Worked solution · Exam P

SOA Exam P Sample Question #121 Solution

This Exam P sample reference tests Mixture Probability. Conditioning on the selected die gives squared six-probabilities 1/36, 1/36, 1/9, and 1. Their equally weighted average is 7/24=0.292, choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Mixture Probability question

Setup

Condition on which die is selected. Given a die, the two rolls are independent with that die's stated six-probability.

p_1=p_2=\frac16,\quad p_3=\frac13,\quad p_4=1

Model

For each die, square its one-roll six-probability to obtain the conditional probability of two sixes. Then average the four conditional probabilities because the die is selected uniformly.

\Pr(\text{two sixes})=\frac14\sum_{j=1}^4p_j^2

Compute

The four conditional probabilities sum to seven sixths. Dividing by four gives seven twenty-fourths, or approximately 0.292.

\Pr(\text{two sixes})=\frac14\left(\frac1{36}+\frac1{36}+\frac19+1\right)=\frac7{24}

Answer

The probability of two sixes is about 0.292.

\boxed{0.292\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.

BThe value 0.250 is only the contribution of selecting the die that always rolls a six, before adding the other three dice's positive contributions.
EThe value 0.417 is the average probability of a six on one roll. It does not square the die-specific probabilities for two rolls.

Original practice · fully worked

Original variant: repeated alerts from one selected sensor

One of three sensors is selected uniformly and used for two independent trials. The sensors have per-trial alert probabilities 0.20, 0.50, and 0.90. Find the probability that both trials produce an alert.

A
0.3000
B
0.3333
C
0.3667
D
0.5333
E
0.6000

Answer and calculation

Conditioning on the sensor gives an average of squared alert probabilities: (0.04+0.25+0.81)/3=0.3667, choice C.

Setup

Condition on the selected sensor. Once a sensor is selected, the two trials are independent with the same sensor-specific alert probability.

p=(0.20,0.50,0.90)

Model

Square each sensor's alert probability to obtain its conditional two-alert probability, then average because the three sensors are equally likely.

\Pr(\text{two alerts})=\frac13(0.20^2+0.50^2+0.90^2)

Compute

The squared probabilities sum to 1.10. Dividing by three gives approximately 0.3667.

\Pr(\text{two alerts})=\frac{1.10}{3}=0.366667

Answer

Both trials alert with probability about 0.3667.

\boxed{0.3667\quad\text{(C)}}

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How to solve this Mixture Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: repeated alerts from one selected sensor

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability