Probability Basics And Listing Outcomes | Worksheets and Revision

# Probability Basics & Listing Outcomes Worksheets, Questions and Revision

Level 1 Level 2 Level 3

## Probability

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

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

## Probability Scale

The probability (i.e. the chance) of something happening is defined on scale from impossible, $0$, where there is a $0\%$ chance of that thing happening, to certain, $1$, where the thing will happen $100\%$ of the time.

Example: The probability that it will rain tomorrow is $65\%$, which is likely. The probability that it will be foggy tomorrow is $\dfrac{1}{6},$ which is unlikely.

Level 1-3

## 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 $0.8$ is, $\text{P(Film)} = 0.8$

Level 1-3

## Calculating Probabilities

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

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

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.

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

Level 1-3

## 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), and indeed we get that,

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

Level 1-3

## Listing Outcomes

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. Question scenarios that will often require drawing a sample space diagram include: flipping two coins, rolling dice, and spinners being spun.

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

a) List all possible outcomes that Megan could find.

To do this it is easiest to create and fill in a two way table to find every possible combination of scores Megan could get when spinning the two spinners. As we can see there a total of $25$ possible outcomes.

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

This question is now much simpler to answer given we have a sample space diagram that visualises all the possible outcomes. All we have to do is count the number of squares on the table that have a score of $8$ or more. There are $6$ possible ways of getting this result. So the probability if given by,

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

Level 4-5

## 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, with $1$ of them vegan, $8$ main courses, with $3$ of them vegan, and $4$ desserts, with $2$ of them vegan.

a) How many different three-course meals can be chosen from the menu?

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

b) How many different ways are there to choose three-course meals that are only vegan?

The number of different vegan three-course meals $= 1 \times 3 \times 2 = 6$

c) What is the probability of only choosing a vegan meal when choosing a three-course meal?

$\text{P (Only vegan)} = \dfrac{\text{no. of vegan three-course meals}}{\text{Total no. of three-course meals}} = \dfrac{6}{96} = \dfrac{1}{16}$

Level 6-7

## 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?

• To find the probability of a draw, we should half the probability that Sarah wins:

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

• Exhausting all options, we know that P(Sarah wins), P(draw), and 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 4-5

### 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.)

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:

$0.22 + 0.56 = 0.78$

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

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:

$1 - 0.25 = 0.75$

b) 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 $\dfrac{5}{12}$.

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 $\dfrac{1}{3}$

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 $1 \div 5 \times 100 = 20\%$

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