Worked solution · Exam P

SOA Exam P Sample Question #150 Solution

This Exam P sample reference tests Conditional Probability. Only family sizes five, six, and seven can produce five riders. Weighting those sizes by their five-rider likelihoods gives posterior probability 35/113 = 0.30973 for size six, choice E.

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 M denote family size and R the number of riders. A five-rider observation restricts M to three possible values.

\Pr(M=m)=\frac{8-m}{28}

\Pr(R=5\mid M=m)=\frac1m,\qquad m\in\{5,6,7\}

Model

Form the three joint weights and normalize the size-six weight.

w_5=\frac3{28}\frac15=\frac3{140},\quad w_6=\frac2{28}\frac16=\frac1{84},\quad w_7=\frac1{28}\frac17=\frac1{196}

\Pr(M=6\mid R=5)=\frac{w_6}{w_5+w_6+w_7}

Compute

Put the weights over a common denominator and simplify the posterior ratio.

w_5+w_6+w_7=\frac{63+35+15}{2940}=\frac{113}{2940}

\Pr(M=6\mid R=5)=\frac{35/2940}{113/2940}=\frac{35}{113}=0.3097345

Answer

The requested conditional probability rounds to 0.31.

\boxed{0.31\quad\text{(E)}}

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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.

AA posterior of 0.17 would imply a total five-rider weight of (1/84)/0.17 = 0.07003, almost twice the correctly enumerated weight 113/2940.
BThe value 0.21 does not normalize the joint size-six weight by all three feasible sizes. The denominator must include the size-five and size-seven contributions.
CThe value 0.24 can arise from mixing unconditioned family-size weights with rider likelihoods. Bayes weighting requires multiplying each feasible prior by its own factor 1/m.
DThe value 0.28 understates the posterior because it uses an incomplete normalizing sum. The exact three-term denominator gives 35/113, not 0.28.

Original practice · fully worked

Original variant: identify a delayed parcel’s route

A parcel network sends 50% of parcels on the North route, 30% on the South route, and 20% on the East route. Their respective delay probabilities are 0.04, 0.10, and 0.20. A randomly selected parcel was delayed. Find the probability that it traveled on the East route.

A
0.20000
B
0.22222
C
0.33333
D
0.44444
E
0.55556

Answer and calculation

The three route contributions to delay are 0.02, 0.03, and 0.04. Dividing the East contribution by their sum gives 4/9 = 0.44444, choice D.

Setup

Multiply each route share by that route’s delay probability.

w_N=(0.50)(0.04)=0.02

w_S=(0.30)(0.10)=0.03

w_E=(0.20)(0.20)=0.04

Model

The total delay probability is the sum of the mutually exclusive route contributions.

\Pr(D)=w_N+w_S+w_E=0.09

Compute

Condition on delay by normalizing the East contribution.

\Pr(E\mid D)=\frac{w_E}{\Pr(D)}=\frac{0.04}{0.09}=\frac49=0.4444444

Answer

A delayed parcel has probability 0.44444 of having used the East route.

\boxed{0.44444\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 delayed parcel’s route

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability