SOA Exam P Sample Question #729. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #729 — Verified Solution

Independent solution

How to solve this Discrete Random Variables question

Answer in brief

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.

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)}}

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

Variant answer in brief

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)}}

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ActuaryProof. “SOA Exam P Sample Question #729 — Verified Solution.” Verified 2026-07-30.

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