After learning about dependent and independent events, it’s time to mention mutually exclusive events.

Events are **mutually exclusive events**, or *disjoint*, if occurrence of one event excludes the occurrence of the other(s). Both can’t happen at the same time, therefore their intersection is empty.

For instance, when we flip a coin, we can get heads OR tails, never both. When we roll a die, we can only get one number at the time. On the other hand, if we roll two dice, those events wouldn’t be mutually exclusive. The outcome on the second die is not connected to the first die in any way.

#### Diagram representation

Let’s say we have two events, $A$ and $B$, in sample space $S$. Mutually exclusive events do not overlap at any point, therefore $P(A \cap B)=0$. Seeing that events don’t have any effect on each other, $P(A|B)=P(B|A)=0$.

General addition formula is $$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$

Since the intersection is equal to $0$, the addition formula for mutually exclusive events is $$P(A \cup B)=P(A)+P(B)$$

#### Example 1

A spinner has $4$ equal sectors colored blue, red, purple and orange. What is the probability of landing on red or purple after spinning the spinner?

Since we only spin the spinner once, all outcomes are mutually exclusive. We can’t land on two different sectors at once.

Let A be the event$\{$landing on red$\}$, and $B=\{$landing on purple$\}$. The probability of $A$ or $B$ is: $$P(A \cup B)=P(A)+P(B)$$

Probability of landing on red is $\frac{1}{4}$. Likewise, probability of landing on purple is also $\frac{1}{4}$.

In conclusion, $$P(A \cup B)=\displaystyle{\frac{1}{4} + \frac{1}{4}= \frac{1}{2}}$$

#### Example 2

The probabilities of three teams A, B and C winning a football competition are $\frac{1}{4}$, $\frac{2}{7}$ and $\frac{1}{10}$ respectively. Calculate the probability that:

a) either A or C will win

b) either A or B or C will win

c) none of these teams will win

d) neither B nor C will win

First and foremost, notice that only one team can win the competition. Winning a competition is mutually exclusive, only one team can be the winner.

**a) either A or C will win**

$\displaystyle{P(A \cup C)=\frac{1}{4} + \frac{1}{10}=\frac{7}{20}}$

**b) either A or B or C will win **

This time we have $3$ events. Their intersection is still $0$, so we just add one more event into the addition formula for mutually exclusive events. $$\displaystyle{P(A \cup B \cup C)=\frac{1}{4} + \frac{2}{7} + \frac{1}{10}=\frac{89}{140}}$$

**c) none of these teams will win **

This is a complement event of the one under **b)**.

$P($none of these teams will win$\displaystyle{)=1-P(A \cup B \cup C)=1-\frac{89}{140}=\frac{51}{140}}$

**d) neither B nor C will win **

Once again it is easier to calculate the probability of the complement of the given event. Complement would be $\{B \cup C\}$.

$\displaystyle{P(B \cup C)=\frac{2}{7} + \frac{1}{10}=\frac{27}{70}}$

Finally,

$P($neither $B$ nor $C \displaystyle{)=1-P(B \cup C)=1-\frac{27}{70}=\frac{43}{70}}$

#### Difference between mutually exclusive and independent events

At first the definitions of mutually exclusive and independent events may sound similar. The biggest difference is that **mutually exclusive** means: if one event happens, the others CAN’T happen. On the other hand, if the events are **independent**, it means the occurrence and the outcome of one event won’t have any affect on the occurrence and outcome of other events.

For instance, outcomes of rolling a die are mutually exclusive events. You can get either $5$ or $6$, but never $5$ and $6$ at the same time. However, outcomes of rolling a die twice are independent events. The number we get on the first roll has no effect on the number we’ll get when we roll the die one more time.

Lets make a chart for easier understanding.

In conclusion, mutually exclusive and independent events are not the same. As a matter of fact, if events are mutually exclusive, then they can’t be independent and vice versa.