Worked solution · Exam P

SOA Exam P Sample Question #329 Solution

This Exam P sample reference tests Bayes' Theorem. Prior-weighting the three conditional rates gives total left-handed probability 221/2400. The no-left-handed-parent contribution is 117/2400, so the requested posterior is 117/221=9/17, approximately 0.529 and 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 Bayes' Theorem question

Setup

Complete the three-category prior distribution before conditioning on the child's handedness.

p_2=\frac1{50},\qquad p_1=\frac15,\qquad p_0=1-p_1-p_2=\frac{39}{50}

Model

Multiply each family-category prior by its corresponding conditional left-handed rate.

w_0=\frac{39}{50}\frac1{16}=\frac{117}{2400},\qquad w_1=\frac15\frac16=\frac{80}{2400},\qquad w_2=\frac1{50}\frac12=\frac{24}{2400}

Compute

Normalize the no-left-handed-parent joint weight by the total weight of all left-handed children.

\Pr(0\mid L)=\frac{w_0}{w_0+w_1+w_2}=\frac{117}{117+80+24}=\frac{117}{221}=\frac9{17}

Answer

The posterior probability is about 0.53.

\boxed{0.5294118\ldots\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 0.09 is the marginal probability that a child is left-handed: 221/2400=0.09208. That is the Bayes denominator, not the posterior share of one family category.
BWith the correct denominator 221/2400, a posterior of 0.42 would require a no-parent joint weight of only 92.82/2400. The stated prior and conditional rate instead give 117/2400, so this choice underweights the target category.
DThe value 0.78 is the prior probability 39/50 of having no left-handed parents. Observing that the child is left-handed changes this category probability.
EThe value 0.91 is the complement of the marginal left-handed probability: 1-221/2400=0.90792. It concerns whether the child is left-handed, not the requested posterior family category.

Original practice · fully worked

Original variant: update from disagreeing message filters

One quarter of incoming messages are malicious and the rest are benign. Two filters operate independently after the message type is fixed. Filter 1 alerts on 80% of malicious and 20% of benign messages; Filter 2 alerts on 50% of malicious and 10% of benign messages. A message triggered Filter 1 but not Filter 2. Calculate the probability that it was malicious.

A
0.1000
B
0.1800
C
0.2500
D
0.4000
E
0.4255

Answer and calculation

Conditional independence makes the observed alert pattern have likelihood 0.80(0.50)=0.40 for malicious messages and 0.20(0.90)=0.18 for benign messages. Prior-weighting gives masses 0.10 and 0.135, so the malicious posterior is 0.10/0.235=20/47, approximately 0.4255 and choice E.

Setup

Let R denote the mixed result in which the first filter alerts and the second does not.

\Pr(M)=0.25,\qquad \Pr(B)=0.75

Model

Multiply the two filter factors within each message type using their stated conditional independence.

\Pr(R\mid M)=0.80(1-0.50)=0.40

\Pr(R\mid B)=0.20(1-0.10)=0.18

Compute

Weight the two likelihoods by their priors and normalize the malicious route.

\Pr(M\cap R)=0.25(0.40)=0.10

\Pr(B\cap R)=0.75(0.18)=0.135

\Pr(M\mid R)=\frac{0.10}{0.10+0.135}=\frac{20}{47}=0.4255319\ldots

Answer

The posterior malicious probability rounds to 0.4255.

\boxed{0.4255\quad\text{(E)}}

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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: update from disagreeing message filters

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P General Probability