**Sets and Venn Diagrams** *Revision and Worksheets*

## What you need to know

In mathematics, a **set** is a collection of things. Sets are often denoted using the curly brackets {}, for example the set of the first 5 odd numbers is \{1, 3, 5, 7, 9\}. There are two definitions that you should know up front. If we have two sets, A and B, then

- A\cup B is the
**union**of A and B. It is the collection of all the things that are contained in A**or**B (or both). - A\cap B is the
**intersection**of A and B. It is the collection of things that are contained in**both**A**and**B.

In a more informal way, \cup means “or” and \cap means “and”.

**Example: **Let A=\{1, 5, 8, 10\} and B=\{2, 3, 5, 7, 8\}. Work out which numbers are in A\cup B and then in A\cap B.

Firstly, A\cup B is the collection of all numbers that are in either of them (or both of them). In other words, all we need to do is combine the two sets, ignoring any repeats. So, we get

A\cup B=\{1, 2, 3, 5, 7, 8, 10\}

Now, for A\cap B we want all numbers that are in both sets. Looking, we can see that there are only two: 5 and 8. Therefore,

A\cap B=\{5, 8\}.

Now we know about our set notation, we can move on to how to display sets using **Venn diagrams**.

A typical Venn diagram looks like the picture on the left. One circle represents the set A, whilst the other represents B. The section where the two circles cross over is for all things that are contained in both sets, i.e. it is A\cap B.

The box is called the **universal set **and is denoted by the Greek letter \xi, and it contains all of the objects we’re concerned with, including objects that aren’t contained within neither A nor B. Such things belong in the section that is inside the rectangle but outside the circle.

**Example: **We have universal set \xi=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}. Let A be the set of odd numbers and let B be the set of prime numbers. Draw a Venn diagram for this information.

Firstly, we must establish sets and . Which of the numbers from the universal set are odd? Answer: 1, 3, 5, 7, 9, and 11. Which of the numbers from the universal set are prime? Answer: 2, 3, 5, 7, and 11. There’s some crossover here – 3, 5, 7, and 11 are contained in __both__ sets, which means they go in the part where the circles intersect.

What about the numbers that are odd but aren’t prime? There are two: 1 and 9. These will go in the part of circle A that doesn’t intersect with B. Similarly, there is 1 number that is prime but isn’t odd: 2. It will go in the part of circle B that doesn’t intersect with A.

Finally, all the numbers that are neither odd nor prime: 4, 6, 8, and 10, will go inside the rectangle but outside either circle. The result is the Venn diagram on the right.

Now, we can discuss how to use Venn diagrams to answer questions of **probability**. Suppose you picked a number at random from the universal set in the example above. We might then ask: “what are the chances it is both odd and prime?”. Looking at the diagram, we see that 4 of the 11 numbers are in both A and B, so the probability of a number being both odd and prime is \frac{4}{11}.

We write

\text{P}(A\cap B)=\dfrac{4}{11}

So, G\cup H means “G or H”, meaning we’re interested firstly in all people to take Geography or History (or both). The total number is 6+12+5=23, and therefore,

\text{P}(G\cup H)=\dfrac{23}{32}

For question b), we need H\text{’} which means all people who don’t take History. From the diagram we can see that 6+9=15 people don’t take it, so we get that

\text{P}(H\text{’})=\dfrac{15}{32}

Part c) is different. It is only for **higher students** and is a question of something called **conditional probability**. The | line means “given”, so \text{P}(G|H) means “the probability of G given H”. In other words, the probability that someone takes Geography **given** that we already know they take History. Venn diagrams are useful for these questions. If we know someone takes History, then they must be one of the 5+12=17 people. Then, there are 12 out of those 17 people who take Geography, or in other words

\text{P}(G|H)=\dfrac{12}{17}

## Example Questions

1) We have:

\xi=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

A\cap B=\{2, 4\}

A\cup B=\{1, 2, 4, 5, 6, 9, 10\}

A=\{1, 2, 4, 9\}

Draw a Venn diagram for this information.

Firstly, we can fill in the intersection between the circles with 2 and 4. Secondly, we can fill in the part of A that doesn’t intersect with B with the remaining square numbers. The square numbers in the universal set are: 1, 4, and 9, but since 4 is in the intersection, the only numbers in the part of A that doesn’t intersect are 1 and 9.

Now we know which numbers are in A, the remaining numbers in A\cup B must all belong to B. They are: 5, 6, and 10.

Finally, any numbers that are in the universal set but not in A\cup B must go outside both circles. So, the resulting Venn diagram is:

2) Kimmy collected data on 40 people’s enjoyment of sport. P is the set of people who play sports regularly, and W is the set of people who watch sports regularly. 32 of the people she asked watch sports regularly. An incomplete Venn diagram for her data is shown below.

a) Complete the Venn diagram.

b) Work out \text{P}(W \cup B) for someone picked at random.

c) (HIGHER ONLY) Work out the probability that someone picked at random watches sports regularly __given__ that they play sports regularly.

a) The question says that 32 of the people watch sports regularly. 24 of those are accounted for in the intersection of the circle, so the missing number inside the W circle must be

32-24=8

The question also says that 40 people were surveyed in total. Therefore, the number of people who neither watched nor played sports regularly is

40-24-8-3=5

Thus, the completed Venn diagram looks like

b) W\cup B means “all people who either watch sport or play sport (or both)”, so the total number of people in this category is

3+24+8=35

There are 40 people in total, so we get

\text{P}(W\cup B)=\dfrac{35}{40}=\dfrac{7}{8}

c) We want the probability that someone watches sports given that they play sports. Looking at the diagram, we can see that there are 24+3=27 people who play sports, and of those people, 24 also watch sports. Therefore, the probability the someone watches sports given that they play sports is

\dfrac{24}{27}=\dfrac{8}{9}.

## Sets and Venn Diagrams Revision and Worksheets

## Sets and Venn Diagrams Teaching Resources

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