Worked solution · Exam P

SOA Exam P Sample Question #506 Solution

This Exam P sample reference tests Conditional Probability. The observed mixed pair has probability 3/8 under replacement and 1/2 without replacement. Equal prior odds and Bayes' rule give a posterior probability of 3/7, so choice C is correct.

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 B and A denote the two possible selectors, and let M denote observing one assignment of each type. Alternating workdays give equal prior probabilities.

\Pr(B)=\Pr(A)=\frac12

Model

Calculate the mixed-pair likelihood under each selection protocol, allowing either order.

\Pr(M\mid B)=2\left(\frac14\right)\left(\frac34\right)=\frac38

\Pr(M\mid A)=\frac14\frac33+\frac34\frac13=\frac12

Compute

Apply Bayes' rule with the two equal prior weights.

\Pr(B\mid M)=\frac{(1/2)(3/8)}{(1/2)(3/8)+(1/2)(1/2)}

\Pr(B\mid M)=\frac{3/8}{3/8+1/2}=\frac37

Answer

The posterior probability that the replacement protocol generated the pair is 3/7.

\boxed{\frac37\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.

AThis computes only the joint numerator P(B and M)=(1/2)(3/8)=3/16 and omits division by P(M).
BThis reports the likelihood P(M | B)=3/8, reversing the desired conditional probability.
DThis reports the marginal evidence probability P(M)=(1/2)(3/8)+(1/2)(1/2)=7/16.
EThis leaves the equal prior P(B)=1/2 unchanged and therefore ignores the different likelihoods 3/8 and 1/2.

Original practice · fully worked

Original variant: archived anomaly attribution

A reliability archive contains 120 reports from production line A and 80 reports from line B. Reviewers marked 30 line-A reports and 40 line-B reports for escalation. One marked report is selected uniformly for a root-cause drill. Calculate the probability that it came from line B.

A
1/4
B
7/20
C
2/5
D
3/7
E
4/7

Answer and calculation

Among the 70 marked reports, 40 came from line B. The requested probability is therefore 40/70=4/7, so choice E is correct.

Setup

Restrict the archive to reports eligible for the root-cause selection.

n_{\mathrm{marked}}=30+40=70

n_{B,\mathrm{marked}}=40

Model

Uniform selection from the marked subset makes the posterior source probability a ratio of counts within that subset.

\Pr(B\mid\mathrm{marked})=\frac{n_{B,\mathrm{marked}}}{n_{\mathrm{marked}}}

Compute

Insert the two marked-report counts and reduce the fraction.

\Pr(B\mid\mathrm{marked})=\frac{40}{30+40}

\Pr(B\mid\mathrm{marked})=\frac47=0.571428\ldots

Answer

The probability that the selected marked report came from line B is 4/7.

\boxed{\frac47\quad\text{(E)}}

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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: archived anomaly attribution

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability