Can anyone explain me elaborately with example what are the differences between two events $A\cap B $ and $A | B $? Can anyone explain me elaborately with example what are the differences between two events $A\cap B $ and $A \mid B $ ?
Thank You in advance.
 A: As events (subsets of the probability space $\Omega$ of $(\Omega, \mathcal A, P)$), they are the same. 
However, (in a bit simplified phrasing, applicable when $P(B) >0$), the probability spaces and measures are different: $(A\mid B)$ lives in $(B, \mathcal A|_B, X\mapsto P(X)/P(B))$. 
A: There is no such thing as the event $A\mid B.$
When one speaks of $\Pr(A\mid B),$ that is not the probability of something called $A\mid B.$ Rather, it is the probability given $B$, of the event $A.$
On the other hand $A\cap B$ is actually an event, and $\Pr(A\cap B)$ is its probability.
A: Suppose you are throwing a six-sided die. Define:
$A$ : You are throwing a 'high' number, i.e. you throw 4,5, or 6)
$B$ : You are throwing an even number, i.e you throw 2,4, or 6
Then:
$A \cap B$ : you throw a number that is high and even, i.e. you throw $4$ or $6$
So: $P(A \cap B) = \frac{2}{6} =\frac{1}{3}$
$A \mid B$ : you throw a number that is high given that it is even ... NOTE: you cannot say that this is the event of throwing a $4$ or $6$! Indeed, $A\mid B$ is not an event! 
But, given that you throw an even number, we know you threw 2,4, or 6, and since two of thse three are high, we have:
$P(A\mid B)=\frac{2}{3}$
A: Related to the anser of Berci.  
The expression $P(A|B)$ is not $P$ applied to an event $A|B$.  Instead, in the expression $P(A|B)$, we have a function "conditional $P$", a function of two arguments, applied to the events $A$ and $B$.
There is a probability measure $P(\cdot)$; we apply it to the event $A \cap B$ to get the number $P(A\cap B)$.
Given $B$ with positive probability, there is a probability measure $P(\cdot|B)$; we apply it to the event $A$ to get the number $P(A|B)$.
A: Consider a sample space $\Omega=\{1,2,3\}$ corresponding to the result of tossing a (fictitious) three-sided die.
As our prototypical examples of events, we will take $A = \{1,2\}$ (rolling the number one or two) and $B = \{1, 3\}$ (rolling the number one or three).
More generally, an event on $\Omega$ is any one of the following:
$$
\emptyset,\left\{ 1\right\} ,\left\{ 2\right\} ,\left\{ 3\right\} ,\left\{ 1,2\right\} ,\left\{ 1,3\right\} ,\left\{ 2,3\right\} ,\left\{ 1,2,3\right\}
$$
(if you are familiar with it, this is nothing than the powerset of $\Omega$).
When we intersect two events, we obtain another event corresponding to both events occurring. In our case, $A\cap B=\{1\}$ corresponds to rolling the number one.
A probability measure $\mathbb{P}$ is responsible for assigning probabilities to particular events. If our three-sided die is "fair", then $\mathbb{P}(A)=2/3$,
$\mathbb{P}(B)=2/3$, and $\mathbb{P}(A\cap B)=1/3$.
Technically speaking, $A\mid B$ is not an event. When we write $A\mid B$, we are alluding to definition of conditional probability:
$$
\mathbb{P}(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}
$$
(note that the above is only well-defined if $\mathbb{P}(B) > 0$).
This should be interpreted as the probability that event $A$ occurs given that event $B$ has occurred.
In our running example,
$$
\mathbb{P}(A\mid B)=\frac{1/3}{2/3}=\frac{1}{2}.
$$
A: As Michael pointed out, $A\cap B$ is an event, while $A|B$ is not. Here we can see it visually: 

Here we can see $A\cap B$ is represented as the intersection, or the purple space. This is the compound event where events $A$ and $B$ happen simultaneously.
$A | B$ is read as $A$ happens given $B$ has already happened beforehand. This is associated with Conditional Probability. We cannot represent this visually as this is not an event. We tend to look more at $P(A|B)$ ; that is, the probability of $A$ happening given $B$ has happened already.
