7.05 Mutually exclusive events

Mutually exclusive events

If events are mutually exclusive, it means they cannot happen at the same time. 

Some examples of experiments that involve mutually exclusive events are:

• tossing a coin - Consider the events 'flipping a head' and 'flipping a tail'. You cannot flip a head and a tail at the same time. 
• rolling a die - Consider the events 'Rolling an even number' and 'rolling an odd number'. We can't roll any number which is both even and odd. 
• picking a card from a deck of cards - Consider the events 'Drawing a 7 card' and 'Drawing a 10 card'. They have no outcomes in common. There is no card that is both a 7 and a 10.

Since these events CANNOT both occur at the same time, they are mutually exclusive events.

However some events can happen at the same time and we call this non-mutually exclusive. For example:

• picking a card from a deck of cards - Consider the events 'drawing a Club card' and 'drawing a 7'. They have outcomes in common. We could pick a card that is a Club and a 7, because I could get the 7 of clubs. 
• rolling a die - Consider the events 'Rolling an even number' and 'Rolling a prime number'. They have outcomes in common, namely the number 2. 

Since these events CAN both occur at the same time,  they are non mutually exclusive events.

 

Worked examples

Example 1

Consider an experiment of drawing one card from a deck of $52$52 cards. And define two events $A$A: 'Drawing a $7$7 card' and $B$B: 'Drawing a 10 card'. What is the probability of drawing a $7$7 or a $10$10?

Think: We are trying to find $P\left(A\cup B\right)$P(A∪B). Are the events mutually exclusive? Yes, as on a single card draw we cannot draw a card that is both a seven and a ten.

Do: Since there are no elements common to both events we can count the 'favourable outcomes' by just adding the outcomes of the individual events and then calculating the probability. Alternatively, we can use the rule to add the probability of the individual events.

There are $4$4 cards of each number in the deck of $52$52 cards, so there are $8$8 favourable outcomes.

$P\left(A\cup B\right)$P(A∪B)	$=$=	$\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}$number of favourable outcomestotal possible outcomes​
	$=$=	$\frac{8}{52}$852​
	$=$=	$\frac{2}{13}$213​

Alternative, calculation:

$P\left(A\cup B\right)$P(A∪B)	$=$=	$P\left(A\right)+P\left(B\right)$P(A)+P(B)
	$=$=	$\frac{4}{52}+\frac{4}{52}$452​+452​
	$=$=	$\frac{8}{52}$852​
	$=$=	$\frac{2}{13}$213​

 

Example 2

Consider an experiment of drawing one card from a deck of $52$52 cards. And define two events $A$A: 'Drawing a Club card' and $B$B: 'Drawing a $7$7 card'. What is the probability of drawing a club or a $7$7?

Think: We are trying to find $P\left(A\cup B\right)$P(A∪B). Are the events mutually exclusive? No, the events share the outcome of the $7$7 of clubs.

If we picture the Venn diagram, we can see if we were to add the number of elements in $A$A to the number of elements in $B$B we would be double counting any elements in the intersection.

 

Do: We can count the 'favourable outcomes' by adding the outcomes of the individual events and subtract the number of elements in the intersection. Alternatively, we can use the rule and the probability of the individual events.

Favourable outcomes $=4+13-1=16$=4+13−1=16

$P\left(A\cup B\right)$P(A∪B)	$=$=	$\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}$number of favourable outcomestotal possible outcomes​
	$=$=	$\frac{16}{52}$1652​
	$=$=	$\frac{4}{13}$413​

Alternative, calculation:

$P\left(A\cup B\right)$P(A∪B)	$=$=	$P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$P(A)+P(B)−P(A∩B)
	$=$=	$\frac{1}{4}+\frac{1}{13}-\frac{1}{52}$14​+113​−152​
	$=$=	$\frac{4}{13}$413​

Practice questions

Question 1

Question 2