Worked solution · Exam P

SOA Exam P Sample Question #94 Solution

This Exam P sample reference tests Conditional Expectation. Given the second roll is the first 6, the first roll is uniformly one of the five non-6 faces. A first-roll 5 occurs with probability 0.2; otherwise memorylessness gives mean 8, so the result is 6.6, choice D.

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.

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Independent solution

How to solve this Conditional Expectation question

Setup

Given that the second roll is the first 6, the first roll is uniformly distributed over the five non-6 faces.

P(X=1\mid Y=2)=\frac15

P(X\ge3\mid Y=2)=\frac45

Model

With conditional probability 1/5 the first roll is already 5. Otherwise no 5 has appeared by roll 2, and the geometric waiting time resumes after that point.

E[X\mid X\ge3]=2+\frac1{1/6}=8

Compute

The residual-case conditional mean is 8. Weighting the immediate and residual cases gives 0.2(1)+0.8(8)=6.6.

E[X\mid Y=2]=\frac15(1)+\frac45(8)=6.6

Answer

The conditional expected waiting time is 6.6 rolls, corresponding to choice D.

\boxed{6.6\quad\text{(D)}}

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.

CThe value 6.0 is the unconditional geometric mean waiting time for a 5. It ignores the information about when the first 6 appears.

Original practice · fully worked

Original variant: waiting for red given blue first appears on draw three

A spinner independently lands red with probability 0.20, blue with probability 0.20, and another color otherwise. Let X be the draw of the first red and Y the draw of the first blue. Find E[X | Y=3].

A
4.600
B
5.000
E
5.125
D
5.800
C
6.200

Answer and calculation

Conditional on no blue in the first two draws, each is red with probability 0.25. Enumerating first red on draw 1, draw 2, or after draw 3 yields the conditional mean.

Setup

Conditioning on no blue in the first two draws changes the red probability on each of those draws to 0.2/0.8=0.25.

P(R\mid\text{not }B)=\frac{0.2}{0.8}=0.25

Model

Partition on first red at draw 1, at draw 2, or after the forced blue at draw 3.

P(X=1\mid Y=3)=0.25

P(X=2\mid Y=3)=0.75(0.25)

P(X\ge4\mid Y=3)=0.75^2

Compute

The three case probabilities are 0.25, 0.1875, and 0.5625. Their conditional waiting-time means combine to 5.125.

E[X\mid Y=3]=0.25(1)+0.1875(2)+0.5625\left(3+\frac1{0.2}\right)=5.125

Answer

Therefore E[X | Y=3]=5.125, corresponding to choice E.

\boxed{5.125\quad\text{(E)}}

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How to solve this Conditional Expectation question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: waiting for red given blue first appears on draw three

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability