In real life, a person may be associated with more than one field. He/she may be a player of tennis as well as a player of soccer. Similarly a person may drink coffee and may also drink tea in different situations. Suppose David is a film actor as well as a politician, he can be referred from either of these two fields, depending on the context. That is, he can be referred as a film actor in some situations and can be referred as a politician in some other situation. Both the statements are correct. But is there a single identity for him that he is a part of both the activities? This thinking has only led to the concept of ‘intersection’ and developed further for various applications, especially in the field of probability. In the present context an intersection is defined as common item/items (or event/events) that exist among various groups. The knowledge of set theory greatly helps us to easily understand the intersection concepts. In set theory the intersection of groups (or sets) $A$ and $B$ are denoted as $A \cap B$.

Intersection of Events

An event means that something which we are looking for has happened. But the description of an event may not be simple. It may be attached with some additional conditions as required in specified situations. For example, a person is to be selected to work in different parts of Europe. The best choice is a candidate who is fluent in all the three languages of English, French and German rather than a person who knows only one or two out of these three languages.  How do we identify such a persons in a given group? Here where the set theory helps. A survey is done on a group to collect the data of persons and their proficiency in the three languages. Assume that sets $A, B$ and $C$ describe the persons who know English, French and German respectively. Then $A \cap B$ represents the intersection of sets $A$ and $B$ meaning the set of people who knows both English and German. The interpretations of $B \cap C$ and $C \cap A$ are accordingly same and the intersection $A \cap B \cap C$ is the set of people who know all the three languages. The following formula helps us to find the unknown set.

$n(A \cup B \cup C)$ = $n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$

The letter '$n$' means the number in the corresponding sets and the symbol $\cup$ between the two sets means a union or sum of those two sets. If the study is limited to only two sets $A$ and $B$, then the formula reduces to, $n(A \cup B)$ = $n(A) + n(B) - n(A \cap B)$. A Venn diagram describing the situation would further help in understanding.


Example 1: 

In a hostel of $175$ students $150$ students liked orange juice and $100$ students liked apple juice. How many students liked both orange and apple juice? How many liked only apple juice?


Let set $A$ be the students who liked orange juice and le set $B$ the students liked apple juice. The $A \cap B$ represents the set of students liked both orange juice and apple juice. Applying the formula

$n(A \cup B)$ = $n(A) + n(B) - n(A \cap B)$,

$175$ = $150 + 100 - n(A \cap B)$ or $n(A \cap B)$ = $75$.

Hence the number students liked both orange and apple juice is $75$.

Now, out of $100$ students liked apple juice, $75$ liked also the orange juice. Therefore, the number of students liked only apple juice is $100 - 75$ = $25$.

The problem can also be solved with the help of a Venn diagram.

Intersection of Events
Let '$x$' '$y$' and '$z$' be the numbers of students who liked only orange juice, apple juice and both respectively. From the Venn diagram, 

$x + z +$ = $150$ and $y + z$ = $100$ and hence $x + y + 2z$ = $250$ ---- (1)

From the problem statement, $x + z + y$ = $175$ ------(2)

$(2) - (1)$ gives $z$ = $75$ and solving further, $x$ = $75$ and $y$ = $25$.

So, $75$ students liked only orange juice, $25$ students liked apple juice and $75$ liked both.
Example 2: 

In a club of $20$ members $12$ play table tennis and $2$ play both table tennis and billiards. How many members play billiards?


Assuming $A$ and $B$ as the sets of members playing table tennis and billiards respectively,

$n(A)$ = $12$ and $n(A \cap B)$ = $2$. 

$n(A \cup B)$ = $n(A) + n(B) - n(A∩B)$

$2$ = $12 - n(B)$ or $n(B)$ = $10$. 

So, $10$ members play billiards.
Example 3: 

A set of cards is numbered from $1$ to $10$ but completely shuffled. If a card is drawn at random, what is the probability that its number is odd and divisible by $3$?


Set $A$ = Odd numbers = $\{1, 3, 5, 7, 9 \}$ and set $B$ = Number divisible by $3$ = $\{3, 6, 9 \}$

So, $A \cap B$ = $\{3, 9 \}$ and hence $n(A \cap B)$ = $2$. The sample space $n(S)$ = $10$

Therefore, the required probability = $\frac{2}{10}$ = $\frac{1}{5}$.
Example 4: 

In a nursing home $200$ patients were admitted for fever. $100$ were found to be infected due to virus $A$, $80$ were found to be affected by virus $B$ and $20$ patients have both the infections. If a patient is picked up at random, what is the probability that patient has fever due to some other cause?


Given $n(A)$ = $100,\ n(B)\ 80$ and $n(A \cap B)$ = $20$

Hence $n(A \cup B)$ = $100 + 80 - 20$ = $160$

It means $160$ patients out of $200$ were affected by virus $A$ and virus $B$. In other words, $40$ patients had fever due to some other cause. Hence the required probability is $\frac{40}{200}$ = $\frac{1}{5}$.