The Central Idea
Let \(A\) and \(B\) be events with \(P(B)>0\).
If knowing that \(B\) occurred leaves the probability of \(A\) unchanged, then
\[ \boxed{P(A\mid B)=P(A)}. \]In that case, \(A\) and \(B\) are independent.
If instead
\[ P(A\mid B)\neq P(A), \]then learning \(B\) changes the probability of \(A\), and the events are dependent.
Independence as “No Probability Update”
Conditional probability asks how a probability changes when information is supplied.
Independence occurs when the update does nothing:
\[ P(A)\longrightarrow P(A\mid B)=P(A). \]Dependence occurs when the update changes the value:
\[ P(A)\longrightarrow P(A\mid B)\neq P(A). \]This makes independence fundamentally an information relationship.
A First Independent Example
Roll a fair six-sided die.
Let
\[ A=\{\text{even result}\} \]and
\[ B=\{\text{result greater than }2\}. \]Unconditionally,
\[ P(A)=\frac{3}{6}=\frac{1}{2}. \]Given \(B\), the possible outcomes are
\[ \{3,4,5,6\}, \]of which \(4\) and \(6\) are even. Therefore,
\[ P(A\mid B)=\frac{2}{4}=\frac{1}{2}. \]Since
\[ P(A\mid B)=P(A), \]\(A\) and \(B\) are independent.
A First Dependent Example
Draw one card from a standard \(52\)-card deck.
Let \(A\) be the event that the card is a king and \(B\) the event that the card is a face card.
Then
\[ P(A)=\frac{4}{52}=\frac{1}{13}. \]Given that the card is a face card, there are \(12\) possible face cards, \(4\) of which are kings:
\[ P(A\mid B)=\frac{4}{12}=\frac{1}{3}. \]Since
\[ \frac{1}{3}\neq\frac{1}{13}, \]the events are dependent.
The Product Criterion for Independence
From the definition of conditional probability,
\[ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. \]If \(A\) and \(B\) are independent, then
\[ P(A\mid B)=P(A). \]Therefore,
\[ \frac{P(A\cap B)}{P(B)}=P(A). \]Multiplying by \(P(B)\) gives
\[ \boxed{P(A\cap B)=P(A)P(B)}. \]This product equality is one of the most useful tests for independence.
Why the Product Criterion Is Especially Useful
The conditional definition
\[ P(A\mid B)=P(A) \]requires \(P(B)>0\).
By contrast, the product criterion
\[ \boxed{P(A\cap B)=P(A)P(B)} \]works as the general definition of independence for any two events, including cases where one event has probability zero.
Equivalent Tests When Probabilities Are Positive
When the relevant conditional probabilities are defined, the following statements are equivalent:
- \(A\) and \(B\) are independent;
- \(P(A\mid B)=P(A)\);
- \(P(B\mid A)=P(B)\);
- \(P(A\cap B)=P(A)P(B)\).
Any one of these equalities can be used to verify independence.
Independence Is Symmetric
If \(A\) is independent of \(B\), then \(B\) is independent of \(A\).
The product criterion makes this immediate:
\[ P(A\cap B)=P(A)P(B). \]Since
\[ A\cap B=B\cap A \]and multiplication is commutative,
\[ P(A)P(B)=P(B)P(A). \]Independence does not have a direction.
Dependence Means Information Changes Probability
Events \(A\) and \(B\) are dependent when
\[ P(A\cap B)\neq P(A)P(B). \]Equivalently, whenever the conditional probability is defined,
\[ P(A\mid B)\neq P(A). \]Dependence can make \(A\) either more likely or less likely after \(B\) is known.
Positive and Negative Association
If
\[ P(A\mid B)>P(A), \]then \(B\) makes \(A\) more likely.
If
\[ P(A\mid B)Both situations represent dependence.
Example: Positive Association
Suppose \(40\%\) of customers purchase a premium plan:
\[ P(P)=0.40. \]Among customers who attend a product demonstration,
\[ P(P\mid D)=0.65. \]Since
\[ 0.65>0.40, \]premium purchase and demonstration attendance are dependent in the observed data, with premium purchase more common among demonstration attendees.
Example: Negative Association
Suppose a machine has an overall failure probability of
\[ P(F)=0.08. \]Among machines that received preventive maintenance,
\[ P(F\mid M)=0.03. \]Since
\[ 0.03<0.08, \]failure and maintenance status are dependent in the observed data.
The probability comparison alone does not establish why the difference occurs.
Independence Does Not Mean Mutual Exclusivity
Independent events can occur together.
In fact, if \(A\) and \(B\) are independent and both have positive probability, then
\[ P(A\cap B)=P(A)P(B)>0. \]Therefore, their intersection must have positive probability.
Mutually Exclusive Events Are Usually Dependent
Suppose \(A\) and \(B\) are mutually exclusive:
\[ A\cap B=\varnothing. \]Then
\[ P(A\cap B)=0. \]If \(P(A)>0\) and \(P(B)>0\), then
\[ P(A)P(B)>0. \]Thus,
\[ P(A\cap B)\neq P(A)P(B), \]so the events are dependent.
Example: Mutually Exclusive but Not Independent
Roll a fair die. Let
\[ A=\{1\} \]and
\[ B=\{2\}. \]These events are mutually exclusive, so
\[ P(A\cap B)=0. \]But
\[ P(A)P(B)=\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}. \]Therefore,
\[ 0\neq\frac{1}{36}, \]and the events are dependent.
Independent Events Can Overlap
Roll a fair die and let
\[ A=\{2,4,6\} \]and
\[ B=\{3,4,5,6\}. \]Their intersection is
\[ A\cap B=\{4,6\}, \]so the events clearly overlap.
Yet
\[ P(A\cap B)=\frac{2}{6}=\frac{1}{3} \]and
\[ P(A)P(B)=\frac{1}{2}\cdot\frac{4}{6}=\frac{1}{3}. \]Therefore they are independent.
Testing Independence from Given Probabilities
Suppose
\[ P(A)=0.60,\qquad P(B)=0.35,\qquad P(A\cap B)=0.21. \]Compute
\[ P(A)P(B)=0.60(0.35)=0.21. \]Since
\[ P(A\cap B)=P(A)P(B), \]\(A\) and \(B\) are independent.
Testing Dependence from Given Probabilities
Suppose
\[ P(A)=0.50,\qquad P(B)=0.40,\qquad P(A\cap B)=0.30. \]If the events were independent, we would have
\[ P(A)P(B)=0.50(0.40)=0.20. \]But
\[ 0.30\neq0.20. \]Therefore the events are dependent.
Testing Independence with Conditional Probability
Suppose
\[ P(A)=0.45 \]and
\[ P(A\mid B)=0.45. \]If \(P(B)>0\), then
\[ P(A\mid B)=P(A), \]so \(A\) and \(B\) are independent.
Testing Dependence with Conditional Probability
Suppose
\[ P(A)=0.45 \]but
\[ P(A\mid B)=0.72. \]The condition changes the probability substantially:
\[ 0.72\neq0.45. \]Therefore \(A\) and \(B\) are dependent.
Independence and Complements
If \(A\) and \(B\) are independent, then their complements inherit independence relationships.
In particular:
- \(A\) and \(B^c\) are independent;
- \(A^c\) and \(B\) are independent;
- \(A^c\) and \(B^c\) are independent.
Why Independence Extends to a Complement
Suppose \(A\) and \(B\) are independent. Then
\[ P(A\cap B)=P(A)P(B). \]Since \(A\) is partitioned by \(B\) and \(B^c\),
\[ P(A)=P(A\cap B)+P(A\cap B^c). \]Therefore,
\[ P(A\cap B^c) = P(A)-P(A)P(B). \]Factor \(P(A)\):
\[ P(A\cap B^c) = P(A)[1-P(B)]. \]Since \(1-P(B)=P(B^c)\),
\[ \boxed{P(A\cap B^c)=P(A)P(B^c)}. \]Thus \(A\) and \(B^c\) are independent.
Independent Repeated Trials
Repeated trials are often modeled as independent when one trial does not alter the probability distribution of another.
For example, if a fair coin is tossed twice, let
\[ A=\{\text{first toss is heads}\} \]and
\[ B=\{\text{second toss is heads}\}. \]Then
\[ P(A)=\frac{1}{2},\qquad P(B)=\frac{1}{2}, \]and
\[ P(A\cap B)=\frac{1}{4}. \]Since
\[ \frac{1}{4} = \frac{1}{2}\cdot\frac{1}{2}, \]the events are independent.
Without Replacement Usually Creates Dependence
A box contains \(5\) red balls and \(3\) blue balls. Two balls are drawn without replacement.
Let \(A\) be the event that the first ball is red and \(B\) the event that the second ball is red.
Initially,
\[ P(B)=\frac{5}{8}. \]Given that the first ball was red,
\[ P(B\mid A)=\frac{4}{7}. \]Since
\[ \frac{4}{7}\neq\frac{5}{8}, \]the two events are dependent.
Replacement Can Restore Independence
Suppose the first ball is returned to the box before the second draw.
Then the composition returns to
\[ 5\text{ red},\qquad3\text{ blue}. \]Therefore,
\[ P(B\mid A)=\frac{5}{8}=P(B). \]Under this replacement process, the color outcomes of the two draws are independent.
Independence Is a Modeling Claim
In theoretical problems, independence may be stated as part of the model.
In applied work, however, independence should not be assumed merely because it would make calculations easier.
A modeler must justify independence through:
- the design of the process;
- physical or structural separation;
- randomization;
- domain knowledge;
- or evidence consistent with the independence condition.
Physical Separation Does Not Automatically Imply Independence
Two events can occur in different places or at different times and still be statistically dependent.
For example, sales in two regions may both respond to the same national economic conditions.
Independence is determined by probability relationships, not by physical distance.
Different Causes Do Not Automatically Imply Independence
Two outcomes can have different immediate causes while still sharing a common influence.
Conversely, events generated by the same system can sometimes be modeled as independent.
The relevant question remains:
\[ \boxed{\text{Does knowing one event change the probability of the other?}} \]Independence from a Two-Way Table
Suppose \(200\) customers are classified by whether they used a trial and whether they purchased:
| Purchased | Did Not Purchase | Total | |
|---|---|---|---|
| Used Trial | 72 | 48 | 120 |
| No Trial | 24 | 56 | 80 |
| Total | 96 | 104 | 200 |
Testing the Table with Conditional Probability
The overall purchase probability is
\[ P(P)=\frac{96}{200}=0.48. \]Among trial users,
\[ P(P\mid T)=\frac{72}{120}=0.60. \]Since
\[ 0.60\neq0.48, \]trial use and purchase are dependent in this table.
Testing the Same Table with the Product Criterion
From the table,
\[ P(T)=\frac{120}{200}=0.60, \] \[ P(P)=\frac{96}{200}=0.48, \]and
\[ P(T\cap P)=\frac{72}{200}=0.36. \]If the events were independent,
\[ P(T)P(P)=0.60(0.48)=0.288. \]But
\[ 0.36\neq0.288. \]Therefore they are dependent.
What an Independent Table Would Look Like
Suppose a different table has
\[ P(A)=0.40 \]and
\[ P(B)=0.30. \]If \(A\) and \(B\) are independent, then the intersection must satisfy
\[ P(A\cap B)=0.40(0.30)=\boxed{0.12}. \]The observed joint proportion would need to be exactly the product of the marginal proportions in an exact probability model.
Pairwise Independence
For three events \(A\), \(B\), and \(C\), we may ask whether each pair is independent:
\[ P(A\cap B)=P(A)P(B), \] \[ P(A\cap C)=P(A)P(C), \] \[ P(B\cap C)=P(B)P(C). \]If all three equalities hold, the events are pairwise independent.
Mutual Independence Is Stronger
Three events are mutually independent only if all pairwise conditions hold and the three-way intersection also satisfies
\[ \boxed{ P(A\cap B\cap C)=P(A)P(B)P(C) }. \]Thus pairwise independence alone does not guarantee full mutual independence.
Example: Pairwise Independent but Not Mutually Independent
Toss two fair coins. Let:
- \(A\): the first coin is heads;
- \(B\): the second coin is heads;
- \(C\): the two coins show the same result.
Each event has probability
\[ \frac{1}{2}. \]Also,
\[ P(A\cap B)=\frac{1}{4} = \frac{1}{2}\cdot\frac{1}{2}, \]and similarly each other pair is independent.
However,
\[ A\cap B\cap C=\{HH\}, \]so
\[ P(A\cap B\cap C)=\frac{1}{4}. \]But
\[ P(A)P(B)P(C)=\frac{1}{8}. \]Therefore the three events are pairwise independent but not mutually independent.
Independence of Several Events
For several events, mutual independence requires the product rule to hold for every nontrivial subcollection of the events.
For example, four events require appropriate equalities for:
- every pair;
- every triple;
- the intersection of all four.
This requirement prevents hidden higher-order dependence from being missed.
Independence and Repeated Bernoulli Trials
Many probability models describe repeated trials with two outcomes, such as success and failure.
A common model assumes that:
- each trial has the same success probability;
- the outcome of one trial does not change the probability on another.
The second assumption is independence.
This structure will later support binomial and related probability models.
Observed Data and Exact Independence
In a theoretical model, independence is expressed by an exact equality:
\[ P(A\cap B)=P(A)P(B). \]In empirical data, observed proportions rarely match this equality perfectly because of sampling variation.
Deciding whether an observed difference is meaningful requires statistical methods beyond simply checking whether two sample proportions are exactly equal.
Dependence Does Not Establish Causation
If \(A\) and \(B\) are dependent, knowing one gives information about the other.
That does not prove that \(A\) causes \(B\), that \(B\) causes \(A\), or even that either directly causes the other.
Dependence may arise from:
- a direct causal effect;
- reverse causation;
- a shared underlying factor;
- selection effects;
- measurement structure;
- or other relationships.
Independence and Conditional Reasoning
The expression
\[ P(A\mid B)=P(A) \]says that conditioning on \(B\) provides no probability-relevant information about \(A\).
This simple equality becomes important throughout probability, statistics, stochastic processes, Bayesian reasoning, and applied modeling.
Common Error: Assuming Separate Events Are Independent
Events may be distinct without being independent.
For example, drawing a red card first and drawing a red card second are different events. If cards are drawn without replacement, however, the first draw changes the composition of the deck and therefore changes the probability of the second.
Distinct does not mean independent.
Common Error: Confusing Independent with Mutually Exclusive
Mutual exclusivity means
\[ P(A\cap B)=0. \]Independence means
\[ P(A\cap B)=P(A)P(B). \]These are fundamentally different conditions.
Common Error: Checking Only Marginal Probabilities
Knowing
\[ P(A) \]and
\[ P(B) \]is not enough to decide whether \(A\) and \(B\) are independent.
We also need information about their intersection or a relevant conditional probability.
Common Error: Using the Product Rule Without Independence
The equality
\[ P(A\cap B)=P(A)P(B) \]is valid precisely when \(A\) and \(B\) are independent.
For dependent events, the correct relationship uses conditional probability:
\[ P(A\cap B)=P(B)P(A\mid B). \]The next lesson develops these multiplication rules systematically.
Common Error: Assuming Independence from Similar Probabilities
Suppose
\[ P(A)=0.50 \]and
\[ P(B)=0.50. \]The fact that the marginal probabilities are equal does not imply independence.
We must still check whether
\[ P(A\cap B)=0.25. \]Common Error: Treating Approximate Data Equality as Exact Proof
If a sample gives
\[ \widehat{P}(A)=0.51 \]and
\[ \widehat{P}(A\mid B)=0.50, \]the values are close, but sample data alone do not establish exact independence.
Statistical inference is required to judge whether an observed difference is compatible with an independence model.
A Reliable Procedure for Two Events
- Identify the two events.
- Determine which probabilities are available.
- Use \(P(A\mid B)=P(A)\) when a conditional probability is convenient.
- Use \(P(A\cap B)=P(A)P(B)\) when joint and marginal probabilities are available.
- If the relevant equality holds, classify the events as independent.
- If it fails, classify them as dependent.
- Interpret dependence as a probability change, not automatically as causation.
Example: Full Workflow with Conditional Probability
Suppose
\[ P(A)=0.30 \]and
\[ P(A\mid B)=0.42. \]Compare the values:
\[ 0.42\neq0.30. \]Therefore, \(A\) and \(B\) are dependent.
Because
\[ P(A\mid B)>P(A), \]observing \(B\) increases the probability of \(A\).
Example: Full Workflow with the Product Criterion
Suppose
\[ P(A)=0.25,\qquad P(B)=0.60,\qquad P(A\cap B)=0.15. \]Compute
\[ P(A)P(B)=0.25(0.60)=0.15. \]Since
\[ P(A\cap B)=P(A)P(B), \]the events are independent.
Practice 1: Conditional Test
Suppose
\[ P(A)=0.40 \]and
\[ P(A\mid B)=0.40. \]Assuming \(P(B)>0\), determine whether \(A\) and \(B\) are independent.
Practice 2: Conditional Dependence
Suppose
\[ P(A)=0.35 \]and
\[ P(A\mid B)=0.52. \]Determine whether \(A\) and \(B\) are independent or dependent.
Practice 3: Product Test
Suppose
\[ P(A)=0.50,\qquad P(B)=0.30,\qquad P(A\cap B)=0.15. \]Determine whether \(A\) and \(B\) are independent.
Practice 4: Product Test for Dependence
Suppose
\[ P(A)=0.50,\qquad P(B)=0.30,\qquad P(A\cap B)=0.20. \]Determine whether \(A\) and \(B\) are independent.
Practice 5: Mutual Exclusivity
Events \(A\) and \(B\) are mutually exclusive with
\[ P(A)=0.20,\qquad P(B)=0.40. \]Are the events independent?
Practice 6: Without Replacement
A bag contains \(7\) black tokens and \(3\) white tokens. Two tokens are drawn without replacement.
Let \(A\) be the event that the first token is black and \(B\) the event that the second token is black.
Determine whether \(A\) and \(B\) are independent.
Practice 7: With Replacement
Repeat Practice 6, but replace the first token before drawing the second. Determine whether \(A\) and \(B\) are independent.
Practice 8: Two-Way Table
A survey of \(100\) people gives:
| Uses App | Does Not Use App | Total | |
|---|---|---|---|
| Subscriber | 30 | 20 | 50 |
| Not Subscriber | 30 | 20 | 50 |
| Total | 60 | 40 | 100 |
Determine whether subscription status and app use are independent.
Practice 9: Complements
Suppose \(A\) and \(B\) are independent, with
\[ P(A)=0.70,\qquad P(B)=0.25. \]Find
\[ P(A\cap B^c). \]Practice 10: Pairwise Versus Mutual Independence
Three events are pairwise independent. Is that information alone sufficient to conclude that
\[ P(A\cap B\cap C)=P(A)P(B)P(C)? \]Explain.
Practice Answers
Practice 1
Since
\[ P(A\mid B)=P(A)=0.40, \]the events are independent.
Practice 2
Since
\[ 0.52\neq0.35, \]the events are dependent.
Practice 3
\[ P(A)P(B)=0.50(0.30)=0.15. \]This equals \(P(A\cap B)\), so the events are independent.
Practice 4
\[ P(A)P(B)=0.50(0.30)=0.15. \]But
\[ P(A\cap B)=0.20. \]Therefore the events are dependent.
Practice 5
Mutual exclusivity gives
\[ P(A\cap B)=0. \]But
\[ P(A)P(B)=0.20(0.40)=0.08. \]Since \(0\neq0.08\), the events are dependent.
Practice 6
Initially,
\[ P(B)=\frac{7}{10}. \]Given that the first token was black,
\[ P(B\mid A)=\frac{6}{9}=\frac{2}{3}. \]Since
\[ \frac{2}{3}\neq\frac{7}{10}, \]the events are dependent.
Practice 7
Replacement restores the original composition before the second draw:
\[ P(B\mid A)=\frac{7}{10}=P(B). \]Therefore the events are independent.
Practice 8
The overall app-use probability is
\[ P(A)=\frac{60}{100}=0.60. \]Among subscribers,
\[ P(A\mid S)=\frac{30}{50}=0.60. \]Since these are equal, subscription status and app use are independent in this probability table.
Practice 9
Since \(A\) is independent of \(B^c\),
\[ P(A\cap B^c)=P(A)P(B^c). \]Now
\[ P(B^c)=1-0.25=0.75. \]Therefore,
\[ \boxed{ P(A\cap B^c)=0.70(0.75)=0.525 }. \]Practice 10
No. Pairwise independence checks only pairs of events. Mutual independence additionally requires
\[ P(A\cap B\cap C)=P(A)P(B)P(C). \]Pairwise independence alone does not guarantee this three-way equality.
What to Remember
Two events are independent when information about one does not change the probability of the other.
When the conditional probability is defined,
\[ \boxed{P(A\mid B)=P(A)}. \]The general product criterion is
\[ \boxed{P(A\cap B)=P(A)P(B)}. \]- Independence is symmetric.
- Dependence means conditioning changes probability.
- Independent events may overlap.
- Mutually exclusive positive-probability events are dependent.
- Independence extends to complements.
- Pairwise independence is weaker than mutual independence.
- Dependence does not by itself prove causation.
The central diagnostic question is: does learning one event change the probability of the other?
Where We Go Next
Independence identifies the special case in which conditioning leaves probabilities unchanged. More generally, intersections can be calculated by combining an initial probability with an appropriate conditional probability.
The next lesson, Multiplication Rules, develops these relationships for independent and dependent events, including sequential probability calculations.