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6.7 Probability and Sets

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1 Chapter 1: ACCUPLACER Suite Overview & Testing Mechanics

1.1 The ACCUPLACER Testing System & Computer-Adaptive Mechanics 1.2 Placement Policies, Score Scales & Institutional Cut Scores 1.3 Test-Taking Strategies, Tool Mechanics & Pacing

2 Chapter 2: Next-Generation Reading

2.1 The Reading Test Blueprint: Set-Based and Discrete Items 2.2 Information and Ideas: Close Reading & Central Themes 2.3 Summarizing and Understanding Relationships 2.4 Rhetoric, Text Structure & Author's Purpose 2.5 Analyzing Point of View and Authorial Purpose 2.6 Analyzing Arguments: Claims, Counterclaims, and Evidence 2.7 Synthesis & Paired Text Comparative Analysis 2.8 Vocabulary in Context & Figurative Language 2.9 Passage Genre, Content Area, Mode, and Text Complexity 2.10 Answer-Choice Traps and Evidence Discipline

3 Chapter 3: Next-Generation Writing: Expression of Ideas

3.1 The Writing Test Blueprint: Passage Sets and Content Distribution 3.2 Development: Proposition and Supporting Evidence 3.3 Focus: Adding, Retaining, and Deleting for Relevance 3.4 Organization: Logical Sequence and Paragraph Order 3.5 Introductions, Conclusions, and Transitional Cohesion 3.6 Effective Language Use: Precision and Word Choice 3.7 Concision: Eliminating Wordiness and Redundancy 3.8 Style, Tone, and Voice Consistency 3.9 Syntax: Sentence Combining for Rhetorical Purpose

4 Chapter 4: Next-Generation Writing: Standard English Conventions

4.1 Sentence Boundaries: Fragments, Run-ons, and Comma Splices 4.2 Subordination and Coordination 4.3 Parallel Structure 4.4 Modifier Placement: Misplaced and Dangling Modifiers 4.5 Inappropriate Shifts in Verb Tense 4.6 Inappropriate Shifts in Verb Voice and Mood 4.7 Inappropriate Shifts in Pronoun Person and Number 4.8 Subject-Verb Agreement, Pronoun Case, and Verb Forms 4.9 Possessive Determiners, Contractions, and Confusable Adverbs 4.10 Pronoun Clarity and Pronoun-Antecedent Agreement 4.11 Frequently Confused Words, Noun Agreement, and Logical Comparison 4.12 Within-Sentence Punctuation: Semicolons, Colons, Dashes, and Apostrophes 4.13 Items in a Series and Nonrestrictive Elements 4.14 End-of-Sentence Punctuation, Hyphenation, and Unnecessary Punctuation

5 Chapter 5: Next-Generation Arithmetic

5.1 Whole Number Operations, Estimation & Multi-Step Applications 5.2 Fractions, Mixed Numbers & Operations 5.3 Decimal Operations & Place-Value Conversions 5.4 Percent Calculations, Increase, Decrease, and Applications 5.5 Number Comparisons and Equivalents Across Formats

6 Chapter 6: Next-Generation Quantitative Reasoning, Algebra, and Statistics (QAS)

6.1 Rational Numbers, Absolute Value, and Order of Operations 6.2 Exponents, Radicals, and Scientific Notation 6.3 Ratio, Proportional Relationships, and Unit Conversions 6.4 Creating and Evaluating Algebraic Expressions 6.5 Linear Equations, Inequalities, and Systems of Two Equations 6.6 Linear Applications, Functions, and Graphs 6.7 Probability and Sets 6.8 Descriptive Statistics: Center, Spread, and Graphical Displays 6.9 Geometry for Pre-Algebra and Algebra 1

7 Chapter 7: Next-Generation Advanced Algebra, Quadratics & Polynomials

7.1 Advanced Linear Equations, Systems, and Coordinate Geometry 7.2 Factoring Methods for Quadratics, Cubics, and Polynomials 7.3 Quadratic Equations, Vertex Form & Parabolic Functions 7.4 Polynomial Equations, Roots, and Graphs 7.5 Radical and Rational Equations and Functions 7.6 Geometry for Algebra 1 and Algebra 2

8 Chapter 8: Functions, Exponentials, Logarithms & Trigonometry

8.1 Function Notation, Transformations, Compositions & Inverses 8.2 Domain, Range, and Interpreting Functions in Context 8.3 Exponential & Logarithmic Functions, Laws & Modeling 8.4 Right-Triangle Trigonometry, the Unit Circle, and Radian Measure 8.5 Law of Sines, Law of Cosines, and Trigonometric Equations 8.6 Choosing Your Math Test: Arithmetic, QAS, or AAF

9 Chapter 9: WritePlacer & WritePlacer ESL Essays

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10 Chapter 10: ACCUPLACER ESL Placement Suite

10.1 The ESL Placement Suite: Structure, Scoring, and Score Interpretation 10.2 ESL Reading Skills & Passage Comprehension 10.3 ESL Sentence Meaning & Context Clues 10.4 ESL Language Use & Grammatical Mechanics 10.5 ESL Listening Comprehension & Spoken Contexts

6.7 Probability and Sets

Key Takeaways Explain This Topic AI

Probability and sets carries 1 to 3 of the 20 QAS questions.

Probability equals favorable outcomes divided by total equally likely outcomes and always falls between 0 and 1.

Independent events multiply; the key test is whether the first outcome changes the second's probability.

Without replacement, both the favorable count and the total shrink, which is the most common compound-probability error.

The addition rule subtracts the overlap, so P(A or B) equals P(A) plus P(B) minus P(A and B).

Last updated: August 2026

A Small Area With Reliable Traps

Probability and sets: Calculating probability (simple, compound, and conditional), and defining sample spaces and events using set notation.

At 1 to 3 of 20 questions this is one of the smaller QAS areas, but its error rate is high because the traps are conceptual rather than computational.

Sets and Sample Spaces

A set is a collection of distinct elements, written in braces: A = { 2 , 4 , 6 , 8 } A = \{2, 4, 6, 8\} A = { 2 , 4 , 6 , 8 } .

The sample space S S S is the set of all possible outcomes. An event is a subset of S S S .

Rolling one die: S = { 1 , 2 , 3 , 4 , 5 , 6 } S = \{1, 2, 3, 4, 5, 6\} S = { 1 , 2 , 3 , 4 , 5 , 6 } , and "rolling even" is the event E = { 2 , 4 , 6 } E = \{2, 4, 6\} E = { 2 , 4 , 6 } .

Set Operations

  • Operation Symbol Meaning
  • Union A ∪ B A \cup B A ∪ B in A or B (or both)
  • Intersection A ∩ B A \cap B A ∩ B in both A and B
  • Complement A ′ A' A ′ or A c A^c A c not in A
  • Empty set ∅ \emptyset ∅ no elements

With A = { 1 , 2 , 3 , 4 } A = \{1,2,3,4\} A = { 1 , 2 , 3 , 4 } and B = { 3 , 4 , 5 , 6 } B = \{3,4,5,6\} B = { 3 , 4 , 5 , 6 } :

  • A ∪ B = { 1 , 2 , 3 , 4 , 5 , 6 } A \cup B = \{1,2,3,4,5,6\} A ∪ B = { 1 , 2 , 3 , 4 , 5 , 6 }
  • A ∩ B = { 3 , 4 } A \cap B = \{3,4\} A ∩ B = { 3 , 4 }

Note that union does not double-count 3 and 4. Sets contain distinct elements, and this non-duplication is exactly why the probability addition rule subtracts an overlap.

Simple Probability P ( E ) = number of favorable outcomes total number of equally likely outcomes P(E) = \frac{\text{number of favorable outcomes}}{\text{total number of equally likely outcomes}} P ( E ) = total number of equally likely outcomes number of favorable outcomes ​

Every probability satisfies 0 ≤ P ( E ) ≤ 1 0 \le P(E) \le 1 0 ≤ P ( E ) ≤ 1 . An answer choice outside that interval can be eliminated on sight.

A bag holds 5 red, 3 blue, and 4 green marbles. P(blue)? Total = 12 = 12 = 12 . P ( blue ) = 3 12 = 1 4 P(\text{blue}) = \dfrac{3}{12} = \dfrac{1}{4} P ( blue ) = 12 3 ​ = 4 1 ​

The Complement Rule P ( not E ) = 1 − P ( E ) P(\text{not } E) = 1 - P(E) P ( not E ) = 1 − P ( E )

This is the single most useful shortcut in the area. Whenever a question says "at least one," compute the complement:

Flip a fair coin 3 times. P(at least one head)?

Directly, you would count 7 favorable outcomes out of 8. Via the complement:

P ( no heads ) = P ( T T T ) = ( 1 2 ) 3 = 1 8 P(\text{no heads}) = P(TTT) = \left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8} P ( no heads ) = P ( TTT ) = ( 2 1 ​ ) 3 = 8 1 ​ P ( at least one head ) = 1 − 1 8 = 7 8 P(\text{at least one head}) = 1 - \tfrac{1}{8} = \tfrac{7}{8} P ( at least one head ) = 1 − 8 1 ​ = 8 7 ​

  • Compound Probability
  • Independent Events — Multiply

Events are independent when the first outcome does not change the second's probability. P ( A and B ) = P ( A ) ⋅ P ( B ) P(A \text{ and } B) = P(A) \cdot P(B) P ( A and B ) = P ( A ) ⋅ P ( B )

Two fair dice. P(both show 4)? 1 6 ⋅ 1 6 = 1 36 \tfrac{1}{6} \cdot \tfrac{1}{6} = \tfrac{1}{36} 6 1 ​ ⋅ 6 1 ​ = 36 1 ​

Dependent Events — With and Without Replacement

This is where most errors occur.

A bag holds 5 red and 3 blue marbles. Two are drawn.

With replacement (first marble returned):

P ( both red ) = 5 8 ⋅ 5 8 = 25 64 P(\text{both red}) = \tfrac{5}{8} \cdot \tfrac{5}{8} = \tfrac{25}{64} P ( both red ) = 8 5 ​ ⋅ 8 5 ​ = 64 25 ​

Without replacement (first marble kept):

P ( both red ) = 5 8 ⋅ 4 7 = 20 56 = 5 14 P(\text{both red}) = \tfrac{5}{8} \cdot \tfrac{4}{7} = \tfrac{20}{56} = \tfrac{5}{14} P ( both red ) = 8 5 ​ ⋅ 7 4 ​ = 56 20 ​ = 14 5 ​

On the second draw both numbers change : one red is gone (5 → 4) and one marble total is gone (8 → 7). The classic error updates only the numerator, giving 5 8 ⋅ 4 8 \tfrac{5}{8} \cdot \tfrac{4}{8} 8 5 ​ ⋅ 8 4 ​ .

Signal words: without replacement , does not replace , keeps , sets aside , eats . Also, any scenario drawing people from a group is inherently without replacement — the same person cannot be chosen twice.

The Addition Rule P ( A or B ) = P ( A ) + P ( B ) − P ( A and B ) P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) P ( A or B ) = P ( A ) + P ( B ) − P ( A and B )

The subtraction removes the double-counted overlap.

Draw one card from a standard 52-card deck. P(heart or face card)?

  • P ( heart ) = 13 52 P(\text{heart}) = \tfrac{13}{52} P ( heart ) = 52 13 ​
  • P ( face card ) = 12 52 P(\text{face card}) = \tfrac{12}{52} P ( face card ) = 52 12 ​ (J, Q, K in four suits)
  • P ( both ) = 3 52 P(\text{both}) = \tfrac{3}{52} P ( both ) = 52 3 ​ (J, Q, K of hearts)

P = 13 52 + 12 52 − 3 52 = 22 52 = 11 26 P = \tfrac{13}{52} + \tfrac{12}{52} - \tfrac{3}{52} = \tfrac{22}{52} = \tfrac{11}{26} P = 52 13 ​ + 52 12 ​ − 52 3 ​ = 52 22 ​ = 26 11 ​

Skipping the subtraction gives 25 52 \tfrac{25}{52} 52 25 ​ — a distractor that will be offered.

When events are mutually exclusive (cannot both occur), the overlap is 0 and the rule reduces to simple addition.

Conditional Probability

P ( B ∣ A ) P(B \mid A) P ( B ∣ A ) is the probability of B given that A has occurred . Conditioning restricts the sample space to the outcomes where A is true.

Of 40 students, 22 take Spanish, 18 take chemistry, and 10 take both. Given that a student takes Spanish, what is the probability they take chemistry?

Restrict to the 22 Spanish students. Of those, 10 also take chemistry:

P ( chem ∣ Spanish ) = 10 22 = 5 11 P(\text{chem} \mid \text{Spanish}) = \dfrac{10}{22} = \dfrac{5}{11} P ( chem ∣ Spanish ) = 22 10 ​ = 11 5 ​

The denominator is 22, not 40 . Using the full group is the defining error in conditional probability, and it is what the item is written to detect.

Two-Way Tables

Many items present a table. Read the denominator from the question's condition:

  • Passed Failed Total
  • Morning 34 6 40
  • Evening 21 9 30
  • Total 55 15 70

P ( passed ) P(\text{passed}) P ( passed ) → denominator 70 → 55 70 = 11 14 \tfrac{55}{70} = \tfrac{11}{14} 70 55 ​ = 14 11 ​

P ( passed ∣ evening ) P(\text{passed} \mid \text{evening}) P ( passed ∣ evening ) → denominator 30 → 21 30 = 7 10 \tfrac{21}{30} = \tfrac{7}{10} 30 21 ​ = 10 7 ​

P ( evening ∣ passed ) P(\text{evening} \mid \text{passed}) P ( evening ∣ passed ) → denominator 55 → 21 55 \tfrac{21}{55} 55 21 ​

The last two use the same cell with different denominators. The word after "given" names the denominator.