7.02 Theoretical probability

Probability terms and notation

In day to day life, there are many places where the language of probability is used. For example:

• The chance of rain today is 75%
• A medication has an effectiveness of 90%
• 4 in 5 people agree that a particular brand of toothpaste is more effective than another
• I will probably go to the gym today

The notation $P\left(A\right)$P(A) means 'the probability that event A occurs. All probabilities lie on the interval $0\le P\left(A\right)\le1$0≤P(A)≤1, where a probability of zero indicates the event cannot possibly occur and a probability of one indicates the event is certain to occur. This can be visualised as follows: 

 

Before we calculate probabilities, let's familiarise ourselves with the language and notation for describing components in this topic. In probability, the sample space is a list of all the possible outcomes of an experiment.

Outcomes are the results of an experiment or trial. For example, think about flipping a coin. There are two possible outcomes - a head or a tail. So when we list (or write out) the sample space, we write:  

$S=\left\{heads,tails\right\}$S={heads,tails}

We can write a sample space using a list, table, set notation as above or a diagram such as a Venn or Tree diagram.

An event is a subset of the of the sample space and is often represented by a capital letter to abbreviate the description of the event in calculations.

For example, we could have the experiment of rolling a six sided dice and let E = the event of rolling an even number, O = the event of rolling an odd number and A = the event of rolling a number less than 3.

Then our sample space is:

$S$S

$=$=

$\left\{1,2,3,4,5,6\right\}${1,2,3,4,5,6}

And the events can be written as the sets:

$E$E	$=$=	$\left\{2,4,6\right\}${2,4,6}
$O$O	$=$=	$\left\{1,3,5\right\}${1,3,5}
$A$A	$=$=	$\left\{1,2\right\}${1,2}

We can calculate the probability of an event occurring as:

$P(event)=\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}$P(event)=number of favourable outcomestotal possible outcomes​

If all the outcomes can be easily listed, then the process of counting favourable and total outcomes is relatively straightforward.

For example, say we wanted to know the probability of rolling a 3 on a die. There are six possible outcomes in total (the sample space is $\left(1,2,3,4,5,6\right)$(1,2,3,4,5,6)). However, there is only one three on a die. So, we write our answer as $P(3)=\frac{1}{6}$P(3)=16​

There is an assumption in the work we are doing that every outcome is equally likely to occur. This allows us to utilise the formula above for the probability of an event occurring. This is also known as theoretical probability, which is the expected probability based on knowledge of the system and determining the number of favourable outcomes and number of total possible outcomes mathematically. 

Practice questions

Question 1

question 2

Complementary events

A complement of an event are all outcomes that are NOT the event. If $A$A is the event then the complement is denoted as $\overline{A}$A or sometimes $A'$A′ .

The following are examples of events and their complements:

• If event $A$A is tossing a coin and getting $\left\{Heads\right\}${Heads}, the complement $\overline{A}$A is $\left\{\text{not a head}\right\}${not a head} which is $\left\{Tails\right\}${Tails}
• If event $B$B is rolling a $6$6 sided die and getting $\left\{2\right\}${2}, then the complement $\overline{B}$B is $\left\{1,3,4,5,6\right\}${1,3,4,5,6}

Complementary events have the property that their probabilities always add to $1$1. Since it is certain either event $A$A occurred or it did not occur. That is:

$P\left(A\right)+P\left(\overline{A}\right)=1$P(A)+P(A)=1

In some cases it may be easier to count the possibilities where an event does not happen and use this fact to find the probability that the event does happen more efficiently. 

Practice questions

Question 3

Question 4

Probability with 2D grids

When listing outcomes a $2$2-dimensional grid can be useful for organising and displaying all the possibilities.  For example, the possible outcomes when rolling two die are:

$\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

While possible to list all outcomes it is not a convenient form to count or find outcomes with particular characteristics. We could instead create a grid to display the outcomes in an organised form, this will also help to not miss any outcomes in the list. For our list above we could create the following array to display the outcomes:

		Die 1
		$1$1	$2$2	$3$3	$4$4	$5$5	$6$6
Die 2	$1$1	$1,1$1,1	$1,2$1,2	$1,3$1,3	$1,4$1,4	$1,5$1,5	$1,6$1,6
	$2$2	$2,1$2,1	$2,2$2,2	$2,3$2,3	$2,4$2,4	$2,5$2,5	$2,6$2,6
	$3$3	$3,1$3,1	$3,2$3,2	$3,3$3,3	$3,4$3,4	$3,5$3,5	$3,6$3,6
	$4$4	$4,1$4,1	$4,2$4,2	$4,3$4,3	$4,4$4,4	$4,5$4,5	$4,6$4,6
	$5$5	$5,1$5,1	$5,2$5,2	$5,3$5,3	$5,4$5,4	$5,5$5,5	$5,6$5,6
	$6$6	$6,1$6,1	$6,2$6,2	$6,3$6,3	$6,4$6,4	$6,5$6,5	$6,6$6,6

This is particularly useful for experiments with two independent events such as tossing two coins, rolling two dice or selecting two marbles from a bag with replacement.

 

Worked example

Example 1

A player is rolling two dice and calculating the sum of the of the two numbers shown.

(a) Create a 2D grid displaying all outcomes of the event.

Think: Create a table with the results of one die shown horizontally and the other die vertically. Find the sum of each corresponding pair and record the sum.

		Die 1
		$1$1	$2$2	$3$3	$4$4	$5$5	$6$6
Die 2	$1$1	$\editable{1+1}$1+1	$\editable{1+2}$1+2	$\editable{1+3}$1+3	$\editable{1+4}$1+4	$\editable{1+5}$1+5	$\editable{1+6}$1+6
	$2$2	$\editable{2+1}$2+1	$\editable{2+2}$2+2	$\editable{2+3}$2+3	$\editable{2+4}$2+4	$\editable{2+5}$2+5	$\editable{2+6}$2+6
	$3$3	$\dots$…
	$4$4
	$5$5
	$6$6

Do:

		Die 1
		$1$1	$2$2	$3$3	$4$4	$5$5	$6$6
Die 2	$1$1	$2$2	$3$3	$4$4	$5$5	$6$6	$7$7
	$2$2	$3$3	$4$4	$5$5	$6$6	$7$7	$8$8
	$3$3	$4$4	$5$5	$6$6	$7$7	$8$8	$9$9
	$4$4	$5$5	$6$6	$7$7	$8$8	$9$9	$10$10
	$5$5	$6$6	$7$7	$8$8	$9$9	$10$10	$11$11
	$6$6	$7$7	$8$8	$9$9	$10$10	$11$11	$12$12

 

(b) State the total number of possible outcomes.

Think: The grid of outcomes has $6$6 rows and $6$6 columns. 

Do: The total number of outcomes from rolling two dice is $6\times6=36$6×6=36.

Reflect: Not all the outcomes are different so our sample space is $S=\left\{2,3,4,5,6,7,8,9,10,11,12\right\}$S={2,3,4,5,6,7,8,9,10,11,12}, however they are not all equally likely.

 

(c) What is the most common outcome and what is its probability?

Think: Which number appears most often in the table and how many times does it appear out of the $36$36 outcomes?

Do: Seven is the most common result from rolling two dice and finding the sum. This occurs $6$6 times and hence:

$P\left(\text{Seven}\right)$P(Seven)	$=$=	$\frac{6}{36}$636​
	$=$=	$\frac{1}{6}$16​

(d) What is the probability of a sum less than $5$5?

Think: Possible results less than five are $2$2, $3$3 and $4$4, how many of these are there?

Do:

$P\left(\text{Result}<5\right)$P(Result<5)	$=$=	$\frac{6}{36}$636​
	$=$=	$\frac{1}{6}$16​

(e) What is the probability of a sum greater than $6$6 and even?

Think: Possible results greater than six that are also even are $8$8, $10$10 and $12$12, how many of these are there?

Do:

$P\left(\text{Result}>6\text{ and even}\right)$P(Result>6 and even)	$=$=	$\frac{9}{36}$936​
	$=$=	$\frac{1}{4}$14​

 

Practice questions

QUESTION 5

Question 6