SOA Exam P Sample Question #646. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #646 — Verified Solution

Independent solution

How to solve this Complement Rule question

Answer in brief

This is a complement-rule calculation for independent days. The chance of a claim-free seven-day period is r to the seventh power, so the chance of at least one claim is 1 minus r to the seventh power and choice D is correct.

Setup

For each of the seven days, the no-claim probability is r, and the daily claim indicators are mutually independent.

\Pr(\text{no claim on one day})=r

Model

The complement of at least one claim is a period with no claim on every day.

\Pr(\text{at least one claim})=1-\Pr(\text{no claims in seven days})

Compute

Independence allows the seven daily no-claim probabilities to be multiplied.

\Pr(\text{no claims in seven days})=\prod_{j=1}^{7}r=r^7

\Pr(\text{at least one claim})=1-r^7

Answer

The required probability is the complement of seven claim-free days.

\boxed{1-r^7\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.

AThe expression r to the seventh power is the probability of no claims during the entire period, which is the complement of the requested event.
BThis is the probability of a claim on one specified day and no claims on the other six days. It does not include every way claims can occur.
CThis is the probability of exactly one claim-day anywhere in the period. The requested event also includes outcomes with claims on multiple days.
EThe expression (1-r) to the seventh power requires a claim on every day. That is only one narrow part of the at-least-one event.

Original practice · fully worked

Original variant: soil-probe self-test

Twelve remote soil probes run independent overnight self-tests. Each probe passes its test with probability 0.96. Calculate the probability that at least one probe requires recalibration after the tests.

A
0.0400
B
0.3064
C
0.3873
D
0.4800
E
0.6127

Variant answer in brief

All twelve probes pass with probability 0.96 to the twelfth power. Taking the complement gives 0.3872902, which rounds to 0.3873 and selects choice C.

Setup

A recalibration is required when a probe does not pass. The tests are independent across the twelve probes.

\Pr(\text{one probe passes})=0.96

Model

Use the complement event in which every probe passes its self-test.

\Pr(\text{at least one recalibration})=1-\Pr(\text{all pass})

Compute

Multiply the twelve pass probabilities, then subtract the result from one.

\Pr(\text{all pass})=0.96^{12}=0.6127097573

1-0.96^{12}=0.3872902427

Answer

The probability that at least one probe needs recalibration is 0.3873 to four decimals.

\boxed{0.3873\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #646 — Verified Solution.” Verified 2026-07-30.

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