Worked solution · Exam P

SOA Exam P Sample Question #559 Solution

This Exam P sample reference tests Exponential Distribution. Exponential cumulative hazard scales with time divided by mean. The second lifetime's five-year hazard is two-thirds of the first lifetime's fifteen-year hazard, giving 0.046656⁽²⁄³⁾=0.1296 and choice D.

Independent analysis by ActuaryProof · reviewed July 30, 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 Exponential Distribution question

Setup

Let beta be the longer mean. The shorter mean is beta divided by two.

E[X_A]=\beta,\qquad E[X_B]=\frac{\beta}{2}

Model

Write the supplied survival probability under the exponential model.

e^{-15/\beta}=0.046656

Compute

Express the required exponent as two-thirds of the known exponent.

\Pr(X_B>5)=e^{-5/(\beta/2)}=e^{-10/\beta}

e^{-10/\beta}=\left(e^{-15/\beta}\right)^{2/3}

\Pr(X_B>5)=0.046656^{2/3}=0.1296

Answer

The required survival probability rounds to 0.13.

\boxed{0.13\quad\text{(D)}}

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 reverses the effective-exposure ratio and computes 0.046656⁽³⁄²⁾=0.01008, which rounds to 0.01.
BThis linearly multiplies the given survival probability by 2/3, obtaining 0.031104, rather than scaling its cumulative hazard.
CThis linearly multiplies 0.046656 by 3/2 to obtain 0.069984, instead of exponentiating by the required exposure ratio.
EA value of 0.18 implies an exponent ln(0.18)/ln(0.046656)=0.559, not the required effective-exposure ratio (5/15)(2)=2/3.

Original practice · fully worked

Original variant: first-alert operating time

Two independent safety sensors trigger after exponentially distributed waiting times with means 8 hours and 12 hours. A system remains in operation only until the first sensor triggers. Calculate the probability that the system operates for more than 4 hours.

A
0.4346
B
0.5654
C
0.6065
D
0.6615
E
0.7165

Answer and calculation

Operating beyond four hours requires both independent sensors to wait beyond four hours. Multiplying their survival probabilities gives exp(-4/8-4/12)=0.4346, so choice A is correct.

Setup

Let T be the earlier of the two trigger times.

T=\min(T_1,T_2)

Model

The system exceeds four hours exactly when both independent trigger times exceed four hours.

\Pr(T>4)=\Pr(T_1>4)\Pr(T_2>4)

Compute

Multiply the two exponential survival probabilities.

\Pr(T>4)=e^{-4/8}e^{-4/12}

\Pr(T>4)=e^{-5/6}=0.4345982085

Answer

The probability of operating beyond four hours is approximately 0.4346.

\boxed{0.4346\quad\text{(A)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: first-alert operating time

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables