Worked solution · Exam P

SOA Exam P Sample Question #423 Solution

This Exam P sample reference tests Conditional Probability. Subtracting the common mass from the overstatement marginal leaves 0.05 in the lower-claim group. Dividing by that group's probability 0.30 gives 1/6, so 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 Conditional Probability question

Setup

Let H denote the higher claimed amount and O denote an overstated amount. Record the supplied marginals and intersection.

\Pr(H)=0.70,\qquad \Pr(O)=0.50,\qquad \Pr(H\cap O)=0.45

Model

Remove the overlap with H from the overstatement marginal, and take the complement of H for the conditioning event.

\Pr(O\cap H^c)=\Pr(O)-\Pr(O\cap H)=0.50-0.45=0.05

\Pr(H^c)=1-0.70=0.30

Compute

Normalize the joint mass by the probability of the conditioning group.

\Pr(O\mid H^c)=\frac{\Pr(O\cap H^c)}{\Pr(H^c)}

\Pr(O\mid H^c)=\frac{0.05}{0.30}=\frac16

Answer

The required conditional probability is one sixth.

\boxed{\frac16\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.

AThe value 1/20=0.05 is the joint mass Pr(O and Hᶜ). It must still be divided by Pr(Hᶜ)=0.30.
BThe calculation 0.05/0.50=0.10 gives Pr(Hᶜ|O), reversing the conditioning direction.
DThe inclusion-exclusion calculation 0.70+0.50-0.45=0.75 gives Pr(H union O). It is an unconditional union probability, not Pr(O given Hᶜ).
EThe value 5/6=0.25/0.30 is Pr(Oᶜ|Hᶜ), the complement of the requested conditional event.

Original practice · fully worked

Original variant: fragile-package scanner

Eight percent of packages in a shipment contain a fragile component. A scanner flags 90% of packages containing a fragile component and 12% of packages without one. Given that a package is flagged, calculate the probability that it contains a fragile component.

A
0.0720
B
0.1200
C
0.3947
D
0.6053
E
0.9000

Answer and calculation

A fragile-and-flagged package has probability 0.08(0.90)=0.072, while the total flag probability is 0.1824. Bayes' rule gives 0.072/0.1824=0.3947 and choice C.

Setup

Let F denote a fragile component and G denote a scanner flag.

\Pr(F)=0.08,\qquad \Pr(G\mid F)=0.90,\qquad \Pr(G\mid F^c)=0.12

Model

Combine the two disjoint ways a package can be flagged.

\Pr(G)=0.08(0.90)+0.92(0.12)=0.1824

Compute

Apply Bayes' rule to the fragile-and-flagged joint probability.

\Pr(F\mid G)=\frac{0.08(0.90)}{0.1824}

\Pr(F\mid G)=0.3947368421\ldots

Answer

About 39.47% of flagged packages contain a fragile component.

\boxed{0.3947\quad\text{(C)}}

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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: fragile-package scanner

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability