SOA Exam P Sample Question #683. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #683 — Verified Solution

Independent solution

How to solve this Conditional Probability question

Answer in brief

The given no-theft probabilities determine a four-cell table: the probability that both households have a theft is 0.10, while the conditioning household has a theft with probability 0.20. Their ratio is 0.50, choice D.

Setup

Write T_A and T_B for the events that the respective households experience at least one theft. Each complement has probability 0.80.

\Pr(T_A)=\Pr(T_B)=0.20

\Pr(T_A^c\cap T_B^c)=0.70

Model

Recover the probability of both theft events through inclusion–exclusion applied to their complements.

\Pr(T_A\cap T_B)=1-\Pr(T_A^c)-\Pr(T_B^c)+\Pr(T_A^c\cap T_B^c)

Compute

Substitute the three supplied probabilities, then divide by the probability of the conditioning event.

\Pr(T_A\cap T_B)=1-0.80-0.80+0.70=0.10

\Pr(T_A\mid T_B)=\frac{0.10}{0.20}=0.50

Answer

Given a theft at household B, the conditional probability of a theft at household A is one-half.

\boxed{0.500\quad\text{(D)}}

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 0.20(0.30)=0.060 treats the chance of at least one theft as though it were B's theft probability and also imposes unsupported independence.
BThe value 0.100 is the joint probability of thefts at both households; it is the numerator before conditioning on B's theft.
CThe value 0.200 is A's marginal theft probability and ignores the information about what happened at household B.
EThe ratio 0.70/0.80=0.875 is a conditional no-theft probability, Pr(T_B^c given T_A^c), for the complementary events.

Original practice · fully worked

Original variant: alpine telemetry links

A control center receives telemetry through a north link and a south link. Each link is available at the audit time with probability 0.85, and both are available with probability 0.75. If the south link is unavailable, calculate the probability that the north link is also unavailable.

A
0.050
B
0.150
C
0.250
D
0.333
E
0.882

Variant answer in brief

Inclusion–exclusion gives a 0.05 probability that both links are unavailable. The south link is unavailable with probability 0.15, so the conditional probability is 0.05/0.15=1/3, choice D.

Setup

Let U_N and U_S denote unavailability of the north and south links.

\Pr(U_N)=\Pr(U_S)=0.15

\Pr(U_N^c\cap U_S^c)=0.75

Model

Use the availability information to find the lower-right cell of the two-link status table.

\Pr(U_N\cap U_S)=1-0.85-0.85+0.75

Compute

Normalize the both-unavailable cell by the south-unavailable marginal probability.

\Pr(U_N\cap U_S)=0.05

\Pr(U_N\mid U_S)=\frac{0.05}{0.15}=\frac13

Answer

The required conditional probability is approximately 0.333.

\boxed{0.333\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #683 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.