SOA Exam P Sample Question #592. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #592

The count is binomial with n=18 and p=0.15. Since (n+1)p=2.85 is not an integer, its unique mode is floor(2.85)=2, so choice C is correct.

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

Answer in brief

The count is binomial with n=18 and p=0.15. Since (n+1)p=2.85 is not an integer, its unique mode is floor(2.85)=2, so choice C is correct.

Setup

Model the number of positive results as a binomial count.

X\sim\operatorname{Binomial}(18,0.15)

Model

A binomial distribution has unique mode floor((n+1)p) when (n+1)p is not an integer.

(n+1)p=19(0.15)=2.85

Compute

Take the integer below 2.85.

\operatorname{mode}(X)=\lfloor2.85\rfloor=2

Answer

The most likely count is 2.

\boxed{2\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 treats the low probability for one trial as implying a zero modal count and ignores the accumulation across eighteen trials.
BThis uses an incorrect mode rule such as floor(np)-1, moving one count below the actual peak.
DThis rounds the mean np=2.7 to 3; a binomial mode is determined by (n+1)p rather than ordinary rounding of the mean.
EThis moves farther into the decreasing upper side of the binomial mass after the peak near the mean.

Original practice · fully worked

Original variant: inferred inspection rate

The number X of flagged items in ten independent inspections is binomial with an unknown flag probability. The probability of exactly four flags is 1.5 times the probability of exactly three flags. Calculate the mode of X.

A
3
B
4
C
5
D
6
E
10

Variant answer in brief

The adjacent-mass ratio gives 7p/[4(1-p)]=1.5, so p=6/13. Then (10+1)p=66/13 lies between 5 and 6, making 5 the unique mode and choice C correct.

Setup

Use the ratio of adjacent binomial masses to recover the unknown probability.

\frac{\Pr(X=4)}{\Pr(X=3)}=\frac{10-3}{4}\frac{p}{1-p}=1.5

Model

Solve the ratio equation for p.

\frac{7p}{4(1-p)}=\frac32

p=\frac6{13}

Compute

Apply the binomial mode rule after inferring the parameter.

(10+1)p=\frac{66}{13}=5.076923\ldots

\operatorname{mode}(X)=5

Answer

The most likely number of flagged items is 5.

\boxed{5\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #592 — Verified Solution.” Verified 2026-07-30.

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