Worked solution · Exam P

SOA Exam P Sample Question #487 Solution

This Exam P sample reference tests Conditional Probability. The relevant nested-event slice has probability 0.48-0.36=0.12, while the conditioning event has probability 1-0.36=0.64. Their ratio is 0.1875, which rounds to 0.19 and selects choice B.

Independent analysis by ActuaryProof · reviewed July 31, 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 A_2 be the event of reaching and passing the second stage, and let A_3 be the event of completing all three stages successfully. The process makes A_3 a subset of A_2.

A_3\subset A_2

\Pr(A_2)=0.48,\qquad \Pr(A_3)=0.36

Model

Within the condition that full completion does not occur, passing the second stage means belonging to A_2 but not A_3.

\Pr(A_2\mid A_3^c)=\frac{\Pr(A_2\cap A_3^c)}{\Pr(A_3^c)}

\Pr(A_2\cap A_3^c)=\Pr(A_2)-\Pr(A_3)

Compute

Evaluate the numerator as the difference of nested probabilities and the denominator as a complement.

\Pr(A_2\cap A_3^c)=0.48-0.36=0.12

\Pr(A_3^c)=1-0.36=0.64

\Pr(A_2\mid A_3^c)=\frac{0.12}{0.64}=0.1875

Answer

The requested conditional probability rounds to 0.19.

\boxed{0.19\quad\text{(B)}}

Test the method on unseen problems

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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.

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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 numerator 0.48-0.36=0.12. It has not been divided by the conditioning probability 0.64.
CThis divides the same numerator by 0.48, giving 0.12/0.48=0.25. That conditions on passing the second stage instead of on not completing all three.
DThis is only the probability of the conditioning event, 1-0.36=0.64, rather than the conditional ratio.
EThis computes 0.36/0.48=0.75, the conditional probability of completing the third stage among those who passed the second.

Original practice · fully worked

Original variant: condition on holding exactly one credential

Among a group of analysts, 55% hold credential A, 40% hold credential B, and 25% hold neither credential. An analyst is selected from those who hold exactly one of the two credentials. Calculate the probability that the selected analyst holds B but not A.

A
0.2000
B
0.2500
C
0.3636
D
0.4000
E
0.5500

Answer and calculation

The union probability is 0.75, so the overlap is 0.20. This leaves 0.20 in B only and 0.55 in exactly one credential, giving 0.20/0.55=0.3636 and choice C.

Setup

Let A and B denote the two credential events. First convert the neither probability into the union probability.

\Pr(A\cup B)=1-0.25=0.75

Model

Use inclusion-exclusion to recover the overlap, then split the union into its A-only, B-only, and both parts.

\Pr(A\cap B)=0.55+0.40-0.75=0.20

\Pr(B\cap A^c)=0.40-0.20=0.20

Compute

The exactly-one event consists of the two nonoverlapping one-credential regions.

\Pr(A\cap B^c)=0.55-0.20=0.35

\Pr(\text{exactly one})=0.35+0.20=0.55

\Pr(B\cap A^c\mid\text{exactly one})=\frac{0.20}{0.55}=\frac{4}{11}=0.363636\ldots

Answer

The conditional probability rounds to 0.3636.

\boxed{0.3636\quad\text{(C)}}

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How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: condition on holding exactly one credential

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability