SOA Exam P Sample Question #602. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #602 — Verified Solution

Independent solution

How to solve this Discrete Random Variables question

Answer in brief

The four relevant count probabilities are proportional to 60, 20, 5, and 1. Their total weight 86 represents probability 0.95, while the first three weights total 85, giving 0.95(85/86)=0.938953 and choice D.

Setup

Let q be the probability assigned to count three. Propagate the supplied ratios downward through counts two, one, and zero.

p_3=q,\qquad p_2=5q

p_1=4p_2=20q,\qquad p_0=3p_1=60q

Model

The cumulative probability through count three fixes the common scale q.

p_0+p_1+p_2+p_3=(60+20+5+1)q=86q

86q=0.95,\qquad q=\frac{0.95}{86}

Compute

For a count no greater than two, retain the first three probability weights.

\Pr(X\le2)=(60+20+5)q

\Pr(X\le2)=0.95\left(\frac{85}{86}\right)=0.9389534884\ldots

Answer

The probability rounds to 0.94.

\boxed{0.94\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.

AThis is much too small because it effectively retains only the dominant low-count mass and discards material probability from the other allowed counts.
BThis can result from reversing or truncating one of the probability-ratio links. Every ratio must be propagated from count three before normalization.
COmitting the zero-count category gives weights 20, 5, and 1; normalizing those to 0.95 and then excluding count three produces about 0.91.
EA probability through count two cannot exceed the given probability through count three. This choice violates nesting of cumulative events.

Original practice · fully worked

Original variant: even-ranked search result

A retrieval system returns a result at rank R in {1,2,3,4}. Before normalization, rank r receives probability weight r(5-r). An analyst is told that the returned rank is even. Calculate the conditional expected rank.

A
0.400
B
0.600
C
2.500
D
2.800
E
4.000

Variant answer in brief

The four rank weights are 4, 6, 6, and 4. Conditioning on an even rank leaves weights 6 and 4 at ranks two and four, so the conditional mean is 2(0.6)+4(0.4)=2.8 and choice D.

Setup

Evaluate the unnormalized weight at each possible rank.

\bigl(w_1,w_2,w_3,w_4\bigr)=\bigl(1(4),2(3),3(2),4(1)\bigr)=(4,6,6,4)

Model

After the even-rank condition, only ranks two and four remain. Normalize their two weights.

\Pr(R=2\mid R\text{ even})=\frac6{6+4}=0.6

\Pr(R=4\mid R\text{ even})=\frac4{6+4}=0.4

Compute

Weight the two remaining ranks by their conditional probabilities.

\operatorname{E}[R\mid R\text{ even}]=2(0.6)+4(0.4)=2.8

Answer

The conditional expected rank is 2.800.

\boxed{2.800\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #602 — Verified Solution.” Verified 2026-07-30.

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