Worked solution · Exam P

SOA Exam P Sample Question #620 Solution

This Exam P sample reference tests Independence. More than two successes among four independent opportunities means exactly three or exactly four. Enumerating the four possible single-failure cases gives 0.27525, while all four successes has probability 0.07425. Their sum is 0.3495, which rounds to choice B.

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 Independence question

Setup

Represent each outcome by an independent Bernoulli indicator with the four supplied success probabilities.

(p_1,p_2,p_3,p_4)=(0.55,0.45,0.50,0.60)

\Pr(S>2)=\Pr(S=3)+\Pr(S=4)

Model

For exactly three successes, sum the four disjoint cases according to which opportunity fails.

\Pr(S=3)=0.45(0.45)(0.50)(0.60)+0.55(0.55)(0.50)(0.60)

\qquad\quad+0.55(0.45)(0.50)(0.60)+0.55(0.45)(0.50)(0.40)=0.27525

Compute

Add the all-success case to the exactly-three probability.

\Pr(S=4)=0.55(0.45)(0.50)(0.60)=0.07425

\Pr(S>2)=0.27525+0.07425=0.34950

Answer

The probability rounds to 0.35.

\boxed{0.35\quad\text{(B)}}

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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.

AThis is the probability of exactly three successes, rounded from 0.27525, and it omits the all-four-success case.
CReplacing all four unequal probabilities by 0.55 gives a binomial upper tail of about 0.391 instead of the required heterogeneous calculation.
DSumming the four three-way success intersections without requiring the remaining outcome to fail gives 0.57225 and counts the all-success outcome four times.
EThis is the probability of at least two successes, 0.72575, which uses a threshold one success too low.

Original practice · fully worked

Original variant: conditional vial audit

Eight sterile culture vials are tested independently, and each vial passes with probability 0.75. An audit report states that at least six of the eight vials passed. Conditional on that report, calculate the probability that exactly seven vials passed.

A
0.148
B
0.267
C
0.393
D
0.459
E
0.679

Answer and calculation

Within the conditioning set, the binomial masses for six, seven, and eight passes have proportional weights 28, 24, and 9. The required conditional probability is therefore 24/61=0.393443, so choice C is correct.

Setup

Let X be the number of passing vials. The common success probability makes X binomial.

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

Model

The audit restricts the possible counts to six, seven, or eight. Factor the same power of 0.75 and 0.25 from their three binomial masses.

\Pr(X=6):\Pr(X=7):\Pr(X=8)=28:24:9

Compute

Normalize the weight for seven passes over the three counts allowed by the report.

\Pr(X=7\mid X\ge6)=\frac{24}{28+24+9}

\Pr(X=7\mid X\ge6)=\frac{24}{61}=0.3934426229\ldots

Answer

The conditional probability rounds to 0.393.

\boxed{0.393\quad\text{(C)}}

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How to solve this Independence question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: conditional vial audit

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability