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$7 card' and 'Drawing a $10$10 card'. They have no outcomes in common. There is no card that is both a $7$7 and a $10$10.
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$7'. They have outcomes in common. We could pick a card that is a Club and a $7$7, because we could get the $7$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$2 .
It's important that we know the difference so that we can accurately answer probability problems that involve the calculation of the union of two events These are the 'either', or the 'or', style of questions we encounter. We have in fact been answering these style of questions already, particularly when we are asked questions about drawing cards from a standard deck. We will now consider two formulas for mutually exclusive and non-mutually exclusive events that might make this process simpler.
The mutually exclusive case: $P(A\cup B)=P(A)+P(B)$P(A∪B)=P(A)+P(B)
Consider a card experiment and the events A: 'Drawing a $7$7 card' and B: 'Drawing a $10$10 card'. What is $P(A\text{ or }B)$P(A or B)?
We know that:
| $P(event)$P(event) | $=$= | $\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}$number of favourable outcomestotal possible outcomes |
| Number of favourable outcomes | $=$= | number of $7$7 cards + number of $10$10 cards |
| $=$= | $4+4$4+4 | |
| $=$= | $8$8 |
Is there any double counting of favourable cards? No, because there are no cards that are both a $7$7 and a $10$10.
| So $\text{P(A or B) }$P(A or B) | $=$= | $\frac{8}{52}$852 |
| $=$= | $\frac{2}{13}$213 | |
| Note: $\text{P(A) }+\text{P(B) }$P(A) +P(B) | $=$= | $\frac{4}{52}+\frac{4}{52}$452+452 |
| $=$= | $\frac{8}{52}$852 | |
| $=$= | $\frac{2}{13}$213 |
The non-mutually exclusive case: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$P(A∪B)=P(A)+P(B)−P(A∩B)
Now, let's consider a card experiment and the events A: 'Drawing a Club card' and B: 'Drawing a $7$7 card'. What is $P(A\text{ or }B)$P(A or B)?
Following our method above, we would get:
| Number of favourable outcomes | $=$= | number of club cards + number of $7$7 cards |
| $=$= | $13+4$13+4 | |
| $=$= | $17$17 |
Do we have any double counting of favourable cards this time? Yes, we do - one card, the $7$7 of clubs. We have counted it twice - once as a club card and once as a $7$7 card. So there are actually only $16$16 favourable outcomes.
It is very easy to spot this when we consider the Venn diagram below.
So:
| $\text{P(A or B) }$P(A or B) | $=$= | $\frac{16}{52}$1652 |
| $=$= | $\frac{4}{13}$413 |
But how can we work this out using the probabilities of the individual events?
Notice:
| $\text{P(A) }+\text{P(B) }$P(A) +P(B) | $=$= | $\frac{13}{52}+\frac{4}{52}$1352+452 |
| $=$= | $\frac{17}{52}$1752 | |
| The overlap of the Venn diagram $P(A\cap B)$P(A∩B) | $=$= | $1/52$1/52 |
| And so, $P(A)+P(B)-P(A\cap B)$P(A)+P(B)−P(A∩B) | $=$= | $\frac{17}{52}-\frac{1}{52}$1752−152 |
| $=$= | $\frac{16}{52}$1652 | |
| $=$= | $\frac{4}{13}$413 |
In actual fact, the relationship above holds for ALL probability events. It's just that when two events are mutually exclusive, $P(A\cap B)=0$P(A∩B)=0 and so we get the shorter formula.
We can work backwards, given probabilities for the union or the intersection of events, to deduce if the events are mutually exclusive. Let's take a look at a question of this type.
Worked example
Example 1
Consider events A and B such that $P(A\cup B)=0.7$P(A∪B)=0.7, $P(\overline{A})=0.4$P(A)=0.4 and $P(\overline{B})=0.9$P(B)=0.9.
Are events A and B mutually exclusive?
Think: The easiest way to work this out is to firstly represent this information with a Venn diagram.
Do: 
Working with the complements, we can establish $P(A)$P(A) and $P(B)$P(B) as $0.6$0.6 and $0.1$0.1. Since $P(A\cup B)$P(A∪B) is $0.7$0.7, we also know that events outside of A and B have a probability of $0.3$0.3.
Now we consider the value in the middle region, the intersection, of the Venn diagram. Since all other regions add to $1$1, the intersection will be $0$0. Therefore, events A and B are mutually exclusive.
Lastly, we might be asked to create a Venn diagram to answer probability questions given information about the number of objects that fall into certain overlapping categories. In these questions, we are given amounts, rather than probabilities. We can use a modified version of the non-mutually exclusive events formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$n(A∪B)=n(A)+n(B)−n(A∩B) to first find out the number of items located in the intersection region. We can then fill the remaining regions in our Venn diagram. Finally, we calculate the probabilities as required.
example 2
In a group of $54$54 athletes, $34$34 play soccer and $38$38 play netball. What is the probability that a randomly selected athlete plays a) both sports and b) only soccer?
Think: We first need to work out how many athletes play both sports. We know that some must, seeing as $34+38>54$34+38>54.
| $n(A\cup B)$n(A∪B) | $=$= | $n(A)+n(B)-n(A\cap B)$n(A)+n(B)−n(A∩B) |
| $54$54 | $=$= | $34+38-n(A\cap B)$34+38−n(A∩B) |
| $n(A\cap B)$n(A∩B) | $=$= | $18$18 |
Do: We now create a Venn diagram with the intersection region first filled with our value for n(soccer and netball) = $18$18. Since n(soccer) = $34$34, we calculate the number of athletes who only play soccer as $16$16, and using the same logic, the number of athletes who only play netball as $20$20.
Thus, the probability that an athlete plays both sports is $\frac{18}{54}=\frac{1}{3}$1854=13 and the probability they only play soccer is $\frac{16}{54}=\frac{8}{27}$1654=827.
Practice questions
Question 1
A random card is picked from a standard deck. Find the probability that the card is:
red or a diamond
an ace or a diamond.
an ace of spades or an ace of clubs
a black or a face card
Question 2
Two events $A$A and $B$B are mutually exclusive.
If P(A) = $0.37$0.37 and P(A or B) = $0.73$0.73, what is P(B)?
Question 3
Question 4
In a music school of 129 students, 83 students play the piano, 80 students play the guitar and 14 students play neither.
Construct a Venn diagram to depict this situation.
A Venn diagram with two overlapping circles inside a rectangle. The circle on the left is labeled "Piano." The circle on the right is labeled "Guitar." The region inside the "Piano" circle but outside the overlapping region, is labeled with the letter "C". The region inside the "Guitar" circle but outside the overlapping region, is labeled with the letter "D". Inside the overlapping region of the two circles is labeled with the letter "B". Outside the overlapping circles but within the rectangle is the letter "A". $A$A $=$= $\editable{}$ $B$B $=$= $\editable{}$ $C$C $=$= $\editable{}$ $D$D $=$= $\editable{}$ What is the probability that a student chosen at random plays both the piano and the guitar?
A Venn diagram with two overlapping circles inside a rectangle. The circle on the left is labeled "Piano." The circle on the right is labeled "Guitar." The region inside the "Piano" circle but outside the overlapping region, is labeled with the number $35$35. The region inside the "Guitar" circle but outside the overlapping region, is labeled with the number $32$32. Inside the overlapping region of the two circles is labeled with the number $48$48. Outside the overlapping circles but within the rectangle is the number $14$14. What is the probability that a student chosen at random plays the piano or the guitar?
A Venn diagram with two overlapping circles inside a rectangle. The circle on the left is labeled "Piano." The circle on the right is labeled "Guitar." The region inside the "Piano" circle but outside the overlapping region, is labeled with the number $35$35. The region inside the "Guitar" circle but outside the overlapping region, is labeled with the number $32$32. Inside the overlapping region of the two circles is labeled with the number $48$48. Outside the overlapping circles but within the rectangle is the number $14$14. What is the probability that a student chosen at random plays neither the piano nor the guitar?
A Venn diagram with two overlapping circles inside a rectangle. The circle on the left is labeled "Piano." The circle on the right is labeled "Guitar." The region inside the "Piano" circle but outside the overlapping region, is labeled with the number $35$35. The region inside the "Guitar" circle but outside the overlapping region, is labeled with the number $32$32. Inside the overlapping region of the two circles is labeled with the number $48$48. Outside the overlapping circles but within the rectangle is the number $14$14.