SOA Exam P Sample Question #594. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #594 — Verified Solution

Independent solution

How to solve this Sampling Without Replacement question

Answer in brief

Favorable hands contain two, three, or four kings, no cards from the two excluded ranks, and all remaining cards from the 40 neutral cards. Summing those combinations over all five-card hands gives probability 0.02402499, so choice C is correct.

Setup

Partition the deck into four target cards, eight excluded cards, and forty neutral cards.

4+8+40=52

Model

Count favorable hands separately for two, three, and four target cards while choosing zero excluded cards.

N_f=\sum_{k=2}^{4}\binom4k\binom80\binom{40}{5-k}

Compute

Divide the favorable count by the number of unordered five-card hands.

\Pr=\frac{\binom42\binom{40}3+\binom43\binom{40}2+\binom44\binom{40}1}{\binom{52}5}

\Pr=0.0240249946\ldots

Answer

The probability rounds to 0.0240.

\boxed{0.0240\quad\text{(C)}}

Error diagnosis

Why the other letters tempt

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

AThis excludes an additional four-card rank along with the two prohibited ranks, leaving only 36 neutral cards and producing about 0.0174.
BThis counts exactly two kings and no prohibited cards but omits favorable hands containing three or four kings.
DThis treats the five selections as independent draws with replacement, producing about 0.0298.
EThis counts hands with at least two kings but ignores the restriction excluding aces and jacks.

Original practice · fully worked

Original variant: conditioned archive tokens

An archive contains 6 bronze, 5 silver, and 4 gold tokens. Five tokens are selected without replacement. Given that exactly two selected tokens are bronze, calculate the probability that at least one selected token is gold.

A
0.119
B
0.444
C
0.476
D
0.881
E
1.000

Variant answer in brief

After conditioning on two bronze tokens, the other three are drawn from five silver and four gold tokens. The probability that all three are silver is 10/84, so the probability of at least one gold is 37/42=0.880952 and choice D is correct.

Setup

Conditioning fixes two bronze selections, leaving three selections from the nine nonbronze tokens.

5\text{ silver}+4\text{ gold}=9\text{ nonbronze}

Model

Use the complement event that all three remaining selections are silver.

\Pr(\text{at least one gold}\mid\text{two bronze})

=1-\frac{\binom53}{\binom93}

Compute

Evaluate the combination ratio.

1-\frac{10}{84}=\frac{74}{84}=\frac{37}{42}

\frac{37}{42}=0.8809523810\ldots

Answer

The conditional probability rounds to 0.881.

\boxed{0.881\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #594 — Verified Solution.” Verified 2026-07-30.

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