Worked solution · Exam P

SOA Exam P Sample Question #551 Solution

This Exam P sample reference tests Binomial Distribution. The number of accident-producing errors is binomial with four trials and probability 0.30. An accident is unreimbursed only when that count exceeds two, so the required tail is 0.0837 and choice B is correct.

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

Let N count the errors that produce accidents. Independence and the common accident probability give a binomial count.

N\sim\operatorname{Binomial}(4,0.30)

Model

At most two resulting accidents are reimbursed, so an unreimbursed accident occurs exactly when the count is three or four.

\Pr(\text{at least one unreimbursed accident})=\Pr(N\ge3)

\Pr(N\ge3)=\Pr(N=3)+\Pr(N=4)

Compute

Evaluate the two upper-tail masses.

\Pr(N\ge3)=\binom43(0.30)^3(0.70)+(0.30)^4

\Pr(N\ge3)=0.0756+0.0081=0.0837

Answer

The probability rounds to 0.08 at the precision of the choices.

\boxed{0.08\quad\text{(B)}}

Test the method on unseen problems

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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.

AThe value 0.03 comes from (0.30)³=0.027, which fixes three particular errors as accidents and omits both their possible locations and the four-accident case.
CThe value 0.24 is (0.70)⁴=0.2401, the probability that none of the four errors produces an accident, not an upper-tail probability.
DThe value 0.32 is close to P(N=0)+P(N=3)=0.2401+0.0756=0.3157; the no-accident case cannot create an unreimbursed accident.
EThe value 0.41 is P(N=1)=4(0.30)(0.70)³=0.4116, which describes exactly one accident and remains entirely within the reimbursement limit.

Original practice · fully worked

Original variant: conditional batch acceptance

Eight independent sensors are tested, and each sensor passes with probability 0.75. A shipment is released only when at least six sensors pass. Given that the shipment is released, calculate the probability that exactly seven sensors passed.

A
0.2670
B
0.3934
C
0.6785
D
0.7273
E
0.7500

Answer and calculation

Condition the seven-pass mass on the release event. Dividing P(X=7) by P(X at least 6) gives 24/61=0.3934, so choice B is correct.

Setup

Let X be the number of sensors that pass the independent tests.

X\sim\operatorname{Binomial}(8,0.75)

Model

Exactly seven passes is contained in the release event, so the conditional probability is a ratio of one mass to an upper tail.

\Pr(X=7\mid X\ge6)=\frac{\Pr(X=7)}{\Pr(X=6)+\Pr(X=7)+\Pr(X=8)}

Compute

Evaluate the three binomial masses and form the ratio.

\Pr(X=6)=0.3114624023,\quad \Pr(X=7)=0.2669677734,\quad \Pr(X=8)=0.1001129150

\Pr(X=7\mid X\ge6)=\frac{0.2669677734}{0.6785430908}=0.3934426230

Answer

Conditional on release, the probability of exactly seven passes is approximately 0.3934.

\boxed{0.3934\quad\text{(B)}}

Keep practicing

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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: conditional batch acceptance

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables