Concept Lesson

Probability Rules: Addition, Multiplication, and Complement Rules

A lesson in addition, multiplication, and complement rules that moves from intuition and definitions to worked reasoning, error correction, and independent practice.

Course status: Revised 2026-27 course
Updated: July 18, 2026
Practice: Easy, Tough and Toughest

Lesson Goals: Probability Rules

Probability rules must preserve the meanings of union, intersection, complement, disjointness, and independence; addition and multiplication formulas are not interchangeable.

Reader tasksample spaces, unions, intersections, disjoint events, and independence

Planned modules8

Mathematics3 expressions

Worked checks60

Boundary: Conditional probability and Bayes extensions belong in P42.

Probability notation

Probability notation in probability rules: $P(A\cup B)=0.43$ and $P(A\mid B)=0.688$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Probability notation in probability rules, For constructed events A and B in a greenhouse germination experiment, $P(A)=0.33$ , $P(B)=0.32$ , and $P(A\cap B)=0.22$ . Apply probability notation.

P(A\cup B)=0.33+0.32-0.22=0.43,\qquad P(A\mid B)=\frac{0.22}{0.32}=0.688.

When the idea is valid

For Probability notation in probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

For Probability notation in probability rules, reject this error: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Sample space

Sample space in probability rules: $P(A\cup B)=0.49$ and $P(A\mid B)=0.722$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Sample space in probability rules, For constructed events A and B in a tutoring-program evaluation, $P(A)=0.39$ , $P(B)=0.36$ , and $P(A\cap B)=0.26$ . Apply sample space.

P(A\cup B)=0.39+0.36-0.26=0.49,\qquad P(A\mid B)=\frac{0.26}{0.36}=0.722.

When the idea is valid

For Sample space in probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

For Sample space in probability rules, reject this error: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Complement rule

Complement rule in probability rules: $P(A\cup B)=0.58$ and $P(A\mid B)=0.730$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Complement rule in probability rules, For constructed events A and B in a package-delivery sample, $P(A)=0.48$ , $P(B)=0.37$ , and $P(A\cap B)=0.27$ . Apply complement rule.

P(A\cup B)=0.48+0.37-0.27=0.58,\qquad P(A\mid B)=\frac{0.27}{0.37}=0.730.

When the idea is valid

For Complement rule in probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

For Complement rule in probability rules, reject this error: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Addition rule

Addition rule in probability rules: $P(A\cup B)=0.53$ and $P(A\mid B)=0.737$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Addition rule in probability rules, For constructed events A and B in a classroom memory study, $P(A)=0.43$ , $P(B)=0.38$ , and $P(A\cap B)=0.28$ . Apply addition rule.

P(A\cup B)=0.43+0.38-0.28=0.53,\qquad P(A\mid B)=\frac{0.28}{0.38}=0.737.

When the idea is valid

For Addition rule in probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

For Addition rule in probability rules, reject this error: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Multiplication rule

Multiplication rule in probability rules: $P(A\cup B)=0.47$ and $P(A\mid B)=0.541$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Multiplication rule in probability rules, For constructed events A and B in a classroom memory study, $P(A)=0.30$ , $P(B)=0.37$ , and $P(A\cap B)=0.20$ . Apply multiplication rule.

P(A\cup B)=0.30+0.37-0.20=0.47,\qquad P(A\mid B)=\frac{0.20}{0.37}=0.541.

When the idea is valid

For Multiplication rule in probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

For Multiplication rule in probability rules, reject this error: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Mutually exclusive events

Mutually exclusive events in probability rules: $P(A\cup B)=0.56$ and $P(A\mid B)=0.655$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Mutually exclusive events in probability rules, For constructed events A and B in a quality-control inspection, $P(A)=0.46$ , $P(B)=0.29$ , and $P(A\cap B)=0.19$ . Apply mutually exclusive events.

P(A\cup B)=0.46+0.29-0.19=0.56,\qquad P(A\mid B)=\frac{0.19}{0.29}=0.655.

When the idea is valid

For Mutually exclusive events in probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

For Mutually exclusive events in probability rules, reject this error: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Worked tables and trees

Worked tables and trees in probability rules: $P(A\cup B)=0.54$ and $P(A\mid B)=0.697$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Worked tables and trees in probability rules, For constructed events A and B in a package-delivery sample, $P(A)=0.44$ , $P(B)=0.33$ , and $P(A\cap B)=0.23$ . Apply worked tables and trees.

P(A\cup B)=0.44+0.33-0.23=0.54,\qquad P(A\mid B)=\frac{0.23}{0.33}=0.697.

When the idea is valid

For Worked tables and trees in probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

For Worked tables and trees in probability rules, reject this error: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Practice

Practice in probability rules: $P(A\cup B)=0.48$ and $P(A\mid B)=0.697$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Practice in probability rules, For constructed events A and B in a water-filtration experiment, $P(A)=0.38$ , $P(B)=0.33$ , and $P(A\cap B)=0.23$ . Apply practice.

P(A\cup B)=0.38+0.33-0.23=0.48,\qquad P(A\mid B)=\frac{0.23}{0.33}=0.697.

When the idea is valid

For Practice in probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

For Practice in probability rules, reject this error: Adding P(A) and P(B) without subtracting the intersection double-counts outcomes in both events.

Formula and Notation Reference

Complement rule

P(A^c)=1-P(A)

Complement rule in Probability Rules: Define the event or random variable first; complements, conditioning, and trial assumptions determine which probability expression applies.

General addition rule

P(A\cup B)=P(A)+P(B)-P(A\cap B)

General addition rule in Probability Rules: Define the event or random variable first; complements, conditioning, and trial assumptions determine which probability expression applies.

General multiplication rule

P(A\cap B)=P(A)P(B\mid A)

General multiplication rule in Probability Rules: Define the event or random variable first; complements, conditioning, and trial assumptions determine which probability expression applies.

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Guided, Independent and Challenge Practice

Every question in Probability Rules: Addition, Multiplication, and Complement Rules is newly written from the revised framework and the logic visible in public College Board materials. Constructed numerical settings are identified as instructional scenarios and are never represented as measurements from a real population. No released or secure question wording is reproduced.

Easy Practice

Easy 1: Sample space

Question P41-Easy-1. For constructed events A and B in a recycling-behavior survey, $P(A)=0.37$ , $P(B)=0.42$ , and $P(A\cap B)=0.27$ . Apply sample space.