Worked solution · Exam P

SOA Exam P Sample Question #500 Solution

This Exam P sample reference tests Independence. Writing the two-trial exactly-one probability as x=2p(1-p), the four-trial exactly-two probability becomes 6p²(1-p)²=(3/2)x². Thus choice D is correct.

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

Let p be the common one-period success probability and let q=1-p. The supplied two-period quantity is the probability of one success and one failure.

x=\binom21pq=2pq

Model

Independence makes the number of successes in four periods binomial. Exactly two successes can occupy any two of the four positions.

\Pr(S_4=2)=\binom42p^2q^2=6p^2q^2

Compute

Replace the product pq by x/2 and simplify.

6p^2q^2=6(pq)^2=6\left(\frac{x}{2}\right)^2

\Pr(S_4=2)=\frac32x^2

Answer

The four-period probability is one and one-half times the square of x.

\boxed{1.5x^2\quad\text{(D)}}

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 target must be quadratic in pq and hence quadratic in x. For example, p=1/2 gives x=1/2 and the true probability 3/8, whereas 0.5x gives 1/4.
BThis linearly doubles the two-period probability. At p=1/2 it gives 2x=1, while the exactly-two count has probability 6/16=3/8.
CRequiring exactly one success in each fixed two-period block gives x²=4p²q². It misses the two arrangements in which one block has two successes and the other has none.
EThe value 3x²=12p²q² doubles all six valid arrangements. Each binary sequence with two successes already has probability p²q² and must be counted once.

Original practice · fully worked

Original variant: conditional online-sensor count

Five sensors operate independently during a diagnostic window. For each sensor, the odds of being online to being offline are 2:3. The diagnostic reports that at least one sensor is online. Calculate the conditional probability that exactly two sensors are online.

A
0.07776
B
0.34560
C
0.37474
D
0.68256
E
0.92224

Answer and calculation

The odds give an online probability of 0.4. Dividing the binomial exactly-two mass 0.34560 by the nonzero-count probability 0.92224 gives 0.37474, so choice C is correct.

Setup

Convert the individual odds into online and offline probabilities.

p=\frac{2}{2+3}=0.4,\qquad q=\frac{3}{5}=0.6

Model

Let K be the online count. Independence gives a binomial model, then the diagnostic conditions on K being positive.

K\sim\operatorname{Binomial}(5,0.4)

\Pr(K=2\mid K\ge1)=\frac{\Pr(K=2)}{\Pr(K\ge1)}

Compute

Calculate the numerator and the conditioning probability separately.

\Pr(K=2)=\binom52(0.4)^2(0.6)^3=0.34560

\Pr(K\ge1)=1-(0.6)^5=0.92224

\Pr(K=2\mid K\ge1)=\frac{0.34560}{0.92224}=0.3747397641\ldots

Answer

The requested conditional probability rounds to 0.37474.

\boxed{0.37474\quad\text{(C)}}

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: conditional online-sensor count

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability