An experiment is something we perform. For example: tossing a coin. The results of the experiments are considered individual events like appearance of head on a coin when it is tossed. These results are known as events. The intersection is an interesting operation on events. Intersection implies that two or more things are taken together.

Intersection of two events means two events occurring together. This means that if there are two events say getting an even number and a multiple of 3 when we throw a dice, they can occur simultaneously only if the number on dice is 6 as it is even as well as a multiple of 3. We represent the intersection with symbol ‘$\cap$’ or even at times by writing logical ‘AND’ in between the events. If A and B are two events then we represent A intersection B as A $\cap$ B or A AND B.

When we find the union of two or more events then we always check for intersection of events as well to avoid duplicity in calculation.
While finding out the probability of intersection of events there are three cases:

The first case is that the given events are mutually exclusive. Mutually exclusive events are the ones in which there is no common outcome possible. In such case if A and B are two mutually exclusive events, then A $\cap$ B = Null, since there is no common occurrence of outcomes in such cases. And hence, P (A $\cap$ B) = 0


The second case is when the events are not mutually exclusive but independent. Independent events are the ones in which occurrence of one event does not change or affect the occurrence of another event at all. For example, if we are taking out two balls from an urn but we chose the second after replacing the first one then the events are independent of each other.

In this case, if A and B are two independent events then the probability of the intersection compound event of A and B is determined by multiplying the individual probabilities of the events, that is, P (A $\cap$ B) = P (A) . P (B)


The third case is when the events are not independent that is, the events are dependent. When events are dependent then the occurrence of first event affects the event occurring after it. For example, we have 52 fair well shuffled cards. We pick out one card. Then we draw another but do not replace back the first one. Now when we determine the probability of second event we would not have 52 cards. We now have 51 after taking the first one out. So the probability changed. This is known as conditional probability as well. If A and B are two dependent events then we evaluate the intersection compound probability of the events A and B as P (A $\cap$ B) = P (A) P (B | A)

In all above, P (A $\cap$ B) is the probability of simultaneous occurrence of two events A and B, P (A) is the probability of the occurrence of first event, P (B) and P (B | A) is the probability of second event if events are independent or dependent respectively. P (B| A) is read as the probability of occurrence of event B when event A has already taken place. Also, P (B) = P (B | A) if the events are dependent and if events are independent then P (B | A) = P (B).

In general we can use the formula P (A $\cap$ B) = P (A) P (B | A)

Depending upon the dependency of the event we can use it either ways as explained above.

It is important to note that intersection of two complement events is always equal to zero. This is because if A is an event, then A’ is the complement of event A. It is clear that A $\cap$ A’ = Null. This implies P (A $\cap$ A’) = 0
Let us see some examples on intersection of events.


Two cards are drawn from a deck of cards without replacement. Find the probability of getting a queen and a face card.


Since, two cards are drawn without replacement then clearly the events are dependent in this case.

Let A be the event of getting a queen

Then P (A) = $\frac{4}{56}$

Let B be the event of second card to be a face card.

Now we are left with 51 cards. Also there are 12 face cards out of which one is already taken out. So we are left with 11 cards.

Then we have the conditional probability of event B that is P (B | A).

So, P (B | A) = 11/51

Hence P (A $\cap$ B) = P (A) . P (B | A) = $\frac{4}{52}$ $\times$ $\frac{11}{51}$ = $\frac{11}{663}$