Conditional Probability¶

Definition¶

Let $A,B$ be events with $\mathbb{P}(B)>0$ $0$ . The conditional probability of $A$ given $B$ is

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

This definition restricts and renormalizes the probability model to $B$ . It is not defined when $\mathbb{P}(B)=0$ $0$ , because the quotient would divide by zero.

Interpretation¶

Conditioning asks how likely $A$ is once we know that $B$ has happened. The outcomes outside $B$ are discarded, and the probabilities of the outcomes in $B$ are rescaled so that they sum to one again.

Two panels with overlapping discs A and B: before conditioning the whole rectangle S counts; after learning B, only the disc B remains as the new sample space, with A and B shown as the part of A inside it.

Before conditioning, the whole rectangle $S$ counts. After learning $B$ , only the disc $B$ remains as the new sample space, and of $A$ only the part $A\cap B$ is still possible. $\mathbb{P}(A\mid B)$ is the share of the probability of $B$ that lies in $A\cap B$ $B$ .

Each part of the formula has a role. The numerator is $\mathbb{P}(A\cap B)$ $B)$ , not $\mathbb{P}(A)$ , because outcomes of $A$ outside $B$ have been ruled out. Dividing by $\mathbb{P}(B)$ makes the new probabilities add up to one, and it requires $\mathbb{P}(B)>0$ $0$ : the formula does not define conditioning on an event of probability zero. With equally likely outcomes, as in naive probability, the formula becomes counting inside $B$ : $\mathbb{P}(A\mid B)=|A\cap B|/|B|$ $B ∣/∣ B ∣$ . For two fair dice, $\mathbb{P}(\text{at least one six})=\tfrac{11}{36}$ $\frac{11}{36}$ , but among the six pairs with sum $7$ exactly two contain a six, so $\mathbb{P}(\text{at least one six}\mid\text{sum}=7)=\tfrac{2}{6}=\tfrac{1}{3}$ $\frac{1}{3}$ .

A common slip is to swap the roles of $A$ and $B$ : in general $\mathbb{P}(A\mid B)\ne\mathbb{P}(B\mid A)$ $A)$ , and Bayes' rule converts one into the other; confusing them is the prosecutor's fallacy. When $\mathbb{P}(A\mid B)=\mathbb{P}(A)$ $P (A)$ , learning $B$ changes nothing, which is the idea of independent events. The law of total probability builds $\mathbb{P}(A)$ back from the conditional probabilities given the cases of a partition.

Key Results¶

$\mathbb{P}(A\cap B)=\mathbb{P}(A\mid B)\mathbb{P}(B)$ $B) P (B)$ .
For fixed $B$ of positive probability, $A\mapsto\mathbb{P}(A\mid B)$ is a probability measure.

Examples¶

For a fair die, $\mathbb{P}(X=6\mid X\text{ is even})=(1/6)/(3/6)=1/3$ $1/3$ .
If $\mathbb{P}(A\cap B)=0.12$ and $\mathbb{P}(B)=0.4$ $0.4$ , then $\mathbb{P}(A\mid B)=0.3$ $0.3$ .

Prerequisites¶

Sources¶

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

Aliases: conditional probability of A given B Keywords: updated probability Symbols: intersection

Last updated: 20 October 2026 03:02