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# 1.3: Mutually Inclusive and Mutually Exclusive Events

Difficulty Level: At Grade Created by: CK-12

When determining the probabilities of events, we must also look at 2 additional terms. These terms are mutually inclusive and mutually exclusive. When we add probability calculations of events described by these terms, we can apply the words and $(\cap)$ and or $(\cup)$ using the Addition Rule. Let’s take another look at Venn diagrams when defining these terms.

2 events $A$ and $B$ that cannot occur at the same time are mutually exclusive events. They have no common outcomes. See in the diagram below that $P(A \ \text{and} \ B) = 0$. Notice that there is no intersection between the possible outcomes of event $A$ and the possible outcomes of event $B$. For example, if you were asked to pick a number between 1 and 10, you cannot pick a number that is both even and odd. These events are mutually exclusive.

To calculate the probability of picking a number from 1 to 10 that is even or picking a number from 1 to 10 that is odd, you would follow the steps below:

$A = \left \{2, 4, 6, 8, 10 \right \}\!\\\\P(A) = \frac{5}{10}\!\\\\B = \left \{1, 3, 5, 7, 9 \right \}\!\\\\P(B) = \frac{5}{10}\!\\\\P(A \ \text{or} \ B) = \frac{5}{10} + \frac{5}{10}\!\\\\P(A \ \text{or} \ B) = \frac{10}{10}\!\\\\P(A \ \text{or} \ B) = 1$

The probability of picking a number from 1 to 10 that is even and picking a number from 1 to 10 that is odd would just be 0, since these are mutually exclusive events. In other words, $P(A \ \text{and} \ B) = 0$.

If events $A$ and $B$ share some overlap in the Venn diagram, they may be considered not mutually exclusive events, but mutually inclusive events. Look at the diagrams below to see how these events can occur. Mutually inclusive events can occur at the same time. Say, for example, in the problem above, you wanted to pick a number from 1 to 10 that is less than 4 and pick an even number.

$A = \left \{1, 2, 3 \right \}\!\\\\P(A) = \frac{3}{10}\!\\\\B = \left \{2, 4, 6, 8, 10 \right \}\!\\\\P(B) = \frac{5}{10}\!\\\\P(A \ \text{and} \ B) = \frac{1}{10} \!\\$

The reason why $P(A \ \text{and} \ B) = \frac{1}{10}$ is because there is only 1 number from 1 to 10 that is both less than 4 and even, and that number is 2.

When representing this on the Venn diagram, we would see something like the following:

Mutually exclusive events, remember, cannot occur at the same time. Mutually inclusive events can. Look at the Venn diagram below. What do you think we need to do in order to calculate the probability of $A \cup B$ just from looking at this diagram?

If you look at the diagram, you see that the calculation involves not only $P(A)$ and $P(B)$, but also $P(A \cap B)$. However, the items in $A \cap B$ are also part of event $A$ and event $B$. To represent the probability of $A$ or $B$, we need to subtract the $P(A \cap B)$; otherwise, we are double counting. In other words:

$P(A \ \text{or} \ B) &= P(A) + P(B) - P(A \ \text{and} \ B)\\& \qquad \qquad \text{or}\\P(A \cup B) &= P(A) + P(B) - P(A \cap B)$

where $\cap$ represents and and $\cup$ represents or.

This is known as the Addition Principle (Rule).

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(A \cap B) = 0$ for mutually exclusive events.

Think about the idea of rolling a die. Suppose event $A$ is rolling an odd number with the die, and event $B$ is rolling a number greater than 2.

Event $A = \{1, 3, 5\}$

Event $B = \{3, 4, 5, 6\}$

Notice that the sets containing the possible outcomes of the events have 2 elements in common. Therefore, the events are mutually inclusive.

What if we said that we were choosing a card from a deck of cards? Suppose event $A$ is choosing an eight and event $B$ is choosing an Ace.

Notice that the sets containing the possible outcomes of the events have no elements in common. Therefore, the events are mutually exclusive.

Now take a look at the example below to understand the concept of double counting.

Example 8

What is the probability of choosing a card from a deck of cards that is a club or a ten?

Solution:

$P(A) &=\text{probability of selecting a club}\\P(A) &= \frac{13}{52}\\P(B) &=\text{probability of selecting a ten}\\P(B) &= \frac{4}{52}\\\\ P(A \cap B) &= \frac{1}{52}\\\\P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\P(A \cup B) &= \frac{13}{52} + \frac{4}{52} - \frac{1}{52}\\ P(A \cup B) &= \frac{16}{52}\\ P(A \cup B) &= \frac{4}{13}$

Example 9

What is the probability of choosing a number from 1 to 10 that is less than 5 or odd?

Solution:

$A = \{1, 2, 3, 4\}\\P(A) &=\text{probability of selecting a number less than 5}\\P(A) &= \frac{4}{10}\\P(A) &= \frac{2}{5}\\B = \{1, 3, 5, 7, 9\}\\P(B) &=\text{probability of selecting a number that is odd}\\P(B) &= \frac{5}{10}\\P(B) &= \frac{1}{2}\\\\ P(A \cap B) &= \frac{2}{10}\\P(A \cap B) &= \frac{1}{5}\\\\P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\P(A \cup B) &= \frac{2}{5} + \frac{1}{2} - \frac{1}{5}\\ P(A \cup B) &= \frac{4}{10} + \frac{5}{10} - \frac{2}{10}\\P(A \cup B) &= \frac{7}{10}$

Notice in the previous 2 examples how the concept of double counting was incorporated into the calculation by subtracting the $P(A \cup B)$. Let’s try a different example where you have 2 events happening.

Example 10

2 fair dice are rolled. What is the probability of getting a sum less than 7 or a sum equal to 10?

Solution:

$P(A) = \text{probability of obtaining a sum less than 7}\!\\\\P(A)=\frac{15}{36}$

$P(B) = \text{probability of obtaining a sum equal to 10}\!\\\\P(B) = \frac{3}{36}$

There are no elements that are common, so the events are mutually exclusive.

$P(A \ \text{or} \ B) &= P(A) + P(B)\\P(A \cup B) &= P(A) + P(B)\\P(A \cup B) &= \frac{15}{36} + \frac{3}{36}\\P(A \cup B) &= \frac{18}{36}\\ P(A \cup B) &= \frac{1}{2}\\P(A \ \text{and} \ B) &= 0$

Example 11

2 fair dice are rolled. What is the probability of getting a sum less than 7 or a sum less than 4?

Solution:

$P(A) = \text{probability of obtaining a sum less than 7}\!\\\\P(A) = \frac{15}{36}$

$P(B) = \text{probability of obtaining a sum less than 4}\!\\\\P(B) = \frac{3}{36}$

Notice that there are 3 elements in common. Therefore, the events are not mutually exclusive, and we must account for the double counting.

$P(A \ \text{and} \ B) &= \frac{3}{36}\\ P(A \cap B) &= \frac{3}{36}\\ P(A \cap B) &= \frac{1}{12}\\\\P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\P(A \cup B) &= \frac{15}{36} + \frac{3}{36}-\frac{1}{12}\\ P(A \cup B) &= \frac{15}{36} + \frac{3}{36}-\frac{3}{36}\\ P(A \cup B) &= \frac{15}{36}\\P(A \cup B) &= \frac{5}{12}\\$

Points to Consider

• What is the difference between the probabilities calculated with the Multiplication Rule versus the Addition Rule?
• Can mutually exclusive events be independent? Can they be dependent?

Vocabulary

With 2 events, the probability of one event occurring or another is given by: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Dependent events
2 or more events whose outcomes affect each other. The probability of occurrence of one event depends on the occurrence of the other.
Independent events
2 or more events whose outcomes do not affect each other.
Multiplication Rule
For 2 events ($A$ and $B$), the probability of $A$ and $B$ is given by: $P(A \ \text{and} \ B) = P(A) \times P(B)$.
Mutually exclusive events
2 events are mutually exclusive when they cannot both occur simultaneously.
Mutually inclusive events
2 events are mutually inclusive when they can both occur simultaneously.
Outcomes
The possible results of 1 trial of a probability experiment.
Probability
The chance that something will happen.
Random
When everyone or everything in a population has an equal chance of being selected.
Sample space
The set of all possible outcomes of an event or group of events.
Venn diagram
A diagram of overlapping circles that shows the relationships among members of different sets.
$\cup$
The union of 2 events, where the sample space contains 2 events, $A$ and $B$, and each member of the set belongs to $A$ or $B$.
$\cap$
The intersection of 2 events, where the sample space contains 2 events, $A$ and $B$, and each member of the set belongs to $A$ and $B$.

Feb 23, 2012

Jan 27, 2015
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