Maths · 9 Probability

9.1 Introduction to probability

Before you start

Cancelling fractions, and changing between fractions, decimals and percentages (1.4).
Subtracting a fraction or a decimal from 1, for example $1 - \frac{3}{8} = \frac{5}{8}$ and $1 - 0.27 = 0.73$ .
Prime numbers, factors and multiples (1.1).
Reading a Venn diagram, and the notation $A^{'}$ for the complement of a set (1.2).
Reading values from a table and from a bar chart.
Solving a simple linear equation such as $0.4 + 3 x = 1$ (2.5).

Syllabus objectives covered (4)

Understand and use the probability scale from 0 to 1.
Understand and use probability notation.
Calculate the probability of a single event.
Understand that the probability of an event not occurring = 1 – the probability of the event occurring.

Part 1 of 6

The probability scale from 0 to 1

A probability is a number that tells you how likely something is to happen. The 'something' is called an event: for example, 'the dice shows a 6' or 'it rains tomorrow'.
Every probability is on a scale from $0$ to $1$ . A probability of $0$ means the event is impossible: it cannot happen. A probability of $1$ means the event is certain: it must happen.
graph diagram — not drawn yet
A horizontal number line from 0 to 1 with ticks at 0, 0.25, 0.5, 0.75 and 1. Below the line: 'impossible' at 0, 'even chance' at 0.5, 'certain' at 1, the word 'unlikely' between 0 and 0.5 and the word 'likely' between 0.5 and 1. Above the line, two labelled arrows: 'fair dice shows 6' at 1/6 (about 0.17) and 'fair coin shows heads' at 0.5.
Every probability is somewhere on this line. The nearer to 1, the more likely the event.
Halfway, at $0.5$ , the event has an even chance: it is as likely to happen as not to happen. A fair coin lands on heads with probability $0.5$ .
Below $0.5$ an event is unlikely, and above $0.5$ it is likely. The nearer the number is to $1$ , the more likely the event. An event with probability $0.9$ is more likely than an event with probability $0.6$ .
To mark a probability on the scale, put it the correct fraction of the way from $0$ to $1$ . Put $\frac{1}{4}$ a quarter of the way along. Put $0.8$ between $0.75$ and $1$ .
No probability is less than $0$ or greater than $1$ . If your answer is $1.2$ , $\frac{9}{7}$ or $- 0.3$ , there is a mistake in your working. Use this as a check on every probability you write.
The words impossible, unlikely, even chance, likely and certain describe a probability. They are not a value. When a question asks for the probability, give a number.

Part 2 of 6

Writing a probability: fraction, decimal or percentage

The syllabus says that probabilities are given as a fraction, a decimal or a percentage. All three forms can give the same value: $\frac{3}{4} = 0.75 = 75 %$ .
Do not write a probability as a ratio, such as $3 : 1$ , or in words, such as '3 out of 4'. Neither is a fraction, a decimal or a percentage. A ratio compares one part with another part. A probability compares the event with all the possible outcomes.
A fraction is exact. If its decimal does not end, such as $\frac{2}{7} = 0.285714 \dots$ , keep the fraction or round the decimal to 3 significant figures: $0.286$ .
Cancel a fraction when you can: $\frac{6}{24} = \frac{1}{4}$ . The 1.4 notes expect you to be able to write a fraction in its simplest form, and a cancelled fraction is easier to compare and check.
Do not mix two forms in one answer. $\frac{3.5}{10}$ puts a decimal inside a fraction. Write $\frac{7}{20}$ or $0.35$ instead.
To compare probabilities in different forms, change them all to decimals. Which is the most likely: $\frac{5}{8}$ , $0.6$ or $63 %$ ? $\frac{5}{8} = 0.625$ and $63 % = 0.63$ . So $63 %$ is the most likely, then $\frac{5}{8}$ , then $0.6$ .

Part 3 of 6

Probability notation: P(A) and P(A′)

You do not have to write 'the probability that the counter is red' each time. Use a capital letter for the event, and the notation $P (\dots)$ .
If $R$ is the event 'the counter is red', then $P (R)$ means **the probability of $R$ **. Read $P (R) = 0.3$ as 'the probability of $R$ is $0.3$ '.
You can also write the event itself inside the brackets, in words or in symbols: $P (red)$ , $P (even number)$ , $P (score > 4)$ .
$A^{'}$ is the event **not $A$ **: the event that $A$ does not happen. So $P (A^{'})$ is the probability of not $A$ . You can read it as 'P of not A'.
This is the same symbol as the complement of a set in 1.2. In a Venn diagram, $A^{'}$ is everything inside the rectangle but outside the circle for $A$ .
The dash goes on the event, inside the brackets: $P (A^{'})$ . It changes which event you mean. $P (A^{'})$ is still an ordinary probability, so it is between $0$ and $1$ .
Write the notation in your working, for example ' $P (R) = \frac{3}{10}$ '. A bare $\frac{3}{10}$ does not say which probability it is. With the notation, each line of your method can be followed.

Part 4 of 6

The probability of a single event

An outcome is one possible result. When you roll an ordinary dice, there are six possible outcomes: $1$ , $2$ , $3$ , $4$ , $5$ and $6$ .
Outcomes are equally likely when each one has the same chance of happening. A question tells you this with words such as a fair dice, coin or spinner, or an item chosen at random. 'At random' means that each item has the same chance of being chosen.
When the outcomes are equally likely, the probability of an event is a count over a count:
$P (event) = \frac{number of outcomes in the event}{total number of possible outcomes}$
Example: a bag contains 5 red, 3 blue and 2 green counters. One counter is taken at random. There are $5 + 3 + 2 = 10$ counters, and 3 of them are blue.
$P (blue) = \frac{3}{10}$
Find the total first, and include every item. The bottom of the fraction is all 10 counters. It is not the 7 counters that are not blue.
When the event is given in words, write a list of its outcomes before you count. On a dice, 'a prime number' is $2, 3, 5$ . The number $1$ is not prime, because a prime has exactly two factors and $1$ has only one. So $P (prime) = \frac{3}{6} = \frac{1}{2}$ .
Read the inequality words carefully. 'Greater than 4' is $5, 6$ , so the probability is $\frac{2}{6} = \frac{1}{3}$ . 'At least 4' includes $4$ : it is $4, 5, 6$ , so the probability is $\frac{3}{6} = \frac{1}{2}$ .
An event with no outcomes has probability $0$ : a 7 on an ordinary dice is impossible, and $P (7) = \frac{0}{6} = 0$ . An event that contains every outcome has probability $1$ : $P (score less than 7) = \frac{6}{6} = 1$ . These are the two ends of the scale.
The counting method needs equally likely outcomes. 'It rains or it does not rain' gives two outcomes, but they do not have the same chance. So you cannot say $P (rain) = \frac{1}{2}$ only because there are two outcomes.
The awkward case: items are added to a bag and you are given the new probability. A bag has 3 red and 7 blue counters. Some red counters are added, and now $P (red) = \frac{1}{2}$ . Call the number added $r$ . Both the top and the bottom change, because the new counters are part of the new total. Then multiply both sides by $2 (10 + r)$ to clear the fractions: $2 (3 + r) = 10 + r$ , so $6 + 2 r = 10 + r$ and $r = 4$ . Check: $\frac{7}{14} = \frac{1}{2}$ .
$\frac{3 + r}{10 + r} = \frac{1}{2}$

Part 5 of 6

Probabilities from tables, bar charts and Venn diagrams

Sometimes the information is a count of people or objects. Each count is a frequency, and the method does not change:
$P (event) = \frac{frequency of the event}{total frequency}$
In a table, find the total first. If the table has a total row or a total column, use it. If it does not, add every entry.
In a two-way table, one cell counts the items that fit both descriptions. A cell in the row 'Year 11' and the column 'Bus' counts only the Year 11 students who take the bus, not all the bus users.
In a bar chart, read the height of each bar. The total is the sum of all the bars. For example, a car park has 14 white, 10 black, 9 silver and 7 red cars. The total is $14 + 10 + 9 + 7 = 40$ , so if one car is chosen at random, $P (silver) = \frac{9}{40}$ .
graph diagram — not drawn yet
A vertical bar chart titled 'Colours of cars in a car park'. The horizontal axis shows four colours: white, black, silver, red. The vertical axis is 'Frequency' from 0 to 16 in steps of 2. Bar heights: white 14, black 10, silver 9, red 7.
Add all four bars to get the total of 40. The silver bar alone gives the 9 on top of the fraction.
In a Venn diagram, each number counts the items in one region. For an event such as 'plays tennis', add every region inside the tennis circle, including the overlap. The total is every number in the rectangle, including the number outside both circles.
Example: 30 students. 12 play football only, 5 play both, 8 play tennis only and 5 play neither. The tennis circle holds $5 + 8 = 13$ students, so $P (tennis) = \frac{13}{30}$ .
graph diagram — not drawn yet
A rectangle labelled U containing two overlapping circles labelled F (football) and T (tennis). The number 12 is in F only, 5 is in the overlap, 8 is in T only, and 5 is inside the rectangle outside both circles.
The T circle is 5 + 8 = 13 students. The total is all four numbers: 12 + 5 + 8 + 5 = 30.
The awkward case: the question chooses from one group only, such as 'a student who plays football is chosen at random'. Then only that group can be chosen, so the total is that group: $12 + 5 = 17$ students. The probability that this student also plays tennis is $\frac{5}{17}$ .

Part 6 of 6

The probability that an event does not happen

When you take one counter from a bag, it is blue or it is not blue. One of these two must happen, and they cannot both happen.
So together the two probabilities make $1$ , which is the probability of something certain:
$P (A) + P (A^{'}) = 1$
Subtract $P (A)$ from both sides to get the rule you will use:
$P (A^{'}) = 1 - P (A)$
Check it with the bag of 5 red, 3 blue and 2 green counters. $P (blue) = \frac{3}{10}$ , so $P (not blue) = 1 - \frac{3}{10} = \frac{7}{10}$ . Counting gives the same: $5 + 2 = 7$ of the 10 counters are not blue.
The rule works in every form. If $P (B) = 0.8$ , then $P (B^{'}) = 1 - 0.8 = 0.2$ . With percentages, certain is $100 %$ : if $P (rain) = 35 %$ , then $P (no rain) = 100 % - 35 % = 65 %$ .
Use the rule when 'not' is quicker than counting. On a fair dice, $P (not a 6) = 1 - \frac{1}{6} = \frac{5}{6}$ . You do not have to list the five other scores.
The same idea applies to a full list of outcomes. If the outcomes cover every possibility and no two can happen together, their probabilities add up to $1$ . A spinner lands on red, blue or green, with $P (red) = 0.5$ and $P (blue) = 0.3$ . Then $P (green) = 1 - 0.5 - 0.3 = 0.2$ .
The trap is the arithmetic, not the idea. $1 - 0.35 = 0.65$ , not $0.75$ . Check by adding back: $0.35 + 0.65 = 1$ .
$P (A^{'})$ is $1 - P (A)$ . It is not $\frac{1}{P (A)}$ , and it is not $- P (A)$ . For $P (A) = 0.8$ those give $1.25$ and $- 0.8$ , which are both off the scale.

Worked examples

Example 1. A bag contains 4 red, 7 blue and 9 yellow counters. One counter is taken at random. (a) Find the probability that it is blue. Give your answer as a fraction, a decimal and a percentage. (b) Find the probability that it is green, and say where this is on the probability scale.

Show the working

'At random' means each counter has the same chance, so you can count. Find the total first.
$4 + 7 + 9 = 20$
(a) 7 of the 20 counters are blue.
$P (blue) = \frac{7}{20}$
To get a decimal, make the denominator 100: multiply the top and the bottom by 5.
$\frac{7}{20} = \frac{35}{100} = 0.35 = 35 %$
Check: $0.35$ is between $0$ and $1$ . It is below $0.5$ , so a blue counter is unlikely. That agrees with the bag: fewer than half of the counters are blue.
(b) There are no green counters, so the event has no outcomes.
$P (green) = \frac{0}{20} = 0$
A probability of $0$ is at the left end of the scale. Taking a green counter is impossible.

Answer: (a) $\frac{7}{20} = 0.35 = 35 %$ . (b) $0$ : the event is impossible.

Example 2. A fair dice has faces numbered 1 to 6. It is rolled once. $A$ is the event 'the score is a prime number' and $B$ is the event 'the score is greater than 4'. Find $P (A)$ , $P (B)$ and $P (B^{'})$ .

Show the working

'Fair' means the six scores are equally likely, so you can count outcomes.
List the outcomes in $A$ . The primes from 1 to 6 are 2, 3 and 5. The number 1 is not prime: it has only one factor.
$P (A) = \frac{3}{6} = \frac{1}{2}$
List the outcomes in $B$ . 'Greater than 4' means 5 and 6. The score 4 is not included.
$P (B) = \frac{2}{6} = \frac{1}{3}$
$B^{'}$ is the event 'the score is not greater than 4'. Use the rule for 'not'.
$P (B^{'}) = 1 - \frac{1}{3} = \frac{2}{3}$
Check by listing: $B^{'}$ is 1, 2, 3, 4. That is 4 outcomes out of 6, and $\frac{4}{6} = \frac{2}{3}$ .

Answer: $P (A) = \frac{1}{2}$ , $P (B) = \frac{1}{3}$ , $P (B^{'}) = \frac{2}{3}$ .

Example 3. The table shows how 120 students travel to school. $\begin{matrix} Walk & Bus & Car & Total \\ Year 10 & 18 & 27 & 15 & 60 \\ Year 11 & 22 & 25 & 13 & 60 \\ Total & 40 & 52 & 28 & 120 \end{matrix}$ (a) A student is chosen at random. Find the probability that the student walks. (b) A student is chosen at random. Find the probability that the student is in Year 11 and travels by car. (c) A Year 10 student is chosen at random. Find the probability that this student takes the bus. (d) A student is chosen at random. Find the probability that the student does not take the bus.

Show the working

(a) The total is the 120 students in the bottom-right cell. The total of the Walk column is 40.
$P (walks) = \frac{40}{120} = \frac{1}{3}$
(b) 'Year 11 and car' is one cell: the 13 students who are in Year 11 and travel by car. The total is still all 120 students.
$P (Year 11 and car) = \frac{13}{120}$
$\frac{13}{120}$ does not cancel, because 13 is prime and is not a factor of 120. As a decimal it is $0.108$ to 3 significant figures, so the fraction is the neater answer.
(c) Now the student is chosen from Year 10 only. So the total is the 60 Year 10 students, not 120. Of these, 27 take the bus.
$P (bus) = \frac{27}{60} = \frac{9}{20} = 0.45$
(d) Use the 'not' rule with the Bus column total, 52.
$P (not bus) = 1 - \frac{52}{120} = \frac{68}{120} = \frac{17}{30}$
Check by counting: the students who walk or go by car are $40 + 28 = 68$ , and $\frac{68}{120} = \frac{17}{30}$ .

Answer: (a) $\frac{1}{3}$ (b) $\frac{13}{120}$ (c) $\frac{9}{20}$ (d) $\frac{17}{30}$

Example 4. A spinner can land on red, blue, green or yellow. The table shows the probability of each colour. $\begin{matrix} Colour & Red & Blue & Green & Yellow \\ Probability & 0.25 & 0.15 & x & 2 x \end{matrix}$ (a) Find the value of $x$ . (b) Find the probability that the spinner does not land on yellow.

Show the working

The spinner lands on exactly one of the four colours. So the four outcomes cover every possibility, no two can happen together, and their probabilities add up to 1.
$0.25 + 0.15 + x + 2 x = 1$
Collect the terms: $0.25 + 0.15 = 0.4$ , and $x + 2 x = 3 x$ .
$0.4 + 3 x = 1$
Subtract 0.4 from both sides, then divide by 3.
$3 x = 0.6 \Rightarrow x = 0.2$
So $P (green) = 0.2$ and $P (yellow) = 2 x = 0.4$ . Check: $0.25 + 0.15 + 0.2 + 0.4 = 1$ , and every value is between 0 and 1.
(b) Use the rule for 'not'.
$P (not yellow) = 1 - 0.4 = 0.6$

Answer: (a) $x = 0.2$ (b) $0.6$

Where people go wrong

Mistake: Saying that an event with two possible results has probability $\frac{1}{2}$ : 'a seed grows or it does not, so $P (grows) = \frac{1}{2}$ '.

Instead: Count outcomes only when they are equally likely. If they are not, you need other information, such as a table of probabilities or a set of results.

The counting method gives each outcome the same share of the probability. Two results with different chances do not each get half.

Mistake: Putting the other items on the bottom of the fraction. With 3 blue counters and 7 other counters, writing $P (blue) = \frac{3}{7}$ .

Instead: Divide by the total number of counters: $P (blue) = \frac{3}{10}$ .

A probability compares the event with all the outcomes. $\frac{3}{7}$ compares blue with not blue, which is the ratio $3 : 7$ , not a probability.

Mistake: Writing a probability as a ratio, $3 : 10$ , or in words, '3 in 10'.

Instead: Write $\frac{3}{10}$ , $0.3$ or $30 %$ .

The syllabus says probabilities are given as a fraction, a decimal or a percentage. A ratio and a phrase in words are neither of these.

Mistake: Using the whole total when the question chooses from one group, for example dividing by all 120 students when 'a Year 10 student is chosen'.

Instead: Divide by the total of the group the choice is made from: here, the 60 Year 10 students.

The possible outcomes are only the items that could be chosen. Students outside the group cannot be chosen, so they are not on the bottom of the fraction.

Mistake: Finding $P (A^{'})$ as $\frac{1}{P (A)}$ , or making a slip such as $1 - 0.35 = 0.75$ .

Instead: $P (A^{'}) = 1 - P (A)$ , so $P (A^{'}) = 1 - 0.35 = 0.65$ . Check by adding: $0.35 + 0.65 = 1$ .

$A$ and not $A$ together are certain, so their probabilities add to 1. $\frac{1}{0.35}$ is about $2.86$ , which is off the scale.

Check yourself

Which of these numbers cannot be probabilities? $0.07$ , $\frac{7}{6}$ , $70 %$ , $- 0.7$ , $0$
Answer
$\frac{7}{6}$ , because it is greater than 1, and $- 0.7$ , because it is less than 0. The others are on the scale; $0$ means impossible.
Write these probabilities in order, least likely first: $0.42$ , $\frac{3}{8}$ , $40 %$ , $\frac{9}{20}$ .
Answer
As decimals they are $0.42$ , $0.375$ , $0.4$ and $0.45$ . Order: $\frac{3}{8}$ , $40 %$ , $0.42$ , $\frac{9}{20}$ .
A fair dice numbered 1 to 6 is rolled once. Find the probability that the score is a multiple of 3.
Answer
The multiples of 3 are 3 and 6. $P = \frac{2}{6} = \frac{1}{3}$ .
A fair spinner has ten equal sections numbered 1 to 10. Find the probability that it lands on a factor of 12.
Answer
The factors of 12 from 1 to 10 are 1, 2, 3, 4 and 6. $P = \frac{5}{10} = \frac{1}{2}$ .
$P (C) = 0.27$ . Write down $P (C^{'})$ , and say in words what it means.
Answer
$P (C^{'}) = 1 - 0.27 = 0.73$ . It is the probability that the event $C$ does not happen.
In a group of 40 people, 14 have a cat only, 6 have a cat and a dog, 11 have a dog only and 9 have neither. One person is chosen at random. Find the probability that this person has a dog.
Answer
Add both regions of the dog circle: $6 + 11 = 17$ . $P (dog) = \frac{17}{40} = 0.425$ .
A team wins, draws or loses each match. $P (win) = 45 %$ and $P (draw) = 20 %$ . Find $P (lose)$ .
Answer
The three results cover every possibility and only one can happen, so they add up to $100 %$ . $P (lose) = 100 % - 45 % - 20 % = 35 %$ .
A box contains 12 pens. 5 are black and the rest are blue. One pen is taken at random. Find the probability that it is not black, and give your answer as a decimal to 3 significant figures.
Answer
$P (not black) = 1 - \frac{5}{12} = \frac{7}{12} = 0.583$ (3 s.f.).

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A bag contains 6 red beads, 9 blue beads and 5 white beads only. One bead is taken from the bag at random.

(a) Write down the probability that the bead is blue. 1 mark No calculator

Mark scheme

B1 $\frac{9}{20}$ oe

(b) Calculate the probability that the bead is not white. 2 marks No calculator

Mark scheme

M1 $1 - \frac{5}{20}$ or $\frac{6 + 9}{20}$ oe
A1 $\frac{3}{4}$ oe

(c) Write down the probability that the bead is green. 1 mark No calculator

Mark scheme

B1 0

(d) Some more white beads are put into the bag. Nothing else is changed. The probability of taking a white bead is now $\frac{2}{5}$ . Work out how many white beads are put into the bag. 3 marks No calculator

Mark scheme

M1 $\frac{5 + w}{20 + w} = \frac{2}{5}$ oe, or the 15 red and blue beads are $\frac{3}{5}$ of the new total
M1 $5 (5 + w) = 2 (20 + w)$ oe, or new total $= 15 \div \frac{3}{5} = 25$
A1 5 cao

Question 2. A spinner has five sections numbered 1 to 5. The table shows the probability of each score. $\begin{matrix} Score & 1 & 2 & 3 & 4 & 5 \\ Probability & 0.12 & 0.25 & 0.23 & x & 3 x \end{matrix}$

(a) Show that $x = 0.1$ . 2 marks No calculator

Mark scheme

M1 $0.12 + 0.25 + 0.23 + x + 3 x = 1$ oe, or $4 x = 1 - 0.6$
A1 $4 x = 0.4$ or $x = \frac{0.4}{4}$ oe seen, leading to $x = 0.1$ with no errors

(b) Write down the score that is most likely. 1 mark No calculator

Mark scheme

B1 5

(c) $A$ is the event 'the score is 2'. Write down $P (A^{'})$ . 1 mark No calculator

Mark scheme

B1 $0.75$ oe

(d) Kofi says, 'The probability that the score is 6 is 0.1.' Explain why Kofi is wrong. 1 mark No calculator

Mark scheme

B1 There is no 6 on the spinner, so a score of 6 is impossible and its probability is 0 oe, or the probabilities of scores 1 to 5 already add up to 1 oe

Question 3. A club has 50 members. 28 members play chess, 21 members play badminton and 9 members play both games. The rest of the members play neither game.

(a) Draw a Venn diagram to show this information. Write the number of members in each of the four regions. 2 marks

Mark scheme

B2 19 chess only, 9 both, 12 badminton only, 10 neither. B1 for two or three of these regions correct

(b) A member is chosen at random. Find the probability that this member plays badminton but not chess. 1 mark

Mark scheme

B1 $\frac{12}{50}$ oe; FT their diagram

(c) A member is chosen at random. Find the probability that this member plays neither game. 1 mark

Mark scheme

B1 $\frac{10}{50}$ oe; FT their diagram

(d) Calculate the probability that a member chosen at random does not play chess. Give your answer as a decimal. 2 marks

Mark scheme

M1 $1 - \frac{28}{50}$ or $\frac{12 + 10}{50}$ oe
A1 0.44 cao

How this is examined

Working out how many items are added to a bag when the new probability is given: A 3-mark Work out: items of one colour are added to a bag, nothing else changes, and you are given the new probability of that colour M1 for an equation with the new count over the new total, such as $\frac{5 + w}{20 + w} = \frac{2}{5}$ . Both the top and the bottom change, because the added items are part of the new total. M1 for clearing the fractions correctly: $5 (5 + w) = 2 (20 + w)$ . A1 for the whole number of items.
Finding a missing probability in a table, using the fact that all the probabilities add up to 1: A 2- or 3-mark Show (that) or Calculate: a table gives the probability of each outcome, and one or more values are missing or written in terms of $x$ An equation that sets the sum of all the probabilities equal to 1, such as $0.6 + 4 x = 1$ , for M1. Then the value of $x$ for A1. For a Show (that), write the line before the given answer (here $4 x = 0.4$ ), so that the given value follows from your working.
The probability of one event from counts: a bag, a dice, a table, a bar chart or a Venn diagram: A 1- or 2-mark Write down or Calculate: one item is chosen at random, and you give the probability that it has a stated property A fraction with the number in the event on top and the total underneath, such as $\frac{9}{20}$ , or the same value as a decimal or a percentage. The total includes every item, or only the stated group when the question chooses from one group. In a Venn diagram, the top is every region inside the circle, including the overlap.
Using P(A′) = 1 – P(A) for an event that does not happen: A 1- or 2-mark Write down or Calculate: $P (A)$ is given or found, and the question asks for $P (A^{'})$ or for the probability of 'not' something In a 2-mark part, the subtraction $1 - P (A)$ in the working for the method mark, then the value for the accuracy mark. In a 1-mark Write down, only the correct value is needed. With percentages, subtract from $100 %$ . Counting the outcomes that are not in the event is an equally valid method.
What a probability can and cannot be: A 1-mark Write down or Explain: give the probability of an impossible or certain event, or say why a given value cannot be correct $0$ for an impossible event and $1$ for a certain event. For an Explain, one clear reason, such as 'there is no 6 on the spinner, so scoring 6 is impossible and its probability is 0', or 'a probability cannot be greater than 1'.

Words to know

Probability: A number from 0 to 1 that tells you how likely an event is. It is written as a fraction, a decimal or a percentage.
Event: Something that can happen or not happen, such as 'the dice shows a 6'.
Outcome: One possible result. A dice has six outcomes: 1, 2, 3, 4, 5 and 6.
Equally likely: Having the same chance of happening. Only then can you find a probability by counting outcomes.
At random: Chosen so that each item has the same chance of being chosen.
Probability scale: The line from 0 (impossible) to 1 (certain), with 0.5 as an even chance.
P(A): The probability of the event A.
P(A′): The probability of not A: the probability that A does not happen. It equals 1 – P(A).

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