SOA Exam P Sample Question #701. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #701

Normalizing the infinite probability sequence gives k=2. Summing the even-indexed terms as a geometric series then yields 1/4, which is choice B.

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Independent solution

How to solve this Discrete Random Variables question

Answer in brief

Normalizing the infinite probability sequence gives k=2. Summing the even-indexed terms as a geometric series then yields 1/4, which is choice B.

Setup

Use exhaustiveness to require the probabilities over all positive indices to sum to one.

1=\sum_{i=1}^{\infty}k\left(\frac13\right)^i

Model

Evaluate the full geometric series to determine its normalizing constant.

\sum_{i=1}^{\infty}\left(\frac13\right)^i=\frac{1/3}{1-1/3}=\frac12

1=\frac{k}{2}\quad\Longrightarrow\quad k=2

Compute

The even indices are 2j for positive integers j, producing a second geometric series with ratio 1/9.

\Pr(\text{even index})=2\sum_{j=1}^{\infty}\left(\frac13\right)^{2j}

\Pr(\text{even index})=2\frac{1/9}{1-1/9}=\frac14

Answer

The total probability carried by the even-indexed outcomes is one quarter.

\boxed{\Pr(\text{even index})=\frac14\quad\text{(B)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThe raw even-term series (1/9)/(1-1/9)=1/8 omits the required normalizing constant k=2.
CUsing k=3, as if the first term alone had to equal one, gives 3(1/8)=3/8 instead of normalizing the entire sequence.
DAssigning probability 1/2 to each parity assumes even and odd indices are symmetric, but the geometrically decreasing masses favor early odd indices.
EThe value 3/4 is the complementary probability of an odd index, not the requested even-index probability.

Original practice · fully worked

Original variant: observatory maintenance tiers

An autonomous observatory assigns each maintenance request to a positive integer tier T. For a normalizing constant c, the assignment probabilities satisfy P(T=n)=c(2/5)^n for n=1,2,3,... . Calculate the probability that the selected tier number is divisible by 3.

A
8/117
B
4/39
C
10/39
D
1/3
E
25/39

Variant answer in brief

The full geometric series gives c=3/2. Applying that constant to the every-third-term series yields 4/39, so choice B is correct.

Setup

Normalize the tier probabilities over all positive integers.

1=c\sum_{n=1}^{\infty}\left(\frac25\right)^n

Model

Evaluate the geometric series and solve for c.

\sum_{n=1}^{\infty}\left(\frac25\right)^n=\frac{2/5}{1-2/5}=\frac23

c=\frac32

Compute

Tier numbers divisible by three have indices n=3j. Their probabilities form a series with ratio (2/5)^3.

\Pr(T\equiv 0\pmod 3)=\frac32\sum_{j=1}^{\infty}\left(\frac25\right)^{3j}

\Pr(T\equiv 0\pmod 3)=\frac32\frac{8/125}{1-8/125}=\frac4{39}

Answer

The probability assigned to every third tier is 4/39.

\boxed{\Pr(T\equiv 0\pmod 3)=\frac4{39}\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #701 — Verified Solution.” Verified 2026-07-30.

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