Worked solution · Exam P

SOA Exam P Sample Question #729 Solution

This Exam P sample reference tests Discrete Random Variables. Normalizing the geometric-form probability function fixes its constant and reveals a continuation ratio of one third. The required conditional probability is 8/9, or approximately 0.888889, which matches choice E.

Independent analysis by ActuaryProof · reviewed July 30, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Discrete Random Variables question

Setup

Normalize the infinite probability sequence before forming the conditional event.

1=6\sum_{n=0}^{\infty}3^{-(n+c)}

Model

Evaluate the geometric series to determine the constant and rewrite the masses in standard geometric form.

1=6\cdot 3^{-c}\frac{1}{1-1/3}=9\cdot 3^{-c}

c=2,\qquad p_n=\frac{2}{3}\left(\frac{1}{3}\right)^n

Compute

The numerator consists of counts three and four, while the denominator is the full tail beginning at three.

p_3+p_4=\frac{2}{81}+\frac{2}{243}=\frac{8}{243}

\Pr(N\ge 3)=\left(\frac{1}{3}\right)^3=\frac{1}{27}

\Pr(3\le N<5\mid N\ge3)=\frac{8/243}{1/27}=\frac{8}{9}

Answer

The conditional probability rounds to the final listed value.

\boxed{\frac{8}{9}=0.888889\ldots\quad\text{(E)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

See the cover and sample pages

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 0.296296 equals 8/27, which is also p_1+p_2; it shifts the two-count window downward and leaves it unconditional.
BWith the correct numerator 0.0329218, a result of 0.434 would require a conditioning denominator near 0.07586 instead of the geometric tail 1/27=0.037037, indicating a mis-summed tail.
CThe value 0.590 is inconsistent with the tail ratio p_{n+1}/p_n=1/3; after conditioning at three, the chance of remaining within the next two masses must be 1-(1/3)²=0.888889.
DA result of 0.704 would use an effective denominator near 0.04676, so it has included probability outside the conditioning event or otherwise failed to renormalize by the exact 1/27 tail.

Original practice · fully worked

Original variant: greenhouse sensor recalibration

A greenhouse controller repeats a sensor calibration until it succeeds. Let K be the number of failed attempts before success, with probability P(K=k)=12/4⁽ᵏ⁺ʳ⁾ for k=0,1,2,..., where r makes this a valid distribution. Given that at least one attempt failed, calculate the probability that no more than two attempts failed.

A
0.234
B
0.750
C
0.938
D
0.984
E
0.996

Answer and calculation

Normalization gives r=2 and a geometric continuation ratio of one fourth. Conditional on at least one failure, allowing one or two failures has probability 15/16=0.9375, so choice C is correct.

Setup

Use the total probability requirement to identify r.

1=12\sum_{k=0}^{\infty}4^{-(k+r)}

Model

Evaluate the geometric sum and express the resulting distribution with its continuation ratio.

1=12\cdot4^{-r}\frac{1}{1-1/4}=16\cdot4^{-r}

r=2,\qquad \Pr(K=k)=\frac{3}{4}\left(\frac{1}{4}\right)^k

Compute

After conditioning on one or more failures, the excess count restarts with the same geometric ratio.

\Pr(1\le K\le2\mid K\ge1)=1-\left(\frac14\right)^2

\Pr(1\le K\le2\mid K\ge1)=\frac{15}{16}=0.9375

Answer

The rounded probability is listed under choice C.

\boxed{0.9375\approx0.938\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Discrete Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: greenhouse sensor recalibration

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables