Worked solution · Exam P

SOA Exam P Sample Question #186 Solution

This Exam P sample reference tests Conditional Probability. The probability of transferring two balls of each color and then selecting blue is 0.20979. The selected transferred ball is blue with marginal probability 6/14. Their ratio is 0.48951, corresponding to choice D.

Independent analysis by ActuaryProof · reviewed July 31, 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 Conditional Probability question

Setup

Let E be the event that the transferred sample contains two balls of each color and let B be the event that the subsequent selection is blue.

\Pr(E)=\frac{\binom82\binom62}{\binom{14}4}

Model

Under E, half of the transferred balls are blue. The marginal selected color matches a uniformly selected original ball.

\Pr(B\mid E)=\frac24=\frac12

\Pr(B)=\frac6{14}=\frac37

Compute

Apply Bayes' rule using the joint numerator.

\Pr(E\cap B)=\frac{\binom82\binom62}{\binom{14}4}\left(\frac12\right)=0.2097902

\Pr(E\mid B)=\frac{0.2097902}{3/7}=0.4895105

Answer

The conditional probability rounds to 0.49.

\boxed{0.49\quad\text{(D)}}

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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 0.21 is the joint probability P(E and B), rounded from 0.20979. It is the Bayes numerator before division by P(B).
BThe value 0.24 results from taking the correct conditional probability near 0.49 and multiplying by the one-half blue share under E a second time.
CThe value 0.43 is close both to the unconditioned transfer probability P(E)=0.41958 and to P(B)=3/7. Neither marginal is the requested conditional probability.
EThe value 0.57 is 8/14, the marginal probability that the subsequently selected ball is red. It uses the complementary color event.

Original practice · fully worked

Original variant: a hypergeometric count after screening

A container holds six silver and four gold tokens. A sample of four is drawn without replacement and discarded unless it includes gold. Among the samples that are kept, what is the chance that the gold count equals two?

A
0.0714
B
0.1143
C
0.3810
D
0.4286
E
0.4615

Answer and calculation

The unconditioned exactly-two probability is 3/7, while the probability of at least one gold token is 13/14. Their ratio is 6/13=0.4615, choice E.

Setup

Let K count gold tokens in the four-token sample.

\Pr(K=k)=\frac{\binom4k\binom6{4-k}}{\binom{10}4}

Model

Form the exactly-two numerator and complement the zero-gold sample for the denominator.

\Pr(K=2)=\frac{\binom42\binom62}{\binom{10}4}=\frac37

\Pr(K\ge1)=1-\frac{\binom64}{\binom{10}4}=\frac{13}{14}

Compute

Normalize the exactly-two mass within the screened samples.

\Pr(K=2\mid K\ge1)=\frac{3/7}{13/14}=\frac6{13}=0.4615385

Answer

The screened conditional probability is about 0.4615.

\boxed{0.4615\quad\text{(E)}}

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How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: a hypergeometric count after screening

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability