# Mutually Exclusive Events

If two events have no elements in common (Their intersection is the empty set.), the events are called
**
mutually exclusive.
**
Thus,
$P\left(A\cap B\right)=0$
. This means that the
probability
of event
$A$
and event
$B$
happening is zero. They cannot both happen.

**
Example 1:
**

A pair of dice is rolled. The events of rolling a
$5$
and rolling a double have NO outcomes in common so the two events are mutually exclusive.

A pair of dice is rolled. The events of rolling a
$4$
and rolling a double have the outcome
$\left(2,2\right)$
in common so the two events are not mutually exclusive.

**
Example 2:
**

From a group of $6$ freshmen and $5$ sophomores, $3$ students are to be selected at random to form a committee. What is the probability that at least $2$ freshmen are selected?

The committee will have at least $2$ freshmen if either $2$ freshmen and $1$ sophomore are selected (event $A$ )or $3$ freshmen are selected (event $B$ ). Since the two events are mutually exclusive$\begin{array}{l}P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)\hfill \\ \begin{array}{l}P\left(A\right)=\frac{{}_{6}{}^{}C{}_{2}\cdot {}_{5}{}^{}C{}_{1}}{{}_{11}{}^{}C{}_{3}}=\frac{15\cdot 5}{165}=\frac{75}{165}=\frac{5}{11}\\ \end{array}\hfill \\ \begin{array}{l}P\left(B\right)=\frac{{}_{6}{}^{}C{}_{3}}{{}_{11}{}^{}C{}_{3}}=\frac{20}{165}=\frac{4}{33}\\ \end{array}\hfill \\ P\left(A\cup B\right)=\frac{5}{11}+\frac{4}{33}=\frac{15}{33}+\frac{4}{33}=\frac{19}{33}\hfill \end{array}$