An event is something that has happened. When we have two or more events then there are chances that some events have common occurrences. To find probabilities we first find the total number of outcomes within an experiment. Then we find the outcomes associated with the event being considered. We divide the outcomes of the event by the total number of outcomes of the experiment to find the probability of the event.

Disjoint events are the ones in which there is not even a single outcome that is common in between the events. Another commonly used term for disjoint events is mutually exclusive events. Disjoint events and mutually exclusive events are the same thing. Example of mutually exclusive events is events like getting even or odd number when we throw a dice. In this case there is no outcome common as no one and odd number is same. Thus these two events are mutually exclusive or disjoint. When there are common outcomes between the events then the events are not mutually exclusive events. For example when we throw a dice, the events of getting an even number or a prime number has a common outcome {2}, that makes these two events non mutually exclusive events.

The probability of occurrence of either of two events say X and Y is given by:

$P (X \cup Y)$ = $P (X) + P (Y) - P (X \cap Y)$

Where
$P (X \cup Y)$ is the probability of occurrence of either X or Y, P (X)
and P (Y) are the probabilities of occurrence of events X & Y
individually, $P (X \cap Y)$ is the probability of occurrence of both X
& Y simultaneously

When two events are disjoint then it is clear that the probability of joint occurrence of the events is equal to zero, that is, $P (X \cap Y)$ = 0 if X and Y are mutually exclusive events.

When
we find combined probability of given disjoint events, we can simply
add up the individual properties of the events to find the combined
probability:

$P (A_1 \cup A_2 \cup …. A_n) = P (A_1) + P (A_2) + … + P (A_n)$

Here,
$A_1, A_2, …, A_n$ are mutually exclusive o disjoint events. $P (A_1), P
(A_2), …, P (A_n)$ are individual probabilities of these disjoint
events.

$P (A_1 \cup A_2 \cup …. A_n)$ is the combined probability of occurrence of wither of the events.

Here S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

$\rightarrow$ n (S) = 8

Let X: event of getting exactly two heads = {HHT, HTH, THH}

$\rightarrow$ P (X) = $\frac{3}{8}$

Let Y: event of getting exactly two tails = {HTT, THT, TTH}

$\rightarrow$ P (Y) = $\frac{3}{8}$

Clearly $X \cap Y$ = $\phi$

$\rightarrow$ P (X \cap Y) = 0

Hence the events are mutually exclusive

Thus the probability of union of two events

= $P (X \cup Y)$

= $P (X) + P (Y)$

= $\frac{3}{8}$ + $\frac{3}{8}$

= $\frac{6}{8}$

= $\frac{3}{4}$

**Example 2:** Eighty tickets are sold for a rock show. Smith buys 10 tickets and Criss buys 20 tickets. One of tickets is randomly select as the winning ticket. What is the probability that Smith or Criss wins?

**Solution:** As smith and Criss cannot win together, so events are disjoint.

Let M = Smith win

N = Criss win

P(M) = $\frac{10}{80}$

P(N) = $\frac{20}{80}$

P(M or N) = P(M) + P(N) = $\frac{10}{80}$ + $\frac{20}{80}$ = $\frac{30}{80}$ = $\frac{3}{8}$

Let M = Smith win

N = Criss win

P(M) = $\frac{10}{80}$

P(N) = $\frac{20}{80}$

P(M or N) = P(M) + P(N) = $\frac{10}{80}$ + $\frac{20}{80}$ = $\frac{30}{80}$ = $\frac{3}{8}$