# $A$ and $B$ are events, is $A|B$ an event?

$A$ and $B$ are events, is $A|B$ an event?

Can I write an event in this form $A|B$ ?

$Pr(A|B)$ means given $B$ happens, what is the probability of $A$ happen. Since $Pr(\cdot)$ is a measurement, so I am wondering $A|B$ is an event too

-
in terms of events, A|B is A. – Sabyasachi Mar 14 '14 at 7:15
$P(A\mid B)$ should not be read as "the probability of the event $A\mid B$" (which doesn't make sense since $A\mid B$ is not an event). It is notation for a conditional probability defined as $P(A\cap B)/P(B)$ when $P(B)>0$. – Stefan Hansen Mar 14 '14 at 7:16
@Sabyasachi It is not, whatever "in terms of events" means. – Did Mar 14 '14 at 7:34

$A|B$ is not an event. It is a notation. Take, for example, the simple roll of a $6$ sided die. The possible outcomes are $A=\{1,2,3,4,5,6\}$, and the possible events are $2^A$, that is all the subsets of $A$. For example, $E=\{2,4,6\}$ is the event "I roll an even number.".
Now, take for example I want to calculate a conditional probability. For example, what is the chance I rolled an even number if I rolled a number, greater than $3$? Well, given that I know i rolled one of the numbers in event $G=\{4,5,6\}$, I now look at the event $E$ (rolling an even number) and calculate $P(E|G)=\frac{P(G\cap E)}{P(G)}$ to get the result. However, writing $P(E|G)$ is purely notation, as there is no event in my original set that corresponds to me rolling even given rolling greater than $3$.
You may think that the event $E|G$ is the set $\{4,6\}$, since these are the even numbers that are greater than $3$, however, this is actually the event "I rolled an even number and I rolled a number, greater than $3$", which is the event $P(E\cap G)$.
"... which is the event $E \cap G$", not ... "which is the event $P(E\cap G)$" (typo) in the last line. – Jimmy R. Mar 15 '14 at 16:33