Concept Lesson

Conditional Probability and Independence: Formulas and Examples

A lesson in conditional probability and independence 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: Conditional Probability Rules

Conditional probability restricts the sample space to the conditioning event, and independence holds only when conditioning leaves the relevant probability unchanged.

Reader taskrestricted sample spaces, trees, tables, product rules, and independence checks

Planned modules8

Mathematics2 expressions

Worked checks60

Boundary: Disjoint events with positive probability are not independent.

Conditional probability

Conditional probability in conditional probability rules: $P(A\cup B)=0.50$ and $P(A\mid B)=0.744$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Conditional probability in conditional probability rules, For constructed events A and B in a package-delivery sample, $P(A)=0.40$ , $P(B)=0.39$ , and $P(A\cap B)=0.29$ . Apply conditional probability.

P(A\cup B)=0.40+0.39-0.29=0.50,\qquad P(A\mid B)=\frac{0.29}{0.39}=0.744.

When the idea is valid

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

Misconception to remove

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

General multiplication rule

General multiplication rule in conditional probability rules: $P(A\cup B)=0.59$ and $P(A\mid B)=0.737$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For General multiplication rule in conditional probability rules, For constructed events A and B in a package-delivery sample, $P(A)=0.49$ , $P(B)=0.38$ , and $P(A\cap B)=0.28$ . Apply general multiplication rule.

P(A\cup B)=0.49+0.38-0.28=0.59,\qquad P(A\mid B)=\frac{0.28}{0.38}=0.737.

When the idea is valid

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

Misconception to remove

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

Independent events

Independent events in conditional probability rules: $P(A\cup B)=0.44$ and $P(A\mid B)=0.630$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Independent events in conditional probability rules, For constructed events A and B in an online-course completion sample, $P(A)=0.34$ , $P(B)=0.27$ , and $P(A\cap B)=0.17$ . Apply independent events.

P(A\cup B)=0.34+0.27-0.17=0.44,\qquad P(A\mid B)=\frac{0.17}{0.27}=0.630.

When the idea is valid

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

Misconception to remove

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

Testing independence

Testing independence in conditional probability rules: $P(A\cup B)=0.43$ and $P(A\mid B)=0.667$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Testing independence in conditional probability rules, For constructed events A and B in a battery-life laboratory trial, $P(A)=0.32$ , $P(B)=0.33$ , and $P(A\cap B)=0.22$ . Apply testing independence.

P(A\cup B)=0.32+0.33-0.22=0.43,\qquad P(A\mid B)=\frac{0.22}{0.33}=0.667.

When the idea is valid

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

Misconception to remove

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

Two-way tables

Two-way tables in conditional probability rules: $P(A\cup B)=0.47$ and $P(A\mid B)=0.667$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Two-way tables in conditional probability rules, For constructed events A and B in a quality-control inspection, $P(A)=0.37$ , $P(B)=0.30$ , and $P(A\cap B)=0.20$ . Apply two-way tables.

P(A\cup B)=0.37+0.30-0.20=0.47,\qquad P(A\mid B)=\frac{0.20}{0.30}=0.667.

When the idea is valid

For Two-way tables in conditional probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

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

Tree diagrams

Tree diagrams in conditional probability rules: $P(A\cup B)=0.59$ and $P(A\mid B)=0.706$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Tree diagrams in conditional probability rules, For constructed events A and B in a package-delivery sample, $P(A)=0.49$ , $P(B)=0.34$ , and $P(A\cap B)=0.24$ . Apply tree diagrams.

P(A\cup B)=0.49+0.34-0.24=0.59,\qquad P(A\mid B)=\frac{0.24}{0.34}=0.706.

When the idea is valid

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

Misconception to remove

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

Bayes rule as enrichment

Bayes rule as enrichment in conditional probability rules: $P(A\cup B)=0.52$ and $P(A\mid B)=0.655$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Bayes rule as enrichment in conditional probability rules, For constructed events A and B in a campus dining survey, $P(A)=0.42$ , $P(B)=0.29$ , and $P(A\cap B)=0.19$ . Apply bayes rule as enrichment.

P(A\cup B)=0.42+0.29-0.19=0.52,\qquad P(A\mid B)=\frac{0.19}{0.29}=0.655.

When the idea is valid

For Bayes rule as enrichment in conditional probability rules, Check that probabilities are coherent and that the denominator event has positive probability.

Misconception to remove

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

Practice

Practice in conditional probability rules: $P(A\cup B)=0.53$ and $P(A\mid B)=0.615$ . The conditional probability restricts the reference group to outcomes where B occurs.

Worked reasoning

For Practice in conditional probability rules, For constructed events A and B in a city bus arrival investigation, $P(A)=0.43$ , $P(B)=0.26$ , and $P(A\cap B)=0.16$ . Apply practice.

P(A\cup B)=0.43+0.26-0.16=0.53,\qquad P(A\mid B)=\frac{0.16}{0.26}=0.615.

When the idea is valid

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

Misconception to remove

For Practice in conditional 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

Conditional probability

P(A\mid B)=\frac{P(A\cap B)}{P(B)}

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

Independence criterion

A\perp B\iff P(A\mid B)=P(A)

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

AdvertisementGoogle AdSense before practice placement reserved here

Guided, Independent and Challenge Practice

Every question in Conditional Probability and Independence: Formulas and Examples 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: Testing independence

Question P42-Easy-1. For constructed events A and B in a reading-speed investigation, $P(A)=0.45$ , $P(B)=0.38$ , and $P(A\cap B)=0.28$ . Apply testing independence.