Verified solution · Exam P

SOA Exam P Sample Question #190 Solution

This Exam P sample reference tests Binomial Distribution. At most 95 error-free applications means at least five applications with errors. For a Binomial(100,0.03) count, subtracting the masses from zero through four from one gives 0.18215, choice E.

Independent analysis by ActuaryProof · reviewed August 4, 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

Let X count applications containing an error and translate the percentage condition to an integer tail.

X\sim\operatorname{Binomial}(100,0.03)

\{\text{error-free count}\le95\}=\{X\ge5\}

Model

Evaluate the upper tail by complementing the first five binomial masses.

\Pr(X\ge5)=1-\sum_{k=0}^{4}\binom{100}{k}(0.03)^k(0.97)^{100-k}

Compute

Compute the lower-tail probability and take its complement.

\Pr(X\le4)=0.8178548061

\Pr(X\ge5)=1-0.8178548061=0.1821451939

Answer

The required probability rounds to 0.18.

\boxed{0.18\quad\text{(E)}}

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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 0.08 is the tail P(X≥6), approximately 0.08084. It treats 95% or less error-free as requiring at least six errors rather than five.
BThe value 0.10 is the single mass P(X=5), approximately 0.10131. The target also includes six or more errors.
CA value near 0.13 can result from a rough normal tail without the needed continuity correction. The exact discrete threshold is evaluated by the binomial sum and gives 0.18215.
DThe value 0.15 is close to P(X=1)=0.14707. It is a lower-count point probability, not the upper tail beginning at five.

Original practice · fully worked

Original variant: calibrate a binomial tail from two equal masses

Six independent checks each raise a flag with probability p. The probability of no flags equals the probability of exactly two flags. Calculate the probability that at least three flags are raised.

A
0.0186
B
0.0868
C
0.1054
D
0.2052
E
0.8946

Variant answer in brief

Equality of the zero- and two-flag masses gives (1-p)²=15p squared, so p=1/(1+√(15)). The resulting upper tail from three through six is 0.10539, choice C.

Setup

Write the two supplied binomial masses and cancel their common factor.

(1-p)^6=\binom62p^2(1-p)^4

(1-p)^2=15p^2

Model

Use the probability range to select the positive square-root relation and recover p.

1-p=\sqrt{15}\,p

p=\frac{1}{1+\sqrt{15}}=0.2052131

Compute

Evaluate the calibrated binomial tail.

\Pr(X\ge3)=1-\sum_{k=0}^{2}\binom6k p^k(1-p)^{6-k}

\Pr(X\ge3)=0.1053897455

Answer

The probability of at least three flags rounds to 0.1054.

\boxed{0.1054\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #190 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Binomial Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: calibrate a binomial tail from two equal masses

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution