Sets and Venn diagrams

​​In a nutshell

A set is a collection of 'things' known as elements. In statistics, sets are usually presented on a Venn diagram. This provides an easy way for the probability of an element, inside a given set, to be worked out.

Set theory

Sets are represented by a list of elements inside two curly brackets: { }

Example 1

The set of even numbers between $$1$$ and $$10$$:

• { $$2,4,6,8$$ }
• {$$x:$$ $$x$$ is an even number between one and ten}​

​

Venn diagrams representing sets

On a Venn diagram, each set is represented by a circle which contains all the elements within it.

​

Set Notation	Definition	Venn diagram Representation
n($$A$$​)	Number of elements in set $$A$$.
​The empty set $$\emptyset$$​​​	A set which has nothing in it.
Universal Set $$\xi$$ or $$E$$ or $$U$$​​	The set containing all elements.
Complement $$A'$$ or $$\overline{A}$$​​	All elements not in $$A$$.
Intersection $$A \cap B$$	All elements in both $$A$$​ and $$B$$.
Union $$A \cup B$$	All elements in either $$A$$ or $$B$$ or both.
Conditional (A given B) $$A|B$$​​	All elements in $$A$$​ given they are in $$B$$.
Disjoint sets	Two sets that share none of the same elements.
​Subset $$A \subset B$$​	​$$A$$​ is a set inside of $$B$$.

Finding probabilities from Venn diagrams

You can use Venn diagrams to work out the probability of an event by counting the number of elements in the relevant set and taking that as a fraction out of the total.

​

Example 2

The Venn diagram shows the numbers of students that said they liked green and blue. 

Find the probability that when selected at random, a pupil would like:

1. The colour green:

 $$\text{N(Year 11 Students)}= 100$$

Work out the number of students that like green.

$$\text{n}(G)= 20+23=43$$

Work out the probability that a student likes green.

$$\text{P(}G)= \frac{43}{100}$$

2. Both blue and green:

Work out the number of students that like both blue and green.

$$\text{n(}B\cap G)= 20$$

Work out the probability that a student likes both blue and green.

$$\text{P(}B \cap G)= \frac{20}{100}=\frac{1}{5}$$

3. Blue given they like green:

Take the number of students that like both blue and green as a fraction out of the number that like green.

$$\text{n(}G)=43 \\\text{ n(}B\cap G)= 20 \\\text{P(}B|G)=\frac{20}{43}$$​​

​​

Note: In the example above $$25$$ is the number of students that did not fit into either category - they did not like blue or green!