An event is something that has happened and is always associated with an experiment.

**For Example:** if we are performing the experiment of throwing a dice then there can be numerous events associated with it, one of which can be the number on top be a multiple of 3.

There are different operations performed on events. Union is one of them. As the name implies union means joining of two things irrespective of their individual behavior. The union of two events gives us a compound event.

Union of events simply mean the condition of occurrence of wither of the events. It is not necessary that all events must hold true. The condition of at least one holds. More than one or even all events under the union operation can be true simultaneously.

Suppose A and B are two events associated with a random experiment. Then the union of A and B is represented by A $\cup$ B.

The probability of union of two events is given by:

$P (A \cup B)$ = $P (A) + P (B)$ – $P (A \cap B)$

Here, P (A) is the probability of event A, P (B) is the probability of event B.

Also, $P (A \cap B)$ is the probability of the intersection of events A and B.

When $A\ and\ B$ are two independent or mutually exclusive events that is the occurrence of event $A$ does not affect the occurrence of event B at all, in such a case, $P (A \cap B)$ = 0 and hence we have,

$P (A \cup B)$ = $P (A) + P (B)$

If we have more than two independent events say A, B & C, then in that case the union probability is given by:

$P (A \cup B \cup C)$ = $P (A) + P (B) + P (C)$

If $A\, B\ and\ C$ are not independent or mutually exclusive then the union probability is given by:

$P (A \cup B \cup C)$ = $P (A) + P (B) + P (C)$ – $P (A \cap B) – P (B \cap C) – P (A \cap C)$

- $P (A \cap B \cap C)$

Here S = {1, 2, 3, 4, 5, 6}, so n (S) = 6

Let A be the event of getting an even number. So A = {2, 4, 6}.

$\rightarrow n (A)$ = 3

Hence we have $P (A)$ = $\frac{3}{6}$

Let B be the event of getting a number that is multiple of 3. So B = {3, 6}

$\rightarrow n (B)$ = 2

Hence we have $P (B)$ = $\frac{2}{6}$

We can clearly see that the events are not mutually exclusive.

That is $A \cap B$ = {6}, => $n (A \cap B)$ = 1 => $P (A \cap B)$ = $\frac{1}{6}$

Thus the compound probability is given by:

$P (A \cap B)$ = $P (A) + P (B) – P (A \cap B)$ = $\frac{3}{6}$ + $\frac{2}{6}$ - $\frac{1}{6}$ = $\frac{4}{6}$ = $\frac{2}{3}$

**Example 2: **Christ rolled a fair die and wished to find the probability of "the number that turns up is odd or divisible by 5".

From the sample space, Let A be the event "an odd number turns up" and the B is the event "the number that turns up is divisible by 5

".

**Solution:** Sample space for a die is: S = {1, 2, 3, 4, 5, 6}.

From the given statement: A = {1, 3, 5} and B = {5}.

Let M = The number that turns up is odd or divisible by 5

So M = A $\cup$ B = {1, 3, 5}

and P(M) = P(A $\cup$ B) = $\frac{n(A \cup B)}{n(S)}$ = $\frac{3}{6}$ = $\frac{1}{2}$ = 0.5

From the sample space, Let A be the event "an odd number turns up" and the B is the event "the number that turns up is divisible by 5

".

From the given statement: A = {1, 3, 5} and B = {5}.

Let M = The number that turns up is odd or divisible by 5

So M = A $\cup$ B = {1, 3, 5}

and P(M) = P(A $\cup$ B) = $\frac{n(A \cup B)}{n(S)}$ = $\frac{3}{6}$ = $\frac{1}{2}$ = 0.5