Independent and Dependent Events

Two events are independent when knowing that one occurred does not change the probability of the other, wherever the relevant conditional probability is defined. Dependence means the probability information changes.

This is not the same as mutual exclusivity. Events can overlap and be independent, or occur at different times and be dependent. The relationship comes from the probability model, not the appearance of the story.

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Test the joint probability

The definition that works even when one event has probability zero is

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

When $P(A)>0$ , this is equivalent to $P(B\mid A)=P(B)$ . When $P(B)>0$ , it is equivalent to $P(A\mid B)=P(A)$ .

A conditional comparison should not use a zero-probability conditioning event. The product definition avoids that difficulty. In the finite examples below, every selected elementary outcome has positive probability.

Independent events can occur in the same trial

Roll one fair die. Let $A$ be an even result and $B$ a result greater than four. Then

P(A)=\frac36=\frac12,\quad P(B)=\frac26=\frac13,\quad P(A\cap B)=\frac16.

Only the face 6 belongs to both events. Since $(1/2)(1/3)=1/6$ , the events are independent. Restricting attention to the even faces $\{2,4,6\}$ leaves one of three faces greater than four, again giving $1/3$ .

Independence can occur within one roll

Each numbered box is one equally likely face of a fair die. Outlines and labels identify the target faces as well as colour.

Diagram explanation and exact data

Among the three even faces, only 6 exceeds four. Among the three odd faces, only 5 does. The event of exceeding four keeps probability 1/3 after conditioning on even or odd.

Independence can occur within one roll: exact reference data
Group	Faces	Greater than 4	Conditional probability
Even	2,4,6	6	1/3
Odd	1,3,5	5	1/3

No separate tosses or devices were needed. Independence is a relationship between events, not a requirement that they happen in physically separate experiments.

Disjoint positive-probability events are dependent

On the same die, let $C=\{1,2\}$ and $D=\{5,6\}$ . They are disjoint, so $P(C\cap D)=0$ . However,

P(C)P(D)=\frac13\cdot\frac13=\frac19\ne0.

They are not independent. Knowing $C$ occurred makes $D$ impossible for that roll. Generally, if disjoint events both have positive probability, their joint probability is zero while the product is positive.

The exception matters: a probability-zero event satisfies the product definition of independence with every event. Introductory statements that disjointness and independence can never coexist need the positive-probability qualification.

Sampling changes conditional probabilities

In a bag of three A and two B tokens, two fresh random selections with replacement have $P(A_2\mid A_1)=3/5=P(A_2)$ . Under the stated independent-selection procedure, the A events are independent.

Without replacement, $P(A_2\mid A_1)=2/4=1/2$ , whereas $P(A_2)=3/5$ . Thus the events are dependent. If the first token was B instead, the conditional chance of A second is $3/4$ .

Both draws still have marginal A probability $3/5$ . That equality describes the two overall distributions, not their relationship. The Multiplication Rule of Probability shows how the second marginal is obtained by adding the two possible first-draw paths.

Do not infer independence from a narrative shortcut

Events occurring on different days can share an unobserved influence. Two devices can be subject to the same environment. Replacing a token can restore its availability without ensuring that the next grab is randomly independent. Conversely, the same die can define independent events, as the earlier example showed.

When a problem stipulates independent trials, use that as a modelling assumption. When data are used to assess independence, an observed match is not proof of exact independence in the wider population. Sampling variability and selection design still matter. This is a point of connection with Sampling Bias and Hypothesis Testing.

Pairwise independence is weaker than mutual independence

Toss two independent fair coins. Let $A$ be a head on the first toss, $B$ a head on the second, and $C$ matching results. Each event has probability $1/2$ .

Every pair has intersection probability $1/4$ : $A\cap B$ is $HH$ , $A\cap C$ is $HH$ , and $B\cap C$ is also $HH$ . Each pair therefore satisfies the product test. But

P(A\cap B\cap C)=\frac14\ne\left(\frac12\right)^3=\frac18.

The three events are pairwise independent but not mutually independent. Knowing both first and second toss results determines whether they match. Multiplying several marginal probabilities requires the stronger mutual-independence condition, or a justified trial model that supplies it.

Independence from a complement

If $A$ and $B$ are independent, then $A$ and $B^c$ are independent as well:

\begin{aligned} P(A\cap B^c)&=P(A)-P(A\cap B)\\ &=P(A)-P(A)P(B)\\ &=P(A)(1-P(B))=P(A)P(B^c). \end{aligned}

The derivation uses a disjoint split of A into the B and not-B parts. It is a general algebraic argument, unlike checking just one numerical example.

Practice questions

If $P(A)=0.4$ , $P(B)=0.5$ and $P(A\cap B)=0.2$ , are A and B independent?
Can two disjoint events, each with positive probability, be independent?
Without replacement from three A and two B tokens, does the unchanged marginal probability $P(A_2)=3/5$ establish independence of the two A events?

Check your answers

Yes. The product is $0.4\times0.5=0.2$ , equal to the joint probability.
No. Their joint probability is zero, but the product of two positive probabilities is positive.
No. $P(A_2\mid A_1)=1/2$ differs from $3/5$ . Marginals can stay the same while conditional probabilities change.