SOA Exam P Sample Question #645. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #645 — Verified Solution

Independent solution

How to solve this Conditional Probability question

Answer in brief

This is a conditional placement problem for two event-years among five independent years. Of the ten equally likely pairs of event-years, three have year two as the first, so the conditional probability is 3/10 and choice C is correct.

Setup

Record only whether at least one event occurs in each year. Conditioning on exactly two event-years makes every pair of positions among the five years equally likely.

\#\{\text{two-year position sets}\}=\binom{5}{2}=10

Model

For year two to be the first event-year, year one must be clear, year two must be selected, and the other selected year can be any of years three through five.

\#\{\text{favorable position sets}\}=\binom{3}{1}=3

Compute

Divide the three favorable placements by all ten placements allowed by the conditioning event.

\Pr(\text{first event-year is 2}\mid\text{two event-years})

=\frac{\binom{3}{1}}{\binom{5}{2}}=\frac{3}{10}=0.30

Answer

The conditional probability is 0.300.

\boxed{0.300\quad\text{(C)}}

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 0.160 is the unconditional product of a clear first year and an event in year two. It ignores both the exact-two-years condition and the remaining three years.
BThe value 0.250 can result from treating the possible first event positions as four equally likely choices. Their probabilities are not uniform once two event-years must fit into five positions.
DThe value 0.384 is the probability of exactly one event-year among the last three years. It omits the required statuses of years one and two and does not divide by the conditioning event.
EThe value 0.400 is the conditional chance that year two is one of the two selected years. It also counts pairs containing year one, for which year two is not the first event-year.

Original practice · fully worked

Original variant: aquaculture alarm nights

An aquaculture sensor independently raises an alarm on each of seven nightly checks with probability 0.35. After the monitoring period, the log shows alarms on exactly two nights. Calculate the conditional probability that the first alarm was on the third night.

A
0.0569
B
0.1905
C
0.2857
D
0.2985
E
0.3500

Variant answer in brief

There are 21 equally likely pairs of alarm nights. Four pairs have night three first—night three paired with one of nights four through seven—so the probability is 4/21, approximately 0.1905, and choice B is correct.

Setup

Given exactly two alarm nights, the selected pair is uniform among all pairs of the seven night positions.

\#\{\text{possible alarm-night pairs}\}=\binom{7}{2}=21

Model

If the first alarm is on night three, nights one and two are not selected, night three is selected, and the other alarm can occur on any later night.

\#\{\text{favorable pairs}\}=\binom{4}{1}=4

Compute

Form the conditional ratio from the favorable and total pair counts.

\Pr(\text{first alarm on night 3}\mid\text{two alarm nights})=\frac{4}{21}

\frac{4}{21}=0.1904761905

Answer

The rounded conditional probability is 0.1905.

\boxed{0.1905\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #645 — Verified Solution.” Verified 2026-07-30.

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