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# Sets

## Use set notation to state set relationships

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Practice Sets

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Union and Intersection of Sets

Freezy's Ice Cream Stand polls its customers on their favorite flavor: chocolate or vanilla? 103 customers said they liked chocolate, 98 customer said they like vanilla, while 27 customers said they liked both chocolate and vanilla. How many customers said they like only chocolate? Use a Venn diagram to help you.

### Guidance

A Venn diagram is shown below.

The diagram illustrates that within some universe of data, there are two subsets, A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, which have some elements in common. The following example relates the use of a Venn diagram to a real world situation.

#### Example A

At a school of 500 students, there are 125 students enrolled in Algebra II, 257 students who play sports and 52 students that are enrolled in Algebra II and play sports. Create a Venn diagram to illustrate this information.

Solution: First, let’s let set A\begin{align*}A\end{align*} represent the students enrolled in Algebra II and set B\begin{align*}B\end{align*} represent the students who play sports. Generally speaking, it is easiest to start in the center or “intersection” of the Venn diagram. Once we place 52 in the intersection, then we can subtract it from the total number of students who play sports and the total number of student who take Algebra II to determine how many just do one or the other. Finally, we can subtract this total from 500 to figure out how many are outside the circles altogether.

There symbols that can be used to describe the number of elements in each region in the diagram as well.

Symbol Description Value for this Problem
n(A)\begin{align*}n(A)\end{align*} The number of elements in set A\begin{align*}A\end{align*} 125
n(AB)\begin{align*}n(A \cap B)\end{align*} The number of elements in the intersection of sets A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} (all the elements that are in both sets-the overlap) 52
n(AB)\begin{align*}n(A \cup B)\end{align*} The number of elements in the union of sets A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} (all the elements that are in one or both of the sets) 330
n(A)\begin{align*}n(A ^\prime)\end{align*} The number of elements in the compliment of A\begin{align*}A\end{align*} (the number of elements outside set A\begin{align*}A\end{align*}) 375
n((AB))\begin{align*}n((A \cup B)^\prime)\end{align*} The number of elements in the compliment of AB\begin{align*}A \cup B\end{align*} (everything outside the union of A\begin{align*} A \end{align*} and B\begin{align*}B\end{align*}) 170
n((AB))\begin{align*}n((A \cap B)^\prime)\end{align*} The number of elements in the compliment of AB\begin{align*}A \cap B\end{align*} (everything outside the intersection of A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}) 448
n(AB)\begin{align*}n(A \cap B^\prime)\end{align*} The number of elements in the intersection of A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}'s compliment (the number of elements in A\begin{align*}A\end{align*} but not in B\begin{align*}B\end{align*}) 73

#### Example B

Create a Venn diagram to illustrate the following information regarding the subsets M\begin{align*}M\end{align*} and N\begin{align*}N\end{align*} in universe U\begin{align*}U\end{align*}:

n(M)=89; n(N)=103; n(MN)=130; n(U)=178\begin{align*}n(M)=89; \ n(N)=103; \ n(M \cup N)=130; \ n(U)=178\end{align*}

Solution: Again, we will start in the middle or intersection. We must determine how many elements are in the intersection. Let’s consider that when we add the elements in M\begin{align*}M\end{align*} to the elements in N\begin{align*}N\end{align*}, we are adding the elements in the intersection twice. This happens because they are counted in set M\begin{align*}M\end{align*} and counted again in set N\begin{align*}N\end{align*}. Did you notice that n(M)+n(N)=89+103=192\begin{align*}n(M)+n(N)=89+103=192\end{align*} while the n(MN)=130\begin{align*}n(M \cup N)=130\end{align*}? We have double counted the 62 (192-130) elements in MN\begin{align*}M \cap N\end{align*}. Now we can put this number in the Venn diagram and work our way out as we did in the previous example.

In general, for two sets, M\begin{align*}M\end{align*} and N\begin{align*}N\end{align*}, we can use the formula: n(M)+n(N)n(MN)=n(MN)\begin{align*}n(M)+n(N)-n(M \cap N)=n(M \cup N)\end{align*} to represent the relationship between the regions in the Venn diagram and to solve problems. In this case, substituting in the given information we can determine the n(MN)\begin{align*}n(M \cap N)\end{align*} as shown below:

89+103n(MN)192n(MN)n(MN)n(MN)=130=130=62=62\begin{align*}89+103-n(M \cap N)&=130 \\ 192-n(M \cap N)&=130 \\ -n(M \cap N)&=-62 \\ n(M \cap N)&=62\end{align*}

#### Example C

Create a Venn diagram to represent the following information and answer the questions that follow.

In a survey of 150 high school students it was found that:

80 students have laptops

110 students have cell phones

125 students have iPods

62 students have both a laptop and a cell phone

58 students have both a laptop and iPod

98 students have both a cell phone and an iPod

50 students have all three items

a. How many students have just a cell phone?

b. How many students have none of the mentioned items?

c. How many students have an iPod and laptop but not a cellphone?

Solution: First we will use the given information to construct the Venn diagram as shown.

We can start by putting 50\begin{align*}{\color{red}50}\end{align*} in the center where students have all three items. Next we can find the values in blue by subtracting 50 from each of the “overlapping” values. For example, there are 62 students with both a laptop and a cell phone and 50 of them also have an iPod. To find the number that have a laptop and cell phone but no iPod, subtract 6250=12\begin{align*}62 -{\color{red}50} = {\color{blue}12}\end{align*}. Once the blue values are found we can find the green values by subtracting the blue and red values in each subset from the total in the subset. For example, the number of students with a cell phone but no other technology item is 110(50+12+48)=0\begin{align*}110 - ({\color{red}50} +{\color{blue}12} +{\color{blue}48}) = {\color{green}0}\end{align*}. Finally we can add up all the values in the circles and subtract this from 150, the total number of students surveyed to determine that 3 students have none of the items.

Now that the Venn diagram is complete, we can use it to answer the questions.

a. There are 0 students that just have a cell phone.

b. There are 3 students with none of the mentioned technology.

c. There are 8 students with an iPod and laptop but no cell phone.

Intro Problem Revisit The number of customers who said they liked both chocolate and vanilla is the intersection of the two circles in the Venn diagram that represents this situation. Since a total of 103 people said they liked chocolate, we must subtract the number who like both chocolate and vanilla to find the number who like only chocolate.

10327=76\begin{align*}103-27= 76\end{align*}

Therefore, 76 of Freezy's customers said they only like chocolate ice cream.

### Guided Practice

Use the Venn diagram to determine the number of elements in each set described in the problems.

1. n(A)\begin{align*}n(A)\end{align*}

2. n(C)\begin{align*}n(C)\end{align*}

3. \begin{align*}n(A^\prime)\end{align*}

4. \begin{align*}n(A \cap B)\end{align*}

5. \begin{align*}n(A \cup B \cup C)\end{align*}

6. \begin{align*}n(A \cap C^\prime)\end{align*}

7. \begin{align*}n(A \cap B \cap C)\end{align*}

8. \begin{align*}n(A^\prime \cap B^\prime \cap C^\prime)\end{align*}

1. \begin{align*}3 + 7 + 8 + 8 =26\end{align*}

2. \begin{align*}8 + 8 + 4 + 12 = 32\end{align*}

3. \begin{align*}8 + 4 + 12 + 6 = 30\end{align*}

4. \begin{align*}7 + 8 = 15\end{align*}

5. \begin{align*}3 + 7 + 8 + 8 + 8 + 4 + 12 = 50\end{align*}

6. \begin{align*}3 + 7 =10\end{align*}.

7. 8

8. 6

### Practice

Use the information below for problems 1-5.

In a survey of 80 households, it was found that:

30 had at least one dog

42 had at least one cat

21 had at least one “other” pet (fish, turtle, reptile, hamster, etc.)

20 had dog(s) and cat(s)

10 had cat(s) and “other” pet(s)

8 had dog(s) and “other” pet(s)

5 had all three types of pets

1. Make a Venn diagram to illustrate the results of the survey.
2. How many have dog(s) and cat(s) but no “other” pet(s)?
3. How many have only dog(s)?
4. How many have no pets at all?
5. How many “other” pet(s) owners also have dog(s) or cat(s) but not both?

Use the letters in the Venn diagram below to describe the region for each of the sets.

1. \begin{align*}A \cap B\end{align*}
2. \begin{align*}A\end{align*}
3. \begin{align*}A \cup B\end{align*}
4. \begin{align*}A \cap B^\prime\end{align*}
5. \begin{align*}(A \cap B)^\prime\end{align*}
6. \begin{align*}(A \cup B)^\prime\end{align*}
7. \begin{align*}A^\prime\end{align*}
8. \begin{align*}B^\prime \cup A\end{align*}

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