In probability, an event is always associated with a random experiment. When something happens we say an event has taken place. For example: In an experiment of flipping a coin there are two possibilities, either getting a tail or a head.

There are different operations that can be performed on events such as union of events, compliment and intersection of events. The intersection of two or more events gives a new compound event.

Intersection of events means that all the events are occurring together. Even if one event holds false all will be false. The intersection of events can only be true if and only if all the events holds true. 


Consider $(M\  and\  N)$ are two events of a random experiment. The intersection of $(M\ and\  N)$ is represented by $(M \cap N)$ or simply $(M\  and\  N)$.
The probability of intersection of the events M and N is given by:

$P (M and N)$ = $P (M) \times P (N)$

Here, $P (M)$ is the probability of occurrence of event $M$ and $P (N)$ is the probability of occurrence of event $N$.

When $M\  and\  N$ are two independent events that is the occurrence of one is not affecting by other, then in such a case, 

$P (M \cap N)$ = 0
This is because in case of independent events the two events have nothing in common at all. And hence M $\cap$ N = {NULL}.

If we have more as compared to two independent events say M, N & O then when this occurs the probability of intersection coming from all three events is due to:

$P (M \cap N \cap O)$ = $P (M) * P (N) * P (O)$

But what do we do when M & N are not at all independent? In such cases the concept of conditional probability arrives. When we have two events M & N where the occurrence of M affects the occurrence of N, in such a case the probability of intersection of the two events is given by:

$P (M \cap N)$ = $P (M) * P (N|M)$

Where $P (N|M)$ is the probability of event $N$ given that event $M$ has already taken place.


Example 1:  Two cards are drawn without replacement from a 52 cards deck given. Find the probability of both the cards being 5’s.

Solution: Here clearly, $n (S)$ = 52.

Let $M$ be the event of getting 5 as first car. So $n (M)$ = 4

Hence we have $P (M)$ = $\frac{4}{52}$

Let $N$ be the event of getting second card as 5 too after drawing the first card as 5, that is, after $M$ has occurred. 

So, $n (N)$ = 3. Also n (S) = 51 after M has occurred.

Hence we have $P (N|M)$ = $\frac{3}{51}$

Since the events are not independent, thus we have

$P (M \cap N)$ = $P (M) * P (N|M)$ = $\frac{4}{52}$ * $\frac{3}{51}$ = $\frac{1}{13}$ * $\frac{1}{17}$ = $\frac{1}{221}$.

Example 2: Roll a fair die. Find the probability of event that, the number that turns in place is even as well as divisible by four.


Let A = an even number turns up, and

B = the number that turns up is divisible by 4

Sample space of equally likely simple events is: S = {1, 2, 3, 4, 5, 6}.

From the given statement: A = {2, 4, 6} and B = {4}.

Now, consider M = A $\cap$ B = {4}

=> P(M) = P(A $\cap$ B) = $\frac{n(A \cap B)}{n(S)}$ = $\frac{1}{6}$