SOA Exam P Sample Question #682. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #682

Condition on the pooled alert by dividing the probability of exactly two affected subjects by the probability of at least one affected subject. The ratio is 0.00999966, so choice C is correct.

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

Independent solution

How to solve this Conditional Probability question

Answer in brief

Condition on the pooled alert by dividing the probability of exactly two affected subjects by the probability of at least one affected subject. The ratio is 0.00999966, so choice C is correct.

Setup

Let X count affected subjects among the three independent trials. The common affected probability is 0.01.

X\sim\operatorname{Bin}(3,0.01)

Model

The observed pooled alert is equivalent to X being positive. Because the event X=2 is contained in that alert event, conditional probability reduces to a ratio.

\Pr(X=2\mid X\ge 1)=\frac{\Pr(X=2)}{\Pr(X\ge 1)}

Compute

Use the binomial mass in the numerator and a complement in the denominator.

\Pr(X=2)=\binom{3}{2}(0.01)^2(0.99)=0.000297

\Pr(X\ge 1)=1-(0.99)^3=0.029701

\frac{0.000297}{0.029701}=0.0099996633\ldots

Answer

The requested conditional probability rounds to 0.0100.

\boxed{0.0100\quad\text{(C)}}

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 0.000297 is the unconditional probability of exactly two affected subjects; it omits division by the observed-alert probability.
BThe value is obtained by retaining only one specified arrangement of two affected and one unaffected subject, thereby omitting the binomial factor of three.
DThe value 2(0.01)(0.99)=0.0198 is an exactly-one calculation for only two trials, so it uses neither the three-subject model nor the conditioning event.
EThe value 1-(0.99)^3=0.029701 is the probability of the pooled alert itself, which is the denominator rather than the conditional result.

Original practice · fully worked

Original variant: mountain relay incident

A monitoring dashboard watches four independently operated mountain relays. Each relay has a 4% chance of missing its midnight heartbeat. The dashboard displays an incident whenever one or more heartbeats are absent. Given that an incident is displayed, calculate the probability that exactly two relays missed the heartbeat.

A
0.00885
B
0.00979
C
0.05873
D
0.06037
E
0.15065

Variant answer in brief

The exactly-two probability is 0.00884736 and the incident probability is 0.15065344. Their ratio is 0.0587266, which selects choice C.

Setup

Let M denote the number of relays that miss the heartbeat.

M\sim\operatorname{Bin}(4,0.04)

Model

An incident means M is at least one. Exactly two misses automatically imply that an incident occurred.

\Pr(M=2\mid M\ge 1)=\frac{\Pr(M=2)}{1-\Pr(M=0)}

Compute

Evaluate the two binomial quantities and form the conditional ratio.

\Pr(M=2)=\binom{4}{2}(0.04)^2(0.96)^2=0.00884736

\Pr(M\ge 1)=1-(0.96)^4=0.15065344

\frac{0.00884736}{0.15065344}=0.0587265714\ldots

Answer

The conditional probability is approximately 0.05873.

\boxed{0.05873\quad\text{(C)}}

Continue by topic

Build a complete proof notebook

The Probability Proof Manual reorganizes verified solutions by syllabus skill and adds formula proofs, error patterns, and original practice.

Explore study products Get the free formula sheet

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #682 — Verified Solution.” Verified 2026-07-30.

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