SOA Exam P Sample Question #640. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #640 — Verified Solution

Independent solution

How to solve this Conditional Probability question

Answer in brief

Let N be the number of adverse outcomes among the three independent trials. Dividing the expanded probability for N=1 by the expanded positive-count probability and cancelling p gives 3(1-p) squared over p squared minus 3p plus 3. This is algebraically equivalent to choice E.

Setup

Represent the number of adverse outcomes among the three independent trials by N.

N\sim\operatorname{Binomial}(3,p)

Model

Because exactly one adverse outcome implies at least one, the conditional numerator reduces to the probability that N equals one.

\Pr(N=1\mid N\ge1)=\frac{\Pr(N=1)}{\Pr(N\ge1)}

Compute

Expand both probabilities as polynomials, then cancel their common factor p.

\Pr(N=1)=3p-6p^2+3p^3

\Pr(N\ge1)=3p-3p^2+p^3

\Pr(N=1\mid N\ge1)=\frac{3-6p+3p^2}{3-3p+p^2}=\frac{3(1-p)^2}{p^2-3p+3}

Answer

The simplified expression is equivalent to the expression in choice E.

\boxed{\frac{3(1-p)^2}{p^2-3p+3}\quad\text{(E)}}

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 chance that the other two trials avoid the adverse outcome after a particular trial is fixed as adverse; it omits both the three possible positions and the conditioning denominator.
BThis is the unconditional probability of exactly one adverse outcome and does not divide by the probability that at least one occurs.
CThis results from dividing the exactly-one probability by (1-p) cubed, the zero-outcome probability, instead of by its complement.
DThis conditions correctly but counts only one specified trial as the sole adverse outcome, omitting the factor of three.

Original practice · fully worked

Original variant: acoustic sensor self-test

Four acoustic sensors run independent self-tests. Each sensor fails its test with probability q, where q is strictly between zero and one. Given that the array reports at least one failure, determine the probability that exactly two sensors fail.

A
q^2(1-q)^2
B
6q^2(1-q)^2
C
6q^2(1-q)^2 / [1-(1-q)^4]
D
[1-(1-q)^4] / [6q^2(1-q)^2]
E
1 - 6q^2(1-q)^2 / [1-(1-q)^4]

Variant answer in brief

The unconditional probability of two failures is 6q squared times (1-q) squared, while at least one failure has probability 1-(1-q) to the fourth power. Their ratio is choice C.

Setup

Let F be the number of failed self-tests in the four-sensor array.

F\sim\operatorname{Binomial}(4,q)

Model

The exactly-two event is contained in the positive-failure event, so use a simple conditional ratio.

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

Compute

The binomial coefficient counts the six possible failed pairs, and a complement gives the denominator.

\Pr(F=2)=\binom42q^2(1-q)^2=6q^2(1-q)^2

\Pr(F\ge1)=1-(1-q)^4

\Pr(F=2\mid F\ge1)=\frac{6q^2(1-q)^2}{1-(1-q)^4}

Answer

The conditional probability is the expression in choice C.

\boxed{\frac{6q^2(1-q)^2}{1-(1-q)^4}\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #640 — Verified Solution.” Verified 2026-07-30.

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