SOA Exam P Sample Question #608. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #608 — Verified Solution

Independent solution

How to solve this Binomial Distribution question

Answer in brief

This is a binomial probability-ratio problem. The zero-to-one mass relation gives failure-to-success odds of 60, and the zero-to-three mass ratio is therefore 60^3 divided by 20, or 10,800, so choice E is correct.

Setup

Let p be the common disease probability and q=1-p. Write the binomial masses at zero and one.

\Pr(N=0)=q^6

\Pr(N=1)=6pq^5

Model

Translate the first probability comparison into an odds equation.

q^6=10(6pq^5)

\frac{q}{p}=60

Compute

Form the requested multiplier by dividing the zero-disease mass by the three-disease mass.

x=\frac{\Pr(N=0)}{\Pr(N=3)}=\frac{q^6}{\binom63p^3q^3}

x=\frac{(q/p)^3}{20}=\frac{60^3}{20}=10800

Answer

The zero-disease probability is 10,800 times the three-disease probability.

\boxed{10800\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.

ADividing 60^3 by 6! instead of by the binomial coefficient C(6,3)=20 gives 300. The count mass uses combinations, not all permutations.
BCubing q/(6p)=10 and omitting the three-success binomial coefficient gives 1000; the factor six belongs only to the one-success comparison.
CThis divides the correct odds cube by an extra factor of six, producing 60^3/[6C(6,3)]=1800.
DUsing 6^2=36 in place of C(6,3)=20 gives 60^3/36=6000.

Original practice · fully worked

Original variant: module pass-count moments

The number X of modules that pass a production diagnostic follows a binomial distribution with unknown trial count and success probability. Historical batches give E[X]=4.8 and Var(X)=1.92. Calculate the probability that at least six modules pass in the next batch.

A
0.0168
B
0.2090
C
0.2787
D
0.3154
E
0.6846

Variant answer in brief

Dividing the variance by the mean gives failure probability 0.40, so the success probability is 0.60 and the trial count is eight. The binomial upper tail from six through eight is 0.31539456, selecting choice D.

Setup

Let the binomial parameters be n and p and write its mean and variance equations.

np=4.8

np(1-p)=1.92

Model

Use the ratio of variance to mean to identify p, then recover the integer trial count.

1-p=\frac{1.92}{4.8}=0.40

p=0.60,\qquad n=\frac{4.8}{0.60}=8

Compute

Add the three binomial masses that satisfy the requested upper-tail event.

\Pr(X\ge6)=\sum_{k=6}^{8}\binom8k(0.60)^k(0.40)^{8-k}

\Pr(X\ge6)=0.20901888+0.08957952+0.01679616

\Pr(X\ge6)=0.31539456

Answer

The probability of at least six passing modules is approximately 0.3154.

\boxed{0.3154\quad\text{(D)}}

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ActuaryProof. “SOA Exam P Sample Question #608 — Verified Solution.” Verified 2026-07-30.

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