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# 1.5: Mutually Inclusive Events

Difficulty Level: Basic Created by: CK-12
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Suppose you're selecting a card from a deck of cards with your eyes closed. What is the probability that you'll select a red card or a Queen? Is this a mutually exclusive event or a mutually inclusive event? Can you explain your answer?

### Watch This

First watch this video to learn about mutually inclusive events.

Then watch this video to see some examples.

Watch this video for more help.

### Guidance

When determining the probabilities of mutually inclusive events, we can apply the words and ()\begin{align*}(\cap)\end{align*} and or ()\begin{align*}(\cup)\end{align*} using the Addition Rule. Let’s take another look at Venn diagrams when defining mutually inclusive events.

If events A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} share some overlap in the Venn diagram, they may be considered 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, you wanted to pick a number from 1 to 10 that is less than 4 and pick an even number. Let event A\begin{align*}A\end{align*} be picking a number less than 4 and event B\begin{align*}B\end{align*} be picking an even number.

A={1,2,3}P(A)=310B={2,4,6,8,10}P(B)=510P(A and B)=110\begin{align*}&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} \\ \end{align*}

The reason why P(A and B)=110\begin{align*}P(A \ \text{and} \ B) = \frac{1}{10}\end{align*} 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 inclusive events, remember, can occur at the same time. Look at the Venn diagram below. What do you think we need to do in order to calculate the probability of AB\begin{align*}A \cup B\end{align*} just from looking at this diagram?

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

P(A or B)P(AB)=P(A)+P(B)P(A and B)or=P(A)+P(B)P(AB)\begin{align*}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)\end{align*}

where \begin{align*}\cap\end{align*} represents and and \begin{align*}\cup\end{align*} represents or.

This is known as the Addition Principle (Rule).

Addition Principle

P(AB)=P(A)+P(B)P(AB)\begin{align*}P(A \cup B) = P(A) + P(B) - P(A \cap B)\end{align*}

Think about the idea of rolling a die. Suppose event A\begin{align*}A\end{align*} is rolling an odd number with the die, and event B\begin{align*}B\end{align*} is rolling a number greater than 2.

Event A={1,3,5}\begin{align*}A = \{1, 3, 5\}\end{align*}

Event B={3,4,5,6}\begin{align*}B = \{3, 4, 5, 6\}\end{align*}

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

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

#### Example A

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

P(A)P(A)P(B)P(B)P(AB)P(AB)P(AB)P(AB)P(AB)=probability of selecting a club=1352=probability of selecting a ten=452=152=P(A)+P(B)P(AB)=1352+452152=1652=413\begin{align*}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}\end{align*}

#### Example B

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

A={1,2,3,4}P(A)P(A)P(A)B={1,3,5,7,9}P(B)P(B)P(B)P(AB)P(AB)P(AB)P(AB)P(AB)P(AB)=probability of selecting a number less than 5=410=25=probability of selecting a number that is odd=510=12=210=15=P(A)+P(B)P(AB)=25+1215=410+510210=710\begin{align*}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}\end{align*}

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

#### Example C

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

P(A)=probability of obtaining a sum less than 7P(A)=1536\begin{align*}&P(A) = \text{probability of obtaining a sum less than 7}\\ \\ &P(A) = \frac{15}{36}\end{align*}

P(B)=probability of obtaining a sum less than 4P(B)=336\begin{align*}&P(B) = \text{probability of obtaining a sum less than 4}\\ \\ &P(B) = \frac{3}{36}\end{align*}

Notice that there are 3 elements in common. Therefore, the events are mutually inclusive, and we must account for the double counting.

P(A and B)P(AB)P(AB)P(AB)P(AB)P(AB)P(AB)P(AB)=336=336=112=P(A)+P(B)P(AB)=1536+336112=1536+336336=1536=512\begin{align*}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}\\\end{align*}

Points to Consider

### Guided Practice

A bag contains 26 tiles with a letter on each, one tile for each letter of the alphabet. What is the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it?

Answer:

The first 10 letters of the alphabet are 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', and 'J'. Also, the vowels are 'A', 'E', 'I', 'O', and 'U'. The only letters that are both in the first 10 letters of the alphabet and vowels are 'A', 'E', and 'I'. Therefore, the probability of reaching into the bag and randomly choosing a tile with one of the first 10 letters of the alphabet on it or randomly choosing a tile with a vowel on it can be calculated as follows:

AP(A)P(A)P(A)BP(B)P(B)P(AB)P(AB)P(AB)P(AB)P(AB)={A, B, C, D, E, F, G, H, I, J}=probability of selecting one of the first 10 letters of the alphabet=1026=513={A, E, I, O, U}=probability of selecting a vowel=526=326=P(A)+P(B)P(AB)=513+526326=1026+526326=1226=613\begin{align*}A &= \{\text{A, B, C, D, E, F, G, H, I, J}\}\\ P(A) &=\text{probability of selecting one of the first 10 letters of the alphabet}\\ P(A) &= \frac{10}{26}\\ P(A) &= \frac{5}{13}\\ B &= \{\text{A, E, I, O, U}\}\\ P(B) &=\text{probability of selecting a vowel}\\ P(B) &= \frac{5}{26}\\ \\ P(A \cap B) &= \frac{3}{26}\\ \\ P(A \cup B) &= P(A) + P(B) - P(A \cap B)\\ P(A \cup B) &= \frac{5}{13} + \frac{5}{26} - \frac{3}{26}\\ P(A \cup B) &= \frac{10}{26} + \frac{5}{26} - \frac{3}{26}\\ P(A \cup B) &= \frac{12}{26} = \frac{6}{13}\end{align*}

### Practice

1. Consider a sample set as \begin{align*}S = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}\end{align*}. Event \begin{align*}A\end{align*} is the multiples of 4, while event \begin{align*}B\end{align*} is the multiples of 5. What is the probability that a number chosen at random will be from both \begin{align*}A\end{align*} and \begin{align*}B\end{align*}?
2. For question 1, what is the probability that a number chosen at random will be from either \begin{align*}A\end{align*} or \begin{align*}B\end{align*}?
3. Jack is a student in Bluenose High School. He noticed that a lot of the students in his math class were also in his chemistry class. In fact, of the 60 students in his grade, 28 students were in his math class, 32 students were in his chemistry class, and 15 students were in both his math class and his chemistry class. He decided to calculate what the probability was of selecting a student at random who was either in his math class or his chemistry class, but not both. Draw a Venn diagram and help Jack with his calculation.
4. Brenda did a survey of the students in her classes about whether they liked to get a candy bar or a new math pencil as their reward for positive behavior. She asked all 71 students she taught, and 32 said they would like a candy bar, 25 said they wanted a new pencil, and 4 said they wanted both. If Brenda were to select a student at random from her classes, what is the probability that the student chosen would want:
1. a candy bar or a pencil?
2. neither a candy bar nor a pencil?
5. A card is chosen at random from a standard deck of cards. What is the probability that the card chosen is a heart or a face card? Are these events mutually inclusive?
6. What is the probability of choosing a number from 1 to 10 that is greater than 5 or even?
7. A bag contains 26 tiles with a letter on each, one tile for each letter of the alphabet. What is the probability of reaching into the bag and randomly choosing a tile with one of the letters in the word ENGLISH on it or randomly choosing a tile with a vowel on it?
8. Are randomly choosing a teacher and randomly choosing a father mutually inclusive events? Explain your answer.
9. Suppose 2 events are mutually inclusive events. If one of the events is passing a test, what could the other event be? Explain your answer.
10. What is the probability of randomly choosing a number from 1 to 10 that is a factor of 8 or a factor of 10?

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### Vocabulary Language: English

Addition Principle

If events A and B are mutually inclusive, then P(A or B) = P(A) + P(B) – P(A and B)

Mutually Exclusive Events

Mutually exclusive events have no common outcomes.

Mutually Inclusive Events

Mutually inclusive events can occur at the same time.

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Feb 24, 2012
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