Worked solution · Exam P

SOA Exam P Sample Question #704 Solution

This Exam P sample reference tests Binomial Distribution. First convert two independent individual survivals into a couple-level success probability of 0.79 squared, or 0.6241. A binomial upper tail for seven couples then gives 0.925226, which selects choice D.

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 Binomial Distribution question

Setup

A couple counts as a success only when both of its members survive. Independence turns that intersection into a product.

q=\Pr(\text{both members survive})=(0.79)^2=0.6241

Model

Let N count successful couples. Independence across all people makes the seven couple indicators independent Bernoulli variables.

N\sim\operatorname{Binomial}(7,0.6241)

\Pr(N\ge3)=1-\Pr(N\le2)

Compute

Evaluate the short lower tail and take its complement.

\Pr(N\le2)=\sum_{k=0}^{2}\binom{7}{k}(0.6241)^k(0.3759)^{7-k}

\Pr(N\le2)=0.07477428837\ldots

\Pr(N\ge3)=0.9252257116\ldots

Answer

The requested probability rounds to 0.925.

\boxed{\Pr(N\ge3)\approx0.925\quad\text{(D)}}

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.

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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.

AAn off-by-one interpretation that effectively requires four or more successful couples pushes the result into this lower range; the event begins at three.
BMixing the individual survival probability 0.79 with the couple probability 0.6241 across different binomial terms understates the coherent upper tail.
CA rough normal approximation without a continuity correction depresses the answer; the seven-trial binomial tail is short enough to evaluate exactly.
ECounting two or more successful couples instead gives 0.986615, near the high distractor after coarse rounding; the threshold is at least three.

Original practice · fully worked

Original variant: observatory cooling clusters

A remote observatory operates eight independent cooling clusters. Each cluster contains three fans, and every fan completes the overnight cycle with probability 0.90 independently of every other fan. A cluster is fully operational only if all three of its fans complete the cycle. Calculate the probability that at least five clusters are fully operational.

A
0.229
B
0.626
C
0.855
D
0.962
E
0.995

Answer and calculation

A cluster succeeds with probability 0.90 cubed, or 0.729. The Binomial(8, 0.729) probability of at least five successes is approximately 0.855108, so choice C is correct.

Setup

Combine the three independent fan outcomes into one cluster-level success probability.

q=(0.90)^3=0.729

Model

Let K count fully operational clusters among the eight independent clusters.

K\sim\operatorname{Binomial}(8,0.729)

\Pr(K\ge5)=1-\Pr(K\le4)

Compute

Sum the lower five binomial masses and subtract from one.

\Pr(K\ge5)=1-\sum_{k=0}^{4}\binom{8}{k}(0.729)^k(0.271)^{8-k}

\Pr(K\ge5)=0.8551076438\ldots

Answer

The probability rounds to 0.855 and selects choice C.

\boxed{\Pr(K\ge5)\approx0.855\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Binomial Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: observatory cooling clusters

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables