Probability is often first introduced as a way of assigning weights to events before an experiment is performed. That is only half of the story. In practice, information arrives after the model is built: a test result is positive, a card is known to be red, a [random variable](/page/Random%20Variable) has fallen in a given range, or a filtration has revealed the past of a stochastic process. Conditional probability is the operation that rebuilds probability inside the world that remains possible after such information is known.

The danger is that new information does not merely add a number to the calculation. It changes the sample space being used. A rare disease can have an accurate test and still be unlikely after a positive result, because most positive tests may come from healthy people when the disease is rare. This is the first lesson of conditioning: the event being conditioned on is not a decoration after the vertical bar. It is the reference universe in which the probability is being measured.

[example: A Positive Test for a Rare Disease] Suppose $1$ percent of a population has a disease. Let $D$ be the event that a randomly chosen person has the disease, and let $T$ be the event that the test is positive. The given data are \begin{align*} \mathbb P(D) &= 0.01, & \mathbb P(T \mid D) &= 0.99, & \mathbb P(T \mid D^c) &= 0.05. \end{align*} Since $D$ and $D^c$ partition the population and $\mathbb P(D^c)=1-\mathbb P(D)=1-0.01=0.99$, *[Law of Total Probability](/theorems/1113)* gives \begin{align*} \mathbb P(T) &= \mathbb P(T \mid D)\mathbb P(D)+\mathbb P(T \mid D^c)\mathbb P(D^c) \\ &= 0.99\cdot 0.01+0.05\cdot 0.99 \\ &= 0.0099+0.0495 \\ &= 0.0594. \end{align*} Now *[Bayes' Formula](/theorems/1114) for a Partition*, applied to the partition $\{D,D^c\}$, gives \begin{align*} \mathbb P(D \mid T) &= \frac{\mathbb P(T \mid D)\mathbb P(D)}{\mathbb P(T)} \\ &= \frac{0.99\cdot 0.01}{0.0594} \\ &= \frac{0.0099}{0.0594} \\ &= \frac{99}{594} \\ &= \frac{1}{6}. \end{align*} The test is highly sensitive, but a positive result gives probability $1/6$, not $0.99$, because the large healthy population contributes $0.0495$ of the total positive-test probability through false positives. [/example]

This example is not a paradox. It is the arithmetic of changing the reference population. Conditional probability formalizes that change and gives a reliable language for every later subject in [probability theory](/page/Cambridge%20IA%20Probability), from independence to martingales.

Before defining conditional probability, we need to isolate the kind of information event on which conditioning is possible by a ratio. If an event has probability zero, then dividing by its probability cannot define a new probability measure. Conditioning on probability-zero information is possible in more advanced settings, but it requires conditional distributions or regular conditional probabilities rather than the elementary ratio.

[definition: Positive Probability Event] Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. An event $B \in \mathcal F$ is a positive probability event if \begin{align*} \mathbb P(B) > 0. \end{align*} [/definition]

This preliminary definition marks the boundary of the elementary theory. Once the conditioning event has positive probability, the updated probability is obtained by restricting attention to that event and renormalizing.

## Definition

After identifying the events whose probability can be used as a normalizing factor, the next problem is to define the updated probability of an arbitrary event $A$. The only outcomes that still matter are those in $B$, so the numerator must be $\mathbb P(A \cap B)$; the denominator must make the conditioned universe $B$ have total probability $1$.

[definition: Conditional Probability] Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space, and let $B \in \mathcal F$ satisfy $\mathbb P(B) > 0$. The conditional probability given $B$ is the map \begin{align*} \mathbb P(\cdot \mid B): \mathcal F &\to [0,1] \\ A &\mapsto \frac{\mathbb P(A \cap B)}{\mathbb P(B)}. \end{align*} [/definition]

The definition says that only the overlap $A \cap B$ matters after $B$ is known. The next useful question is how to recover the probability of a joint event from a conditional probability. This is the form needed when an experiment is performed in stages and we want to multiply probabilities along the path.

[quotetheorem:4995]

The multiplication rule is the bridge between intersections and updates. It also warns that $\mathbb P(A \mid B)$ and $\mathbb P(B \mid A)$ are different ratios with the same numerator but different denominators. A finite sample-space computation makes this change of denominator visible.

[example: Two Dice and a Restricted Universe] Roll two fair six-sided dice, so the sample space has $36$ equally likely ordered pairs. Let $A$ be the event that the sum is $8$, and let $B$ be the event that the first die is $3$. The conditioning event is \begin{align*} B=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\}, \end{align*} so $|B|=6$ and \begin{align*} \mathbb P(B)=\frac{6}{36}=\frac{1}{6}>0. \end{align*} The event $A$ is \begin{align*} A=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}, \end{align*} so \begin{align*} A\cap B=\{(3,5)\} \end{align*} and therefore \begin{align*} \mathbb P(A\cap B)=\frac{1}{36}. \end{align*} Using the definition of conditional probability, \begin{align*} \mathbb P(A\mid B) &=\frac{\mathbb P(A\cap B)}{\mathbb P(B)} \\ &=\frac{1/36}{6/36} \\ &=\frac{1}{36}\cdot \frac{36}{6} \\ &=\frac{1}{6}. \end{align*} Without the information $B$, the sum $8$ has five favorable outcomes among the thirty-six equally likely outcomes, so \begin{align*} \mathbb P(A)=\frac{5}{36}. \end{align*} Conditioning has changed the reference set from all $36$ outcomes to the six outcomes compatible with the first die being $3$. [/example]

## Conditioning on Events

### Probability After Information

The formula for $\mathbb P(A \mid B)$ resembles an ordinary probability measure in the variable $A$. We need this resemblance to be genuine, because once information has arrived we still want to take complements, count disjoint alternatives, and reason with the probability axioms inside the new universe.

[quotetheorem:4972]

This theorem justifies treating conditional probabilities like ordinary probabilities once the conditioning event has been fixed. The next problem is to handle several pieces of information arriving in sequence, where each new event is conditioned on all previous events.

[quotetheorem:4996]