Worked solution · Exam P

SOA Exam P Sample Question #641 Solution

This Exam P sample reference tests Conditional Probability. List the three mutually exclusive ways that exactly one loss category can occur. The target category alone has probability 0.072 and all exactly-one cases total 0.306, so the conditional probability is 0.235294 and choice B is correct.

Independent analysis by ActuaryProof · reviewed July 30, 2026

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Independent solution

How to solve this Conditional Probability question

Setup

Write F, W, and T for occurrence of a loss in the first, second, and third categories. Their loss probabilities are the complements of the given no-loss probabilities.

\Pr(F)=0.10,\qquad \Pr(W)=0.10,\qquad \Pr(T)=0.20

Model

Under independence, multiply across categories for each of the three disjoint exactly-one cases.

\Pr(F\cap W^c\cap T^c)=0.10(0.90)(0.80)=0.072

\Pr(F^c\cap W\cap T^c)=0.90(0.10)(0.80)=0.072

\Pr(F^c\cap W^c\cap T)=0.90(0.90)(0.20)=0.162

Compute

Add the exactly-one cases for the conditioning probability, then divide the target case by that total.

\Pr(\text{exactly one category})=0.072+0.072+0.162=0.306

\Pr(F\mid\text{exactly one category})=\frac{0.072}{0.306}=0.2352941176\ldots

Answer

The conditional probability rounds to 0.235.

\boxed{0.235\quad\text{(B)}}

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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 is the joint probability that the target category is the sole loss category, before division by the exactly-one conditioning probability.
CNormalizing the three marginal loss probabilities 0.10, 0.10, and 0.20 gives 0.10/0.40=0.25, but the exactly-one cases require the accompanying no-loss factors.
DThis is the probability of exactly one loss category, which belongs in the denominator of the conditional probability.
ENormalizing the no-loss probabilities gives 0.90/(0.90+0.90+0.80)=0.346; the conditioning event concerns loss occurrences, not no-loss weights.

Original practice · fully worked

Original variant: greenhouse alert diagnosis

A greenhouse controller monitors four independent alert channels: moisture, frost, pests, and backup power. Their probabilities of triggering during a shift are 0.20, 0.10, 0.05, and 0.02, respectively. The shift log shows that exactly one channel triggered. Calculate the conditional probability that it was the moisture channel.

A
0.1676
B
0.2000
C
0.2910
D
0.5405
E
0.5758

Answer and calculation

The moisture-only probability is 0.16758, and the four mutually exclusive one-alert probabilities total 0.29102. Their ratio is 0.575837, so the correct choice is E.

Setup

Denote the four trigger probabilities by 0.20, 0.10, 0.05, and 0.02 in the order listed.

(p_M,p_F,p_P,p_B)=(0.20,0.10,0.05,0.02)

Model

For a named channel to be the sole trigger, that channel must trigger while all other independent channels remain quiet.

w_i=p_i\prod_{j\ne i}(1-p_j)

Compute

Evaluate each one-trigger cell and normalize the moisture cell by their sum.

(w_M,w_F,w_P,w_B)=(0.16758,0.07448,0.03528,0.01368)

\Pr(\text{exactly one trigger})=0.29102

\Pr(M\mid\text{exactly one trigger})=\frac{0.16758}{0.29102}=0.5758367123\ldots

Answer

Given a single alert, the moisture channel accounts for about 57.58% of cases.

\boxed{0.5758\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: greenhouse alert diagnosis

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P General Probability