SOA Exam P Sample Question #644. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #644 — Verified Solution

Independent solution

How to solve this Conditional Probability question

Answer in brief

This is a conditional exchangeability problem. Once exactly three of six equally exposed units are known to be damaged, each three-unit subset is equally likely; four of the twenty subsets avoid both units in the distinguished pair, giving 0.20 and choice D.

Setup

Condition on exactly three damaged units among the six. Because every unit has the same independent damage probability, the identity of the damaged three is uniformly distributed over all three-unit subsets.

\#\{\text{possible damaged subsets}\}=\binom{6}{3}=20

Model

For neither member of the two-unit distinguished group to be damaged, all three damaged units must come from the other four.

\#\{\text{favorable damaged subsets}\}=\binom{4}{3}=4

Compute

Take the ratio of favorable conditional subsets to all possible conditional subsets.

\Pr(\text{distinguished pair undamaged}\mid\text{three damaged})

=\frac{\binom{4}{3}}{\binom{6}{3}}=\frac{4}{20}=0.20

Answer

The required conditional probability is 0.2000.

\boxed{0.2000\quad\text{(D)}}

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 unconditional joint probability that exactly three of the four nondistinguished units are damaged and every other unit is undamaged. It is the numerator before conditioning.
BThis result can be produced by retaining only one undamaged factor after selecting three damaged units from the group of four. It fails to account for both members of the distinguished pair.
CThis is the unconditional probability that exactly three of all six units are damaged. It is the denominator event, not the requested conditional ratio.
EThis is the unconditional probability that the two distinguished units are both undamaged. Knowing that three of the six are damaged changes that probability.

Original practice · fully worked

Original variant: navigation-beacon shock test

Eight sealed navigation beacons ride through the same shock test. Five use a standard casing and three use a ceramic casing. Each beacon independently fails with probability 0.20. After learning that exactly two beacons failed, calculate the probability that both failed beacons used the standard casing.

A
0.1049
B
0.2936
C
0.3571
D
0.3906
E
0.6250

Variant answer in brief

Given two failures among equally exposed beacons, every pair is equally likely. Ten of the 28 pairs come from the five standard-casing beacons, so the conditional probability is 5/14, approximately 0.3571, and choice C is correct.

Setup

Condition on the observed total of two failed beacons. Equal independent failure probabilities make the failed pair uniform among all pairs of the eight labeled beacons.

\#\{\text{possible failed pairs}\}=\binom{8}{2}=28

Model

A favorable pair contains two of the five standard-casing beacons.

\#\{\text{favorable pairs}\}=\binom{5}{2}=10

Compute

The conditional probability is the favorable-pair count divided by the total pair count.

\Pr(\text{both standard}\mid\text{two failures})=\frac{10}{28}

\frac{10}{28}=\frac{5}{14}=0.3571428571

Answer

Rounded to four decimal places, the result is 0.3571.

\boxed{0.3571\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #644 — Verified Solution.” Verified 2026-07-30.

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