Worked solution · Exam P

SOA Exam P Sample Question #496 Solution

This Exam P sample reference tests Survival Functions. A unit survives through the second-year endpoint with probability 2⁻¹⁄⁴. Independence makes the three-unit no-failure probability 2⁻³⁄⁴, so the expected 1000 payment is 1000(1-2⁻³⁄⁴)=405.396 and choice C.

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 Survival Functions question

Setup

Let T_i be the failure age of unit i. The supplied distribution gives the survival probability through age 2 directly.

\Pr(T_i>2)=1-F(2)=2^{-1/4}

\Pr(T_1>2,T_2>2,T_3>2)=\left(2^{-1/4}\right)^3

Model

The payment indicator is one unless every unit avoids failure during the covered year.

I=\mathbf{1}\{\text{at least one covered failure}\}

\Pr(I=1)=1-\Pr(T_1>2,T_2>2,T_3>2)

Compute

Use independence, take the complement, and multiply by the fixed benefit.

\Pr(I=1)=1-2^{-3/4}=0.4053964425

\operatorname{E}[1000I]=1000(0.4053964425)=405.3964425

Answer

The expected payment is approximately 405.

\boxed{405\quad\text{(C)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables 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 pays only when all three units fail: with q=1-2⁻¹⁄⁴, the calculation 1000q³=4.028 rounds to 4. The contract triggers on at least one failure.
BThis considers only one specified unit, giving 1000q=159.104. There are three independent opportunities for the covered event.
DThis multiplies 1000 by the probability of no covered failure, 2⁻³⁄⁴=0.594604, rather than by its complement.
EThis uses 1-q³=0.995972, the complement of all three units failing. The needed complement is of zero failures, not of three failures.

Original practice · fully worked

Original variant: infer a batch-discard probability

Four sensors are calibrated independently and have a common probability p of producing an acceptable reading. The probability that all four readings are acceptable is 0.4096. A batch is discarded when at least two readings are unacceptable. Calculate the discard probability.

A
0.0016
B
0.1808
C
0.4096
D
0.5904
E
0.8192

Answer and calculation

The all-acceptable probability gives p⁴=0.4096 and hence p=0.80. For K, the number unacceptable, P(K≥2)=1-P(K=0)-P(K=1)=1-0.4096-0.4096=0.1808, so choice B.

Setup

Recover the one-sensor acceptable probability from the four-sensor joint probability.

p^4=0.4096

p=0.80,\qquad 1-p=0.20

Model

Let K count unacceptable readings. The discard event is K at least two, whose complement contains K=0 and K=1.

K\sim\operatorname{Binomial}(4,0.20)

\Pr(K\ge 2)=1-\Pr(K=0)-\Pr(K=1)

Compute

Evaluate the two complement terms using p=0.80.

\Pr(K=0)=0.8^4=0.4096

\Pr(K=1)=\binom{4}{1}(0.2)(0.8)^3=0.4096

\Pr(K\ge2)=1-0.4096-0.4096=0.1808

Answer

The batch-discard probability is 0.1808.

\boxed{0.1808\quad\text{(B)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Survival Functions question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: infer a batch-discard probability

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables