent solution

How to solve this Conditional Probability question

Setup

Split the portfolio by sex and by whether age exceeds the cutoff.

N_M=0.54(900)=486,\qquad N_F=414

N_{\le25}=900-395=505

Model

Within the female group, the probability of being at or below the cutoff is the complement of the supplied older-age probability.

N_{F,\le25}=414(1-0.43)=235.98

Compute

Subtract the female portion from the full under-25 group and condition on that group.

N_{M,\le25}=505-235.98=269.02

\Pr(M\mid \le25)=\frac{269.02}{505}=0.5327129

Answer

The conditional male proportion rounds to 0.53.

\boxed{0.53\quad\text{(B)}}

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.

See the cover and sample pages

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 0.47 is the female share within the under-25 group, approximately 235.98/505. It is the complement of the requested conditional probability.
CThe value 0.54 is the unconditional male proportion in the entire portfolio. Conditioning on age changes that proportion.
DThe value 0.55 is approximately 269.02/486, which is the under-25 proportion among males, P(under 25|male), with the conditioning direction reversed.
EThe value 0.56 is the overall under-25 proportion, 505/900. It does not identify sex within that age group.

Original practice · fully worked

Original variant: supplier share among accepted components

A component inventory comes 40% from supplier A, 35% from supplier B, and 25% from supplier C. Defect rates are 2% for A and 4% for B, while the inventory-wide defect rate is 3.25%. A nondefective component is selected. Determine the probability that it came from supplier C.

A
0.2200
B
0.2395
C
0.2475
D
0.2500
E
0.9580

Answer and calculation

The mixture defect rate implies supplier C has defect rate 0.042. Its nondefective mass is 0.25(0.958)=0.2395, and dividing by the total nondefective mass 0.9675 gives 0.2475, choice C.

Setup

Let q be the unknown defect probability for supplier C.

0.0325=0.40(0.02)+0.35(0.04)+0.25q

Model

Solve the mixture equation and complement the resulting supplier-specific defect rate.

q=0.042,\qquad \Pr(N\mid C)=1-q=0.958

Compute

Apply Bayes' rule to the nondefective selection.

\Pr(C\cap N)=0.25(0.958)=0.2395

\Pr(N)=1-0.0325=0.9675

\Pr(C\mid N)=\frac{0.2395}{0.9675}=0.2475452

Answer

About 24.75% of accepted components come from supplier C.

\boxed{0.2475\quad\text{(C)}}

Keep practicing

Next steps for Exam P General Probability

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P General Probability practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: supplier share among accepted components

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability