5.06 The addition rule
Previously, we looked at probabilities that involved the intersection of two or more events. Now, we will examine probabilities involving the union of two or more events.
| Intersection | Union |
|---|---|
| $P(A\text{ and }B)$P(A and B) | $P(A\text{ or }B)$P(A or B) |
Identifying mutually exclusive events
To find probabilities involving the union of two events, we must first know whether or not the two events are related. If events are mutually exclusive, it means they cannot happen at the same time.
Some examples of experiments that involve mutually exclusive events are:
- 'Flipping a head' and 'flipping a tail' with a single flip of a coin. You cannot flip a head and a tail at the same time.
- 'Rolling an even number' and 'rolling an odd number' from a single roll of a die. We can't roll any number which is both even and odd.
- 'Drawing a $7$7 card' and 'Drawing a $10$10 card' when picking a single card from a deck. 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:
- 'Drawing a Club card' and 'drawing a $7$7' when picking a single card from a deck. We could pick a card that is a Club and a $7$7, because we could get the $7$7 of clubs.
- 'Rolling an even number' and 'Rolling a prime number' from a single roll of a die. They have outcomes in common, namely, the number $2$2 .
Probability of mutually exclusive events
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) | $=$= | $\frac{\text{number of favorable outcomes}}{\text{total possible outcomes}}$number of favorable outcomestotal possible outcomes |
| Number of favorable outcomes | $=$= | number of $7$7 cards + number of $10$10 cards |
| $=$= | $4+4$4+4 | |
| $=$= | $8$8 |
Is there any double-counting of favorable 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 |
This example illustrates the first of two addition rules for probability
Probability of non-mutually exclusive events
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 favorable outcomes | $=$= | number of club cards + number of $7$7 cards |
| $=$= | $13+4$13+4 | |
| $=$= | $17$17 |
Do we have any double-counting of favorable 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 favorable 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 $P(A\text{ and }B)$P(A and B) | $=$= | $1/52$1/52 |
| And so, $P(A)+P(B)-P(A\text{ and }B)$P(A)+P(B)−P(A and B) | $=$= | $\frac{17}{52}-\frac{1}{52}$1752−152 |
| $=$= | $\frac{16}{52}$1652 | |
| $=$= | $\frac{4}{13}$413 |
This leads to the second of the addition rules for probability.
In fact, the relationship above holds for all probability events. It's just that when two events are mutually exclusive, $P(A\text{ or }B)=0$P(A or B)=0 and so we get the shorter formula.
We can work backward, 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 $P(A\text{ or }B)=0.7$P(A or B)=0.7, $P(\text{not }A)=0.4$P(not A)=0.4 and $P(\text{not }B)=0.9$P(not B)=0.9.
Are events $A$A and $B$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 $P(A\text{ or }B)$P(A or 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.
Practice questions
Question 2
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 3
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 4
Question 5
Use the formula $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$P(A or B)=P(A)+P(B)−P(A and B) to solve the following question.
If $P\left(A\right)=\frac{1}{7}$P(A)=17, $P\left(B\right)=\frac{1}{2}$P(B)=12 and $P\left(A\andword B\right)=\frac{1}{14}$P(AandB)=114, find $P(A\text{ or }B)$P(A or B).