# Probability Basics and Listing Outcomes

GCSEKS3Level 1-3Level 4-5Level 6-7AQACambridge iGCSEEdexcelEdexcel iGCSEOCRWJEC

## Probability

Probability is the study of how likely things are to happen. We express the probability of an unknown event happening on a scale from certain to impossible, as a decimal, fraction or percentage.

Make sure you are happy with the following topics before continuing:

Level 1-3GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE

## Probability Scale

The probability (i.e. the chance) of something happening is defined on a scale from:

Impossible, $0$, there is a $0\%$ chance of that thing happening.

Certain, $1$, there is a $100\%$ chance of that thing happening.

E.g. The probability that it will rain tomorrow is $65\%$, this would be classed as likely

The probability that it will be foggy tomorrow is $\dfrac{1}{6},$ this would be classed as unlikely.

Level 1-3GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE

## Probability Notation

To express “the probability of $\text{X}$ happening” in a mathematical way, we write $\text{P(X)}$.

Example: The probability that I will watch a film tonight is $0.8$

This can be written as $\text{P(Film)} = 0.8$

Level 1-3GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE

@mmerevise

Level 1-3GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE

## Calculating Probabilities

To calculate the probability, when all possible outcomes are equally likely, we can use the following formula:

$\textcolor{Blue}{\text{Probability}} = \textcolor{Blue}{\text{P(X)}} = \dfrac{\text{number of ways an outcome can happen}}{\text{total number of possible outcomes}}$

Example: Calculate the probability of rolling an even number on a fair $\textcolor{black}{6}$ sided die.

• There are $3$ even numbers, so there are $3$ different, equally likely ways of rolling an even number.
• There are $6$ numbers on a die, so there are a total of $6$ possible outcomes.

$\textcolor{blue}{\text{P(even)}} = \dfrac{\text{even numbers on a die}}{\text{total numbers on a die}} = \dfrac{3}{6} = 0.5$

Level 1-3GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE

## Probabilities Add Up to 1

If you consider all possible outcomes of an event (known as exhausting all options) then the probabilities must add up to $1$

Example: The chance of flipping a coin and landing on heads is one half, and the chance of landing on tails is the same, one half. These are the only two possibilities (we’ve exhausted all options).

Adding these two up we get,

$\text{P(heads)}+\text{P(tails)}=\dfrac{1}{2}+\dfrac{1}{2}=1$

Level 1-3GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
Level 1-3GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE

## Sample Space Diagrams

Listing outcomes is exactly what it sounds like – given a scenario, list every possible outcome. When there are two events taking place, we make use of a sample space diagram to help keep track of all the possible outcomes. This takes the form of a two way table.

Example: Megan spins two spinners numbered $1$ to $5$ and records the sum of the values each spinner lands on.

a) List all possible outcomes that Megan could find.

To do this we create and fill in a two way table to find every possible combination of scores from two spinners. As we can see there a total of $25$ possible outcomes.

b) What is the probability of Megan recording a score of $8$ or more?

All we have to do is count the number of squares on the table that have a score of $8$ or more and make a fraction out of the total possible number of outcomes.  There are $6$ ways of getting this result.

This gives,

$\text{P(8 or more)} = \dfrac{6}{25}$

Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE

## Systematic Listing

You may be asked to consider a scenario and list all of the possible outcomes there could be. The best way of doing this is systematically, or using systematic listing.

Listing options in a systematic way ensures you don’t miss out any possible outcomes.

Example:

A maths student is randomly selecting a three digit number containing the digits $1, 5$ and $9$.

List all possible outcomes of this selection.

As we are listing systematically, we need to list in a specific order. We will start with numbers beginning with $1$,

$159, 195$

And now starting with $5$,

$519, 591$

Finally, starting with $9$,

$915, 951$

So the possible outcomes are,

$159, 195, 519, 591, 915, 951$

Level 4-5GCSEAQAEdexcelOCRWJEC
Level 6-7GCSEAQAEdexcelOCRWJEC

## Product Rule

When there are a large number of outcomes or there are more than $2$ events occurring, we can use the product rule to count the number of outcomes instead of listing all the possible outcomes, as this can take too long.

The product rule states that:

The total number of outcomes for $2$ or more events is equal to the number of outcomes for each event multiplied together.

Example: A restaurant offers a set menu, that contains $3$ starters, $8$ main courses, and $4$ desserts.

How many different ways are there to choose a three-course meal?

To do this, all we need to do is multiply the number of options together.

The number of different three-course meals $= 3 \times 8 \times 4 = 96$

Level 6-7GCSEAQAEdexcelOCRWJEC
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE

## Example:

Sarah and Charlie are playing a game, there are $3$ possible outcomes for the game: Sarah wins, Charlie wins, or it’s a draw.

The chance of Sarah winning is $0.6$, a draw is half as likely as Sarah winning.

What is the probability that Charlie wins?

[2 marks]

Probability of a draw = half the probability that Sarah wins:

$\text{P(draw)}=0.6\div 2=0.3$

Exhausting all options, we know that $\text{P(Sarah wins)}$, $\text{P(draw)}$, and $\text{P(Charlie wins)}$ must all add up to $1$, so

\begin{aligned} 0.6+0.3+\text{P(Charlie wins)}&=1 \\ 0.9+\text{P(Charlie wins)}&=1\end{aligned}

Then, subtracting $0.9$ from both sides of the equation, we get

$\text{P(Charlie wins)}=1-0.9=0.1$

Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE

## Probability Basics and Listing Outcomes Example Questions

Since $A$, $B$, and $C$ are the only possible outcomes, their probabilities must add up to $1$.

Therefore, if we take both $\text{P(A)}$ and $\text{P(B)}$ away from $100\%$ or $1$, we can work out $\text{P(C)}$, the probability of the wheel stopping section $C$.

The only issue presented to us in this question is that the probability of the wheel stopping in section $A$ is given as a percentage, whereas for section $B$ it is expressed as a fraction. We will therefore need to convert either the fraction to the percentage or the percentage to the fraction so that the two probabilities are expressed in the same way.

Although you can turn a percentage to a fraction quite easily, in this question it is probably easier to convert the fraction $\frac{2}{5}$ to a percentage (provided you know what the percentage value of $\frac{1}{5}$ is):

$\frac{1}{5} = 20\%$

so

$\frac{2}{5} = 40\%$

Since the probability of the wheel stopping in section $A$ is $55\%$, and the probability of the wheel stopping in section $B$ is $40\%$, that means we can calculate the probability of the wheel stopping in either section $A$ or $B$:

$\text{P(A)}+\text{P(B)}= 55\% + 40\% = 95\%$

This therefore means that chance of the wheel stopping in section $C$ can be calculated as follows:

$100\% - 95\% = 5\%$

(An answer expressed as a decimal ($0.05$) or as a fraction $\frac{1}{20}$ is also acceptable.)

Gold Standard Education

a)  We know that the probability of Jimmy watching a romantic comedy is $0.56$.  Therefore we can easily calculate the probability of Jimmy not watching a romantic comedy:

$1 - 0.56 = 0.44$

This figure of $0.44$, the probability of Jimmy not watching a romantic comedy, is the same probability as Jimmy watching either a horror film or a sci-fi movie.

Since the probability of Jimmy watching a sci-fi movie or a horror film is equal, then the probability of Jimmy watching a sci-fi movie must be half of this amount:

$0.44 \div2 = 0.22$

b)  From part a), we know that the probability of Jimmy watching a sci-fi movie is $0.22$.

The probability of Jimmy watching a romantic comedy is $0.56$.

In order to calculate the probability of Jimmy watching either a romantic comedy or a sci-fi movie we need to add the probabilities of each, since this is an either / or scenario:

$0.22 + 0.56 = 0.78$

Gold Standard Education

If Macy chooses black trousers, then she has $3$ choices for the colour of her jumper, so the $3$ possible outcomes are:

BO,  BY,  BW

If she chooses navy trousers, then the $3$ possible outcomes are:

NO,  NY,  NW

Finally, if she chooses purple trousers, then the remaining possible outcomes are:

PO,  PY,  PW

BY was already given in the question, so the full list of other possible outcomes is:

BO,  BW,  NO,  NY, NW,  PO,  PY,  PW

Gold Standard Education

a)  We know that the probability of selecting a red, blue or green bead must add up to $1$.  We know that the probability of selecting a red bead is $0.25$, so the probability of selecting either a blue or a green bead must be:

$1 - 0.25 = 0.75$

The problem in this question is that the probability of selecting a blue bead and the probability of selecting a green bead is not the same.  Selecting a blue bead has a probability $5x$ and selecting a green bead has a probability of $4x$.  This means that the probability of selecting a blue or green bead is:

$5x + 4x = 9x$

Since we know that the probability of selecting a blue or green bead is $0.75$, we can therefore conclude that $9x = 0.75$

If

$9x = 0.75$

then

$x = 0.75 \div 9 = \dfrac{1}{12}$

If the probability of selecting a blue bead is $5x$, and $x = \dfrac{1}{12}$, then the probability of selecting a blue bead is $\dfrac{5}{12}$.

b) If the probability of selecting a green bead is $4x$, and $x = \dfrac{1}{12}$, then the probability of selecting a blue bead is $\dfrac{4}{12}$.  This fraction can then be simplified to $\dfrac{1}{3}$

Gold Standard Education

a)  We know that the tickets are only numbered between $1$ and $50$, so we need to work out how many of these fall into the category of being either a multiple of $5$ or an odd number.

There are $25$ odd numbers in total between $1$ and $50$:

$1, \, 3, \, 5, \, 7, \, 9, \, 11,\, 13,\, 15,$ $17, \, 19, \, 21, \, 23, \, 25, \, 27, \, 29, \, 31,$ $33, \, 35, \, 37, \, 39, \, 41, \, 43, \, 45, \, 47, \, 49$

There are $10$ multiples of $5$ between $1$ and $50$: $5, \, 10, \, 15, \, 20, \, 25, \, 30, \, 35, \, 40, \, 45, \, 50$

However, some of these multiples of $5$ also feature on the odd number list, so cannot be counted twice.  So, ignoring the odd multiples of $5$, there are only $5$ multiples of $5$ remaining.

$25 \text{ odd numbers} + 5 \text{ (even) multiples of } 5 = 30 \text{ numbers in total}$

If $30$ of the $50$ numbers fall into the category of being either odd or a multiple of $5$, then as a fraction we can express this as:

$\dfrac{30}{50}$ which can be simplified to $\dfrac{3}{5}$

It is also perfectly acceptable to express the probability as a decimal or as a percentage:

$\dfrac{3}{5}$ as a decimal is $3 \div 5 = 0.6$

$\dfrac{3}{5}$ as a percentage is $3 \div 5 \times 100 = 60\%$

b)  The factors of $48$ are as follows:

$1$ and $48$

$2$ and $24$

$3$ and $16$

$4$ and $12$

$6$ and $8$

(If you are ever working out the factors of a number, write them in pairs, and start at $1$.)

This means that $10$ numbers out of the $50$ in the hat are factors of $48$.  We can express this as a fraction:

$\dfrac{10}{50}$ which can be simplified to $\dfrac{1}{5}$

It is also perfectly acceptable to express the probability as a decimal or as a percentage:

$\dfrac{1}{5}$ as a decimal is $1 \div 5 = 0.2$

$\dfrac{1}{5}$ as a percentage is $1 \div 5 \times 100 = 20\%$

Gold Standard Education

## Probability Basics and Listing Outcomes Worksheet and Example Questions

### (NEW) Probability Basics and Listing Outcomes Exam Style Questions - MME

Level 1-3GCSE

Gold Standard Education

Level 1-3GCSEKS3