Verified solution · Exam P

SOA Exam P Sample Question #178 Solution

This Exam P sample reference tests Conditional Expectation. The joint expectation of the advanced count times the early-stage condition indicator is 6(0.1)(1-0.8⁵)=0.403392. Dividing by P(at least one early)=1-0.8⁶ gives 0.546708, choice C.

Independent analysis by ActuaryProof · reviewed August 4, 2026

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Independent solution

How to solve this Conditional Expectation question

Setup

Let X count early-stage selections, Y count advanced-stage selections, and A be the event X≥1.

\Pr(A)=1-(0.8)^6=0.737856

Model

For each patient to contribute to Y while A occurs, that patient must be advanced and at least one of the other five must be early-stage.

E[Y\mathbf{1}_A]=6(0.1)\left[1-(0.8)^5\right]

Compute

Evaluate the joint expectation and normalize by the conditioning probability.

E[Y\mathbf{1}_A]=0.403392

E[Y\mid A]=\frac{0.403392}{0.737856}=0.5467083

Answer

The conditional expected advanced-stage count rounds to 0.547.

\boxed{0.547\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.

AThe value 0.403 is the unnormalized joint expectation E[Y 1_A]. It must be divided by P(A)=0.737856.
BThe value 0.500 treats the condition as exactly one fixed early-stage patient, leaving five patients each with unconditional advanced probability 0.10.
DThe value 0.600 is the unconditional expectation 6(0.10), ignoring the information about the early-stage count.
EThe value 0.625 is 5(0.10/0.80), combining an exactly-one-early assumption with an inappropriate re-normalization of each remaining patient.

Original practice · fully worked

Original variant: conditional count from Poisson splitting

Independent event counts A and B are Poisson with means 2 and 3, respectively. Given that their combined count equals 4, calculate the conditional expected value of A.

A
1.2
B
1.6
C
2.0
D
2.4
E
4.0

Variant answer in brief

Conditioned on a combined count of four, the type-A count is Binomial(4,2/5). Its expectation is 4(2/5)=1.6, choice B.

Setup

Combine the two independent Poisson streams.

A+B\sim\operatorname{Poisson}(2+3)

Model

Given the total, each event is classified as type A with probability proportional to its rate.

A\mid(A+B=4)\sim\operatorname{Binomial}\left(4,\frac{2}{5}\right)

Compute

Apply the binomial mean formula.

E[A\mid A+B=4]=4\left(\frac25\right)=\frac85=1.6

Answer

The conditional expected type-A count is 1.6.

\boxed{1.6\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #178 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Conditional Expectation question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: conditional count from Poisson splitting

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution