Worked solution · Exam P

SOA Exam P Sample Question #455 Solution

This Exam P sample reference tests Conditional Probability. This is a Bayesian update using the zero-count likelihood of each Poisson class. Normalizing the two prior-weighted likelihoods gives 0.749753, which rounds to 0.750 and choice D.

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 Conditional Probability question

Setup

Let L and H identify the two risk classes, and let Z be the observed zero-count event. A Poisson variable with mean λ is zero with probability exp(-λ).

\Pr(Z\mid L)=e^{-0.05}

\Pr(Z\mid H)=e^{-0.30}

Model

Multiply each zero-count likelihood by its class prior and normalize the low-risk weight.

\Pr(L\mid Z)=\frac{0.70e^{-0.05}}{0.70e^{-0.05}+0.30e^{-0.30}}

Compute

Evaluate the two joint weights and their ratio.

w_L=0.70e^{-0.05}=0.665860598

w_H=0.30e^{-0.30}=0.222245466

\Pr(L\mid Z)=\frac{0.665860598}{0.888106064}=0.749753464

Answer

The posterior low-risk probability rounds to 0.750.

\boxed{0.750\quad\text{(D)}}

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.

AUsing each Poisson mean itself as its zero-count likelihood gives 0.70(0.05)/[0.70(0.05)+0.30(0.30)]=0.280. The zero probability is exponential.
BA posterior of 0.666 has odds about 1.99, below the prior odds 7/3. That moves probability away from the lower-rate class even though observing zero favors that class.
CThe value 0.700 is the prior probability of the low-risk class and ignores the information supplied by the observed count.
EReplacing exp(−λ) by the coarse linear approximation 1-λ gives 0.70(0.95)/[0.70(0.95)+0.30(0.70)]=0.760.

Original practice · fully worked

Original variant: identify a kit's workshop from pass results

Sealed three-component kits come from workshops A, B, and C in proportions 0.50, 0.30, and 0.20. Within a kit, components independently pass inspection with probabilities 0.80, 0.50, and 0.20 for the respective workshops. A selected kit has exactly two passing components. Calculate the probability it came from workshop A.

A
0.192
B
0.384
C
0.500
D
0.593
E
0.631

Answer and calculation

The three prior-weighted exactly-two likelihoods are 0.1920, 0.1125, and 0.0192. Workshop A's normalized share is 0.1920/0.3237=0.593142, so choice D.

Setup

Let E denote exactly two passes. Compute its binomial likelihood under each possible workshop.

\Pr(E\mid A)=\binom32(0.80)^2(0.20)=0.384

\Pr(E\mid B)=\binom32(0.50)^2(0.50)=0.375

\Pr(E\mid C)=\binom32(0.20)^2(0.80)=0.096

Model

Weight each likelihood by the corresponding production share.

w_A=0.50(0.384)=0.1920

w_B=0.30(0.375)=0.1125

w_C=0.20(0.096)=0.0192

Compute

Normalize the workshop-A weight over all three ways the evidence can arise.

\Pr(A\mid E)=\frac{0.1920}{0.1920+0.1125+0.0192}

\Pr(A\mid E)=\frac{0.1920}{0.3237}=0.593141798

Answer

The posterior probability of workshop A is approximately 0.593.

\boxed{0.593\quad\text{(D)}}

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How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: identify a kit's workshop from pass results

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P General Probability