# Probability and Venn Diagrams

## Probability and Venn Diagrams Revision

**Probability and Venn Diagrams**

**Probability** is a measure of how **likely** something is to happen.

It always falls** between \mathbf{0} and \mathbf{1}**, with 0 being impossible and 1 being certain.

**Notation:Â **The probability of an event A is \mathbb{P}(A).

**Calculating Probabilities**

To calculate the probability of an event, the formula is:

\dfrac{\color{red}\text{number of desired outcomes}}{\color{blue}\text{total number of possible outcomes}}

**Example:** What is the probability of obtaining an even number when rolling a 6-sided die?

Our possible outcomes are \color{blue}1,2,3,4,5,6, of which \color{red}2,4,6 are even. This gives \color{red}3 desired outcomes, out of \color{blue}6 possible outcomes, so:

\mathbb{P}(\text{roll an even number})=\dfrac{\color{red}3}{\color{blue}6}=\dfrac{\color{red}1}{\color{blue}2}

**AND & OR**

A\cap B means **\mathbb(A)****Â AND \mathbb(B)** – that both event A and event B happen.

A\cup B means **\mathbb(A)****Â OR \mathbb(B)** – that event A happens or event B happens (or both).

The probabilities of A\cap B and A\cup B are related by the formula:

\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)

These probabilities can be calculated with a **Venn Diagram**.

**Example:** considering dice again, if A is “rolls a multiple of 3” and B is “rolls 3 or 4”, then we have:

By looking at where the circles overlap, we notice there is one value, so A\cap B contains one value.

By looking at the circles in their entirety, we notice that there are 3 values contained within circle A or circle B, so A\cup B contains three values.

Since there are six values overall, we can conclude:

\mathbb{P}(A\cap B)=\dfrac{1}{6}

\mathbb{P}(A\cup B)=\dfrac{3}{6}=\dfrac{1}{2}

**Complement of an Event**

The **complement** of an event is the event that it doesn’t happen, for example if A is rolls a 3 or 4 then the complement of A, written A', is does not roll a 3 or 4, i.e. rolls a 1,2,5 or 6.

Since any event definitely either does or does not happen,** the probability of an event and its complement must add to \mathbf{1}.**

\mathbb{P}(A)+\mathbb{P}(A')=1

On a **Venn Diagram**, the complement of A is everything not contained within the circle representing A.

**Two-Way Tables**

**Two way tables** model two events, A and B, by displaying every combination of whether or not each one happens. They can either display **frequency** (which should be familiar from GCSE) or **probability**. If it displays **probability**, then the numbers in it excluding row totals and column totals **must add to \mathbf{1}**, and the final total in the bottom right must be 1. An example is given below.

**Note:** the totals are probabilities for one event, e.g. the total in the A column is the probability of event A. Using this, we can create the most general form of a two-way table:

## Probability and Venn Diagrams Example Questions

**Question 1:** Inside a bag are 3 red marbles and 8 green marbles. Cate picks one marble at random. What is the probability that she has picked a green marble?

**[1 mark]**

There are 11 marbles, representing 11 possible outcomes, and 8 marbles that give the desired outcome, so the probability is \dfrac{8}{11}.

**Question 2:Â **If A has probability 0.4 and B has probability 0.3, what is \mathbb{P}(A\cup B)+\mathbb{P}(A\cap B)?

**[2 marks]**

\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)

\mathbb{P}(A)=0.4

\mathbb{P}(B)=0.3

\begin{aligned}\mathbb{P}(A\cup B)&=0.4+0.3-\mathbb{P}(A\cap B)\\[1.2em]&=0.7-\mathbb{P}(A\cap B)\end{aligned}

\mathbb{P}(A\cup B)+\mathbb{P}(A\cap B)=0.7

**Question 3:** Consider flipping a coin three times. Event A is that the second flip is tails. Event B is that there are at least two heads.

a) Represent the events A and B on a Venn Diagram.

b) Find \mathbb{P}(A)

c) Find \mathbb{P}(B)

d) Find \mathbb{P}(A\cap B)

e) Find \mathbb{P}(A\cup B)

f) How many possibilities do not fall within the Venn Diagram at all?

**[6 marks]**

a)

b) 8 total events, 4 of them in circle A, so probability is \dfrac{4}{8}=\dfrac{1}{2}

c) 8 total events, 4 of them in circle B, so probability is \dfrac{4}{8}=\dfrac{1}{2}

d) This corresponds to where the circles overlap, which contains one event, so probability is \dfrac{1}{8}

e) This corresponds to both circles, which contain 7 events, so probability is \dfrac{7}{8}

f) 1 possibility does not lie within the Venn Diagram

**Question 4:** An event has probability 0.932. What is the probability of its complement?

**[1 mark]**

**Question 5:** Complete the two-way table and use it to determine \mathbb{P}(A\cap B')

**[4 marks]**

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