Worked solution · Exam P

SOA Exam P Sample Question #306 Solution

This Exam P sample reference tests Bayes' Theorem. This is Bayes' theorem with category-specific exponential survival likelihoods. The two prior-weighted survivor masses are 0.4 exp(−0.25) and 0.6 exp(−0.5); normalizing the first mass gives 0.4612, so choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Bayes' Theorem question

Setup

Let H denote the home category, A the apartment category, and S survival beyond one year.

\Pr(H)=0.4,\qquad \Pr(A)=0.6

Model

For an exponential lifetime, survival through time t is exp(-t divided by the mean).

\Pr(S\mid H)=e^{-1/4},\qquad \Pr(S\mid A)=e^{-1/2}

Compute

Weight each survival likelihood by its category prior and normalize the home contribution.

\Pr(H\mid S)=\frac{0.4e^{-1/4}}{0.4e^{-1/4}+0.6e^{-1/2}}

=\frac{0.3115203132}{0.6754387090}=0.4612118154

Answer

The conditional home-category probability rounds to 0.46.

\boxed{0.46\quad\text{(C)}}

Test the method on unseen problems

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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 0.08 results if the stated exponential means 4 and 2 are misread as rate parameters: normalizing 0.4 exp(−4) and 0.6 exp(−2) gives about 0.083.
BThe value 0.27 is the posterior home share among policies canceled during the first year, using 0.4(1-exp(−0.25)) over the two cancellation contributions. It conditions on the opposite event.
DThe value 0.56 is approximately exp(−0.25)/[exp(−0.25)+exp(−0.5)]. It normalizes the two survival likelihoods without the category priors.
EThe value 0.66 is approximately 4/(4+2). It treats the two mean lifetimes as posterior weights instead of using exponential survival likelihoods and the category priors.

Original practice · fully worked

Original variant: device type after a second-hour failure

A device is type A with probability 0.37 and type B with probability 0.63. Conditional on type, its lifetime is exponential. The respective probabilities of surviving one hour are 0.50 and 0.80. A device is observed to fail after hour one but no later than hour two. Calculate the conditional probability that it is type A.

A
0.1600
B
0.2500
C
0.3700
D
0.4785
E
0.5215

Answer and calculation

Exponential survival implies S(2)=S(1)², so the second-hour failure likelihoods are 0.50-0.25=0.25 for A and 0.80-0.64=0.16 for B. The prior-weighted masses are 0.0925 and 0.1008, giving posterior probability 0.478531 and choice D.

Setup

Let F be the event that failure occurs during the second hour. Exponential survival over two hours is the square of one-hour survival.

S_A(2)=0.50^2=0.25,\qquad S_B(2)=0.80^2=0.64

Model

Subtract two-hour survival from one-hour survival to obtain each interval likelihood.

\Pr(F\mid A)=0.50-0.25=0.25

\Pr(F\mid B)=0.80-0.64=0.16

Compute

Apply Bayes' theorem to the two prior-weighted interval probabilities.

\Pr(A\mid F)=\frac{0.37(0.25)}{0.37(0.25)+0.63(0.16)}

=\frac{0.0925}{0.1933}=\frac{925}{1933}=0.4785307812

Answer

The posterior probability of type A rounds to 0.4785.

\boxed{0.4785\quad\text{(D)}}

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How to solve this Bayes' Theorem question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: device type after a second-hour failure

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P General Probability