Venn Diagrams Worksheets | Questions and Revision | MME

# Venn Diagram Worksheets, Questions and Revision

Level 6 Level 7

## What you need to know

### Venn Diagrams

In this topic you will learn the basics on set notation, how to construct a Venn diagram and then how to use a Venn diagram to determine probabilities. Some students may refer to the notation we use as Venn diagram symbols, which basically is set notation.

Here we will focus more on Venn diagrams, so for the full picture on sets, visit the set notation revision page. Understanding Venn diagrams will also require a good understanding of the following topics:

### Venn Diagram Symbols

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”. Understanding sets is essential for Venn diagrams.

### Venn Diagrams

Now we know more 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 1: Understanding Sets

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,

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

## Example 2: Drawing Venn Diagrams

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.

## Example 3: Using Venn Diagrams

The Venn diagram below shows $A$ is a set of odd numbers and $B$ is a set of prime numbers. Use the Venn diagram to determine the following:

a) The probability of a number that is selected at random being both odd and prime?

b) The probability of selecting an odd number that isn’t a prime number?

a) Looking at the Venn 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 $\frac{4}{11}$.

b) We are looking for all numbers that are in set A but not set B, 1 and 9, which is $\frac{2}{11}$.

## Example 4: Venn Diagrams and Conditional Probability

The Venn diagram below shows $G$ represents people selecting Geography and $B$ is people selecting History. Use the Venn diagram to determine the following:

a) Find $G\cup H$

b) Find $H\text{’}$

c) $\text{P}(G|H)$

part a) 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,

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

Part 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) 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

$A\cap B$ means the intersection of subset $A$ and subset $B$, in other words, any values that both subsets $A$ and $B$ share. In this case, the numbers that they have in common are the numbers 2 and 4, so we can input these numbers in the intersection of the two circles.

We have input two of the four numbers that are in subset $A$, leaving two other numbers, 1 and 9, that are in subset $A$, but which are not in the intersection of subsets $A$ and $B$, This means that we can write these numbers inside the $A$ circle (but not inside the intersection).

We are told that subsets $A$ and $B$ contain the numbers 1, 2, 4, 5, 6, 9 and 10 $(A\cap B$ means all the numbers in $A$ and in $B)$, so that means that if 1, 2, 4 and 9 are in subset $A$ or the intersection of subset $A$ and subset $B$, then the remaining numbers, numbers 5, 6 and 10, must be in subset $B$.

All the other numbers of the universal set $\xi$ which have not yet been placed, numbers 3, 7, 8 and 11, also need to placed in the Venn diagram. Since these numbers do not belong in either subset $A$ or subset $B$, they must be placed outside the two circles, but inside the rectangle.

Your completed Venn Diagram should be similar to the below:

a) The question states that 32 of the people surveyed watch sport regularly. 24 of these are accounted for in the intersection of the circle, so the missing number inside the $W$ circle is:

$32-24=8$

The question also states that 40 people were surveyed in total. Of these 40 people, 35 are accounted for in the Venn diagram. Therefore, the number of people who neither watch nor play sport regularly is:

$40-35=5$

Therefore, your completed Venn diagram should look like the below:

b) $W\cup B$ means “all people who either watch sport or play sport (or both)”. There is a total of 35 people in this category $(3+24+8)$.

Since there are 40 people in total, of which 35 belong to the category of $W\cup B$, we can write this as the following fraction:

$\dfrac{35}{40}$

This can be simplified to:

$\dfrac{7}{8}$

Therefore, $\text{P}(W \cup B)=\dfrac{7}{8}$

c) The phrase ‘given that they play sport regularly’ means that we should only looking at the section of people that play sport regularly. 3 people play sport regularly and 24 people both play and watch sport regularly. Therefore, the total number of people who play sport regularly is 27.

The question we need to ask ourselves is of these 27, how many of these watch sport regularly? We have already answered this question above: 24 people who play sport regularly also watch sport regularly. We can express this as the following fraction:

$\dfrac{24}{27}$

This fraction can be simplified to:

$\dfrac{8}{9}$

a) The first thing we need to do is draw a rectangle labelled $\xi$ to represent the universal set (the set of year 9 students). Then we need to draw two circles which overlap, one labelled ‘football’ and the other labelled ‘rugby’.

There is some simple information which we can input immediately. We are told that 14 students do not play rugby or football, so we can write ‘14’ outside the circles, but still inside the rectangle. We are also told that 42 students play both sports, so we can write ‘42’ in the intersection of the two circles.

The only information which we need to complete is the number of students who play football only and the number of students who play rugby only. Since there are 126 students in total, of which 14 play neither sport and of which 42 play both sports, we can work out how many students play football or rugby only:

$126-42-14 = 70\text{ students}$

Now that we know that a total of 70 students play only football or only rugby, we are now in a better position to work out exactly how many students fall into each category. We are told that the ratio of the number of students who play football only to the number of students who play rugby only is $7 : 3$. This means that $\frac{7}{10}$ of the 70 students play football only and $\frac{3}{10}$ of the 70 students play rugby only. (We are dealing in tenths here because the sum of the ratio is 10.).

The number of students who play football only can be calculated as follows:

$\dfrac{7}{10}\times70 = 49\text{ students}$

The number of students who play rugby only can be calculated as follows:

$\dfrac{3}{10}\times70 = 21\text{ students}$

Therefore we need to insert ‘49’ inside the football circle (but not in the intersection) and ‘21’ inside the rugby circle (but not in the intersection).

Our completed Venn diagram should be similar to the below:

b) We know that 21 students play rugby only out of the total 126 students. We can write this as the following fraction:

$\dfrac{21}{126}$

This fraction can be simplified to:

$\dfrac{1}{6}$

(Since $\frac{21}{126}$ is not the easiest fraction to simplify, try typing $21\div126$ into your calculator and it should display the answer of $\frac{1}{6}$.)

a) The number of students in year 10 is the total of:

• All the students that study geography only (40)
• All the students that study history only (52)
• All the students that study both geography and history (14)
• All the students that study neither history nor geography (36)

$40+52+14+36=142\text{ students}$

b) We know that there are 102 students in total and, of these, 14 study both history and geography. We can express this as the following fraction:

$\dfrac{14}{102}$

This fraction can be simplified to:

$\dfrac{7}{51}$

(You could give the probability as a decimal or as a percentage (the question doesn’t state how the answer should be given), but since both the decimal (0.137….) and the percentage (13.725…%) have several decimal places, leaving the probability as a fraction would probably be the most sensible option here, but the decimal and percentage answers are not incorrect.)

c) The phrase ‘given that the student selected does not study geography’ tells us that we need to work out how many students do not study geography. The students that don’t study geography are those who study history only (52 students) or those who study neither subject (36 students). The total that do not study geography is therefore $52+36 = 88$ students.

Of these 88 students, we need to find how many do not study history. The 36 that study neither subject are the ones that do not study history, so 36 out of the 88 who do not study geography do not study history. We can write this as the following fraction:

$\dfrac{36}{88}$

This fraction can be simplified to:

$\dfrac{18}{44}$

This fraction can be simplified again to:

$\dfrac{9}{22}$

(The decimal answer of 0.409….. or the percentage answer of 40.909….% are both acceptable ways to express the probability also.)

a) The first thing we need to do is draw a rectangle labelled $\xi$ to represent the universal set of the 80 students, and inside the rectangle draw 3 circles that overlap. The three circles should be labelled ‘sharks’, ‘crocodiles’ and ‘hippos’. The three circles should be drawn in such a way that there is a small section in the middle where all three circles overlap.

INSERT BASIC VENN DIAGRAM

What we need to do next is to input the data. Some of the data we are going to input will be easy, other parts may be more confusing. Let’s start with one fact at a time:

• All 80 students like at least 1 of the animals. There is not much we can do with this information. However, this is a key piece of information since it tells us that once we have filled in the Venn diagram, the numbers should add up to 80.

• 15 students like all 3 animals. If 15 students like all 3 animals, then we can insert ’15’ in the intersection of all 3 circles:

• 14 students like sharks and crocodiles but do not like hippos. With this information we can insert ‘14’ in the section where sharks and crocodiles intersect (which should be outside the hippos circle):

• 23 students like crocodiles and hippos. This is the piece of information which can cause confusion. Although that statement says that 23 students like crocodiles and hippos, this doesn’t mean that 23 students like crocodiles and hippos only. This figure of 23 also includes students who like sharks as well as crocodiles and hippos. This is the slightly confusing part of the question! 15 students like sharks, crocodiles and hippos, so that means that these 15 students definitely like crocodiles and hippos. So, if there are 23 students in total who like crocodiles and hippos, that means that there are 8 students who like crocodiles and hippos only. Therefore we can now insert ‘8’ in the intersection of crocodiles and hippos (but not in the intersection of sharks, crocodiles and hippos). You may need to read this paragraph a few times: it will sink in and make sense eventually!

• 21 students like sharks and hippos. This is a very similar piece of information to the above piece of information about crocodiles and hippos. 15 students like sharks, crocodiles and hippos, therefore that means that there are at least 15 students who like sharks and hippos. Since we have been told that the total number of students who like sharks and hippos is 21, then this means that 6 students like sharks and hippos, but not crocodiles. Therefore we can insert ‘6’ in the intersection of sharks and hippos.

• 44 students like crocodiles.  We know that 14 students like crocodiles and sharks, that 15 students like sharks, crocodiles and hippos, and that a further 8 that like crocodiles and hippos, so if we add these numbers up we know that 37 students like crocodiles. Since we are told that 44 students like crocodiles in total, this means that there are 7 students who like crocodiles only. Therefore we can insert ‘7’ in the part of the crocodile circle that does not overlap with any other section.

• 12 students only like sharks. All we need to do is insert ‘12’ in the sharks only section.

INSERT VENN DIAGRAM WITH NUMBER 12 ADDED?

All we need to do is work out how many students only like hippos. We know that there are 80 students in total, and we have filled in every section of the Venn diagram except this final hippos only section. If we subtract all of the numbers from our total of 80, we will be able to work out how many students only like hippos:

$80-12-14-7-6-15-8=18$

Therefore our completed Venn diagram will look similar to the below:

b) There are 80 students in total, of which 18 only like hippos. This means that 18 out of the 80 only like hippos which can be written as the following fraction:

$\dfrac{18}{80}$

This fraction can be simplified to:

$\dfrac{9}{40}$

So, the probability of a person being selected at random and only liking hippos is $\frac{9}{40}$ (or $9\div40=0.225$ if you prefer to give your answer as a decimal. The answer of 22.5% is also acceptable.)

c) There are 44 students in total that like crocodiles. There are 14 students that like crocodiles and sharks and 8 students that like crocodiles and hippos, so there are 22 students that like crocodiles and one other animal.

This means that 22 out of the 80 only like crocodiles and one other type of animal, which can be written as the following fraction:

$\dfrac{22}{80}$

This fraction can be simplified to:

$\dfrac{11}{40}$

So, the probability of a person being selected at random and only liking hippos is $\frac{11}{40}$ or $9\div40=0.275$ if you prefer to give your answer as a decimal. The answer of 27.5% is also acceptable.

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Level 6-7

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