Independent Events¶

Definition¶

Events $A$ and $B$ are independent if

\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B).

For a collection of events, mutual independence requires the product identity for every finite subcollection; pairwise independence alone is insufficient.

Interpretation¶

Two events are independent when knowing that one of them happened leaves the probability of the other unchanged. In formula form, the probability that both happen is the product of the two probabilities.

A unit square split vertically for A with probability 0.6 and horizontally for B with probability 0.5; their overlap has area 0.3, the product of the two probabilities.

The square $S$ has area $1$ , the band $A$ has width $0.6$ and the band $B$ has height $0.5$ , so the overlap has area $\mathbb{P}(A\cap B)=0.6\cdot0.5=0.3=\mathbb{P}(A)\mathbb{P}(B)$ $P (A) P (B)$ . Equivalently, $\mathbb{P}(A\mid B)=0.3/0.5=0.6=\mathbb{P}(A)$ $P (A)$ : inside $B$ , the band $A$ still takes up $60\%$ , just as it does in the whole square.

The product form is the definition because it is symmetric in $A$ and $B$ and also makes sense when $\mathbb{P}(B)=0$ $0$ ; the form $\mathbb{P}(A\mid B)=\mathbb{P}(A)$ from the Key Results needs $\mathbb{P}(B)>0$ $0$ . Independence is a property of the probability function, not of the sets: on a fair die, $A=\{1,2\}$ and $B=\{2,4,6\}$ overlap, yet $\mathbb{P}(A\cap B)=\tfrac{1}{6}=\tfrac{1}{3}\cdot\tfrac{1}{2}$ $\frac{1}{2}$ , so they are independent; on a loaded die the same sets need not be.

Two slips are common. Disjoint events with positive probabilities are never independent, since $\mathbb{P}(A\cap B)=0<\mathbb{P}(A)\mathbb{P}(B)$ $P (A) P (B)$ . And for three or more events, pairwise independence is not enough: for two fair coin tosses, "first toss heads", "second toss heads" and "both tosses equal" are pairwise independent, but all three together have probability $\tfrac{1}{4}$ , not $\tfrac{1}{8}$ . The same idea is later applied to random variables.

Key Results¶

If $\mathbb{P}(B)>0$ $0$ , independence is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$ $P (A)$ .
Independence of $A,B$ implies independence of $A^c,B$ and of $A^c,B^c$ .

Examples¶

Heads on two independent fair tosses have intersection probability $1/4=(1/2)(1/2)$ $(1/2) (1/2)$ .
On one fair die, $\{1\}$ and $\{2\}$ are disjoint but not independent: $0\ne(1/6)(1/6)$ $(1/6) (1/6)$ .

Prerequisites¶

Sources¶

Probability Theory for EOR · Week 2
Nicky van Foreest · Probability Distributions · Refresher

Aliases: independence of events, statistically independent events Keywords: mutually independent events Symbols: intersection

Last updated: 20 October 2026 03:02