Worked solution · Exam P

SOA Exam P Sample Question #365 Solution

This Exam P sample reference tests Conditional Probability. The probability of having neither procedure is the complement of 0.35, namely 0.65. Dividing this joint no-procedure probability by the 0.70 probability of no fillings gives 13/14, approximately 0.93, and 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.

Read the solution Open official source

Independent solution

How to solve this Conditional Probability question

Setup

Let F and R denote having at least one filling and at least one root canal. Complement the reported union.

\Pr(F^c\cap R^c)=1-\Pr(F\cup R)=1-0.35=0.65

Model

The requested event conditions the joint absence of both procedures on having no fillings.

\Pr(R^c\mid F^c)=\frac{\Pr(F^c\cap R^c)}{\Pr(F^c)}

Compute

Substitute the joint and marginal probabilities.

\Pr(R^c\mid F^c)=\frac{0.65}{0.70}=\frac{13}{14}=0.928571\ldots

Answer

Given no fillings, the probability of no root canals is approximately 0.93.

\boxed{0.93\quad\text{(E)}}

Test the method on unseen problems

General Probability Practice Workbook

30 original general probability problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

See the cover and sample pages

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 0.50 divides the probability of at least one procedure by the probability of no fillings. Those events are neither nested nor the requested pair.
BThe value 0.65 is the unconditional probability of having neither procedure. It has not been normalized by the no-fillings event.
CThe value 0.72 is approximately 0.65 divided by 0.90. That reverses the conditioning and calculates no fillings given no root canals.
DThe value 0.78 is approximately 0.70 divided by 0.90. It divides two marginals instead of using their joint probability.

Original practice · fully worked

Original variant: exactly one server alert

During a monitoring interval, a server has probability 0.30 of raising a capacity alert and probability 0.25 of raising a latency alert. The probability of raising neither alert is 0.55. Given that at least one alert is raised, calculate the probability that exactly one alert is raised.

A
0.100
B
0.350
C
0.450
D
0.550
E
0.778

Answer and calculation

At least one alert occurs with probability 0.45. Inclusion–exclusion gives overlap 0.10, so exactly one alert occurs with probability 0.35. Dividing by the conditioning probability gives approximately 0.778 and choice E.

Setup

Complement the no-alert probability.

\Pr(A\cup B)=1-0.55=0.45

Model

Use inclusion–exclusion to recover the overlap.

\Pr(A\cap B)=0.30+0.25-0.45=0.10

Compute

Remove the overlap twice from the marginal sum, then condition on the union.

\Pr(\text{exactly one})=0.30+0.25-2(0.10)=0.35

\Pr(\text{exactly one}\mid A\cup B)=\frac{0.35}{0.45}=\frac79

Answer

Given at least one alert, exactly one is raised with probability approximately 0.778.

\boxed{0.778\quad\text{(E)}}

Keep practicing

Next steps for Exam P General Probability

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P General Probability practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: exactly one server alert

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability