Probability Basics And Listing Outcomes | Worksheets and Revision

# Probability Basics & Listing Outcomes Worksheets, Question sand Revision

Level 1 Level 2 Level 3

## What you need to know

### Probability

Probability is the study of how likely things are to happen.

Probability is a large topic and the basic concepts feed into many different areas such as:

There are other topics that also benefit from having basic probability knowledge so it is definitely one of the core topics you need to understand well.

### Probability

All probabilities must fall between 0 and 1, where 0 means ‘impossible’ and 1 means ‘certain’. If the probability of an outcome is closer to 0 then it’s unlikely to happen, but if it’s closer to 1 then it’s likely to happen.

We can express probabilities as fractions, decimals or percentages.

## Listing Outcomes

Listing outcomes is exactly what it sounds like – given a scenario, list every possible outcome. Usually, to save writing time, we’ll use single letters to denote outcomes, e.g. when flipping a coin, we’ll say H = heads and T = tails.

## Probability Notation

To say “the probability of X happening” we write P(X). This notation will be used throughout the rest of this page.

If you consider all possible outcomes of an event – this is known as exhausting all options – then their probabilities must add up to 1. For example, the chance of flipping and 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

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

## The And/Or Rule

Suppose we’re considering two outcomes that might happen: A and B. Then, the and/or rule of probability is

$\text{P(both A and B)}=\text{P(A)}\times\text{P(B)}$

$\text{P(Either A or B)}=\text{P(A)}+\text{P(B)}$

The rule can be summarised as: if it’s and, we times, but if it’s or, we add.

### Example 1: Listing Outcomes

Megan flips a coin 3 times and records the results of each 3 flips, in order. List all possible outcomes.

As mentioned, we’re going to use H = heads and T = tails. This means that to represent the outcome of getting “tails, heads, tails”, we’ll write THT.

Furthermore, she’s recording the order of the flips too, meanings THT is different to TTH.

Therefore, the complete list of all possible outcomes is

HHH,HHT,HTT,HTH,TTT,THT,THH,TTH

The main aspect of listing outcomes is making sure you don’t miss any out.

### Example 2: Basic Probability

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. Written as an equation, we get

\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

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

### Example 3: The And Rule

Fae and Sunny play badminton together. It’s impossible to draw in badminton. The chance of Fae winning any match against Sunny is $\frac{1}{6}$. Calculating the probability that Sunny wins 2 matches of badminton against Fae.

Firstly, the only two possible outcomes are Sunny wins or Fae wins. Therefore, their probabilities must add up to 1. In other words, if we subtract P(Fae wins) from 1, we get

$\text{P(Sunny wins)}=1-\dfrac{1}{6}=\dfrac{5}{6}$

Now, how do we think about Sunny winning two matches? We have to split it up into her winning one, and then winning another. The key word there is ‘and’ – we’re finding the probability of 1 win and then another win, which means we need to multiply the chance of her winning 1 game by the chance of her winning another game. Doing this, we get

$\text{P(Sunny wins 2 games)}=\dfrac{5}{6}\times\dfrac{5}{6}=\dfrac{25}{36}$

### Example 4: The Or Rule

Calculate the probability of rolling a 2,3 or 4 on a 6 sided unbiased die.

The probability of rolling any number is $\dfrac{1}{6}$ therefore the probability of rolling a 2 or 3 or 4 is

$\dfrac{1}{6}$ + $\dfrac{1}{6}$ + $\dfrac{1}{6}$ = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

### Example Questions

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

Therefore, if we take both P(A) and P(B) away from 100% or 1, we can work out 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 $\dfrac{1}{4}$

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 and 49

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