Disjoint Events and Independence, introduction to probability lesson

Roll a fair die and consider the events “even” and “odd”. They have no outcome in common, so it is tempting to call them unrelated and therefore independent. Yet learning that the roll is odd rules out an even roll completely, which is the opposite of having no influence. Disjointness describes which outcomes the events contain, while independence describes how the model assigns probability to them. This lesson compares the two properties and settles exactly when they can hold together.

Key terms

Disjoint events with positive probability

Definition Events $A$ and $B$ in the same probability model are disjoint, or mutually exclusive, if $A \cap B = \varnothing$ , so they cannot both occur in one trial. They are independent if

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

If $A$ and $B$ are disjoint with $P(A) > 0$ and $P(B) > 0$ , then they are dependent, because

\begin{aligned} P(A \cap B) &= P(\varnothing) = 0 \\[0.4em] &< P(A)\,P(B). \end{aligned}

In this case $P(A \mid B) = 0$ , which differs from $P(A)$ .

In plain words Disjointness is a fact about sets, and independence is a fact about probabilities. When both events have a positive chance, learning that one occurred makes the other impossible, and that is a strong dependence, not a lack of connection. The two properties also come with different rules. For disjoint events the probability of the union is the sum $P(A) + P(B)$ , and for independent events the probability of the intersection is the product $P(A)\,P(B)$ .

Example Roll one fair die, and let $A = \{2, 4, 6\}$ be “even” and $B = \{1, 3, 5\}$ be “odd”. Each event has probability $\tfrac12$ , so the product is $\tfrac14$ , but $P(A \cap B) = 0$ . The events are dependent. Before any information the probability of an even roll is $\tfrac12$ , and after learning that $B$ occurred it is $P(A \mid B) = 0$ .

Overlapping independent events

Definition Events $A$ and $B$ overlap if $A \cap B \ne \varnothing$ , so at least one outcome belongs to both. Overlapping events are independent when they pass the product test $P(A \cap B) = P(A)\,P(B)$ . For independent events the probability of the union still needs the inclusion-exclusion term,

\begin{aligned} P(A \cup B) &= P(A) + P(B) \\[0.4em] &\quad - P(A)\,P(B). \end{aligned}

In plain words Independence does not say that the events never happen together. It says that the probability that they happen together equals the product of their probabilities. So whenever $P(B) > 0$ , learning that $B$ occurred leaves the probability of $A$ unchanged.

Example Roll a red die and a blue die, both fair and independent, so all $36$ ordered pairs are equally likely. Let $C$ be “the red die is even” and $D$ be “the blue die is even”. The event $C$ has $3 \cdot 6 = 18$ outcomes, the event $D$ also has $18$ , and $C \cap D$ has $3 \cdot 3 = 9$ , among them red $2$ with blue $4$ . So

\begin{aligned} P(C \cap D) &= \frac{9}{36} = \frac14 \\[0.4em] &= P(C)\,P(D), \end{aligned}

and the events are independent although they overlap. Their union has probability $\tfrac12 + \tfrac12 - \tfrac14 = \tfrac34$ , which matches the $27$ pairs with at least one even die.

Overlapping dependent events

Definition Overlapping events $A$ and $B$ are dependent when $P(A \cap B) \ne P(A)\,P(B)$ . Whether two overlapping events are independent is decided by the product test under the stated probability model, not by the overlap itself. A Venn diagram whose areas are not drawn to scale shows the overlap but not the probabilities, so by itself it cannot settle independence.

In plain words Sharing outcomes does not settle independence either way. Two overlapping events can still influence each other, and only the probabilities can tell.

Example On one fair die, let $A = \{1, 2, 3\}$ and $B = \{2, 4, 6\}$ . They overlap at $2$ , and $P(A \cap B) = \tfrac16$ , while $P(A)\,P(B) = \tfrac14$ . They are dependent, and learning that the roll is even lowers the probability of $A$ from $\tfrac12$ to $P(A \mid B) = \tfrac13$ . On the same die the events $\{1, 2\}$ and $\{2, 4, 6\}$ also meet exactly in $\{2\}$ , yet $\tfrac16 = \tfrac13 \cdot \tfrac12$ , so they are independent. The intersection is the same in both pairs, and only the probabilities decide.

Zero-probability event

Definition A zero-probability event is an event $Z$ with $P(Z) = 0$ . Such an event is independent of every event $B$ . Since $Z \cap B \subseteq Z$ ,

0 \le P(Z \cap B) \le P(Z) = 0,

so $P(Z \cap B) = 0 = P(Z)\,P(B)$ . Together with the result for disjoint events with positive probability, this shows that two disjoint events are independent exactly when at least one of them has probability $0$ . A zero-probability event need not be empty. The conditional probability $P(B \mid Z)$ is not defined by the ratio $P(B \cap Z)/P(Z)$ , since that ratio would divide by zero.

In plain words An event that carries no probability can never make the product test fail, because both sides of the test are zero. That is the only way two disjoint events can be independent. Independence here is a statement about the product, not about updating on the event, since the ratio formula cannot condition on it.

Example Give three outcomes $a$ , $b$ and $c$ the probabilities $0$ , $0.4$ and $0.6$ . The event $Z = \{a\}$ is not empty, but $P(Z) = 0$ . It overlaps $\{a, b\}$ at $a$ , and

P(Z \cap \{a, b\}) = 0 = 0 \cdot 0.4,

so the two events are independent. The event $Z$ is also disjoint from $\{b\}$ and independent of it. The empty event $\varnothing$ is the simplest zero-probability event, and it is both disjoint from and independent of every event.

Nested events

Definition Two events are nested when one contains the other, say $A \subseteq B$ . Then $A \cap B = A$ , and the product test reads $P(A) = P(A)\,P(B)$ , that is, $P(A)\bigl(1 - P(B)\bigr) = 0$ . So nested events with $A \subseteq B$ are independent exactly when $P(A) = 0$ or $P(B) = 1$ . If $P(A) > 0$ and $P(B) < 1$ , they are dependent, and

P(A \mid B) = \frac{P(A)}{P(B)} > P(A).

In plain words Nested events overlap as much as possible. Whenever the smaller event occurs, the larger one occurs too. So learning that the larger event occurred raises the probability of the smaller one, except in the corner cases where the smaller event has probability zero or the larger one was certain anyway.

Example On one fair die, let $E = \{1\}$ and $F = \{1, 2\}$ , so $E \subseteq F$ . Then $P(E \cap F) = P(E) = \tfrac16$ , while $P(E)\,P(F) = \tfrac16 \cdot \tfrac13 = \tfrac1{18}$ . The events are dependent, and learning that $F$ occurred raises the probability of $E$ from $\tfrac16$ to $P(E \mid F) = \tfrac12$ . By contrast, the event $E$ and the whole sample space $\Omega$ are nested and independent, since $P(\Omega) = 1$ .

Check yourself

A card is picked uniformly at random from eight cards numbered $1$ to $8$ . Let $A = \{1, 2, 3, 4\}$ , $B = \{5, 6, 7, 8\}$ and $C = \{2, 4, 6, 8\}$ . Classify the pair $A$ , $B$ and the pair $A$ , $C$ as disjoint or overlapping and as independent or dependent.

Show the answer

Each event has probability $\tfrac48 = \tfrac12$ , so each product of two of them is $\tfrac14$ . The events $A$ and $B$ share no card, so they are disjoint, and $P(A \cap B) = 0 \ne \tfrac14$ makes them dependent. The events $A$ and $C$ overlap in $\{2, 4\}$ , and $P(A \cap C) = \tfrac28 = \tfrac14$ , so they are independent, even though both describe the same card.

In the same model let $D = \{1, 2\}$ . Are $D$ and $A$ independent? Which events that contain $D$ are independent of $D$ ?

Show the answer

The events are nested, since $D \subseteq A$ . So $P(D \cap A) = P(D) = \tfrac14$ , while $P(D)\,P(A) = \tfrac14 \cdot \tfrac12 = \tfrac18$ , and the events are dependent. Learning that $A$ occurred raises the probability of $D$ from $\tfrac14$ to $P(D \mid A) = \tfrac12$ .

An event $G$ with $D \subseteq G$ is independent of $D$ exactly when $P(D) = 0$ or $P(G) = 1$ . Here $P(D) = \tfrac14 > 0$ , so $G$ needs probability $1$ . Every card has probability $\tfrac18 > 0$ , so the only event of probability $1$ is the whole set $\{1, 2, \dots, 8\}$ , and that is the only event containing $D$ that is independent of $D$ .

Events $A$ and $B$ in some probability model are disjoint and independent, and $P(A) = 0.3$ . What is $P(B)$ ? Must $B$ be the empty event?

Show the answer

Disjointness gives $P(A \cap B) = 0$ , and independence gives $P(A \cap B) = P(A)\,P(B) = 0.3\,P(B)$ . So $0.3\,P(B) = 0$ and $P(B) = 0$ .

The event $B$ need not be empty. Take three outcomes $a$ , $b$ and $c$ with probabilities $0$ , $0.3$ and $0.7$ , and let $A = \{b\}$ and $B = \{a\}$ . They are disjoint, and $P(A \cap B) = 0 = 0.3 \cdot 0$ , so they are independent while $B$ is nonempty. In a model where every outcome has positive probability, such as a fair die, the only event of probability $0$ is $\varnothing$ , so there $B$ must be empty.

Disjoint Events and Independence

Key terms

Disjoint events with positive probability

Overlapping independent events

Overlapping dependent events

Zero-probability event

Nested events

Check yourself

How this lesson connects

Builds on