Worked solution · Exam P

SOA Exam P Sample Question #351 Solution

This Exam P sample reference tests Independence. The probability of no accidents is 0.2916. Exactly one accident contributes 0.3888 from a high-risk employee and 0.0648 from a low-risk employee. Adding these disjoint cases gives 0.7452 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 Independence question

Setup

Convert the no-accident probabilities into accident probabilities for the two employee classes.

p_H=0.40,\qquad p_L=0.10

Model

Calculate the probability that no employee has an accident.

P_0=(0.60)^2(0.90)^2=0.2916

Compute

Separate the exactly-one case according to whether the accident belongs to a high-risk or low-risk employee.

P_{1,H}=2(0.40)(0.60)(0.90)^2=0.3888

P_{1,L}=2(0.10)(0.90)(0.60)^2=0.0648

\Pr(N\le1)=0.2916+0.3888+0.0648=0.7452

Answer

At most one employee has an accident with probability 0.7452.

\boxed{0.7452\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.

AThis is the probability of exactly one accident. It omits the allowed case in which nobody has an accident.
BThis is the probability that at least one high-risk employee has an accident while both low-risk employees avoid accidents. It does not describe every at-most-one case.
CThis includes no accidents and exactly one high-risk accident but omits the cases in which exactly one low-risk employee has an accident.
DThis is the complement of the no-accident probability, or the probability of at least one accident. It also includes cases with two or more accidents.

Original practice · fully worked

Original variant: parity of four sensor outages

Four independent sensors have outage probabilities 0.15, 0.25, 0.35, and 0.45 during a test window. Calculate the probability that an even number of sensors are offline.

A
0.0105
B
0.2279
C
0.4948
D
0.5053
E
0.7721

Answer and calculation

For independent Bernoulli indicators, the even-parity probability is one-half of one plus the product of one minus twice each outage probability. The product is 0.0105, giving 0.50525 and choice D.

Setup

Encode outage-count parity through the expectation of minus one raised to the count.

\Pr(N\text{ even})-\Pr(N\text{ odd})=\operatorname{E}[(-1)^N]

Model

Independence makes the parity expectation a product of four Bernoulli factors.

\operatorname{E}[(-1)^N]=\prod_{j=1}^4(1-2p_j)

Compute

Evaluate the product and combine it with the probabilities summing to one.

(0.70)(0.50)(0.30)(0.10)=0.0105

\Pr(N\text{ even})=\frac{1+0.0105}{2}=0.50525

Answer

An even number of sensors are offline with probability approximately 0.5053.

\boxed{0.5053\quad\text{(D)}}

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 Independence question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: parity of four sensor outages

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability