Conditional Probability

What Is Conditional Probability?

In many real situations, the occurrence of one event changes what we know about the likelihood of another. For instance, knowing that a card drawn is red immediately changes the chance that it is a heart. Conditional probability captures this idea precisely: given two events $A$ and $B$ of a sample space $S$, with $P(B)>0$, the conditional probability of $A$ given that $B$ has already occurred is defined as

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

The notation $P(A\mid B)$ is read "the probability of $A$ given $B$." Similarly, when $P(A)>0$,

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

Why the Definition Makes Sense

Once it is known that $B$ has occurred, the sample space effectively shrinks from $S$ to $B$ -- every outcome outside $B$ is now impossible. Within this reduced sample space, the event $A$ can only happen through the outcomes that lie in both $A$ and $B$, that is, through $A\cap B$. Dividing $P(A\cap B)$ by $P(B)$ simply re-scales these probabilities so that $B$ itself has probability $1$ in the new, conditional world -- exactly matching the requirement $P(B\mid B)=P(B\cap B)/P(B)=P(B)/P(B)=1$.

Basic Properties of Conditional Probability

Conditional probability, for a fixed event $B$ with $P(B)>0$, obeys the same three axioms as an ordinary probability function, so all the familiar probability laws continue to hold when every probability is replaced by a conditional probability given $B$:

• $P(S\mid B)=1$ and $P(B\mid B)=1$.
• $0\le P(A\mid B)\le 1$ for every event $A$.
• $P(A\mid B)=1-P(A'\mid B)$, where $A'$ is the complement of $A$.
• For any two events $A_1,A_2$: $P((A_1\cup A_2)\mid B)=P(A_1\mid B)+P(A_2\mid B)-P((A_1\cap A_2)\mid B)$.

Worked Illustration

A fair die is rolled once. Let $E=\{2,4,6\}$ (an even number) and $F=\{4,5,6\}$ (a number at least $4$). Here $P(F)=\tfrac36=\tfrac12$ and $E\cap F=\{4,6\}$, so $P(E\cap F)=\tfrac26=\tfrac13$. Hence

$P(E\mid F) = \frac{P(E\cap F)}{P(F)} = \frac{1/3}{1/2} = \frac23.$

Notice that $P(E)=\tfrac12$ while $P(E\mid F)=\tfrac23$: knowing that the outcome is at least $4$ has genuinely increased the chance that it is even, because two of the three outcomes in $F$ happen to be even.