Chapter 23 · Higher tier

GCSE Maths Probability (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Use the probability scale from 0 to 1 and the language of chance, and write probabilities as fractions, decimals and percentages
Calculate the theoretical probability of an event when outcomes are equally likely, and use P(not A) = 1 − P(A)
Use relative frequency from experiments to estimate probabilities, find expected numbers, and judge whether something is fair or biased
List outcomes systematically and use sample-space tables for two events
Use the multiplication rule for independent events and the addition rule for mutually exclusive events
Draw and use tree diagrams, with and without replacement
Use sets, Venn diagrams, two-way tables and frequency trees to organise information and find probabilities
Higher: calculate conditional probabilities and solve probability problems that need algebra

You should already know

Fractions, decimals and percentages, and simplifying fractions (Chapters 3, 4 and 5)
Multiplying and adding fractions (Chapter 3)
Ratio and sharing in a ratio (Chapter 15)
Solving linear equations (Chapter 9)

Where this leads

Statistics: charts, averages and sampling (Chapter 24)
Probability in real-life risk, games and insurance

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: P1, P2, P3, P4, P5, P6, P7, P8, P9.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

23.1 The probability scale and probability of single events: Worksheet (PDF) · Answers (PDF) · lesson page
23.2 Probabilities add up to 1: the probability of “not A”: Worksheet (PDF) · Answers (PDF) · lesson page
23.3 Relative frequency, expected outcomes and fairness: Worksheet (PDF) · Answers (PDF) · lesson page
23.4 Listing outcomes and sample-space tables: Worksheet (PDF) · Answers (PDF) · lesson page
23.5 Independent events: the AND rule and the OR rule: Worksheet (PDF) · Answers (PDF) · lesson page
23.6 Tree diagrams: with and without replacement: Worksheet (PDF) · Answers (PDF) · lesson page
23.7 Sets and Venn diagrams: Worksheet (PDF) · Answers (PDF) · lesson page
23.8 Two-way tables and frequency trees: Worksheet (PDF) · Answers (PDF) · lesson page
23.9 Conditional probability and probability with algebra: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWrite the fraction 1230 in its simplest form.

Done

FWork out 25 of 30.

Done

FWrite 35% as a fraction in its simplest form.

Done

FWork out 23 × 34.

Done

FWork out 1 − 38.

Done

23.1

The probability scale and probability of single events

Edexcel 1MA1: P3, P7
Grade 1-4

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: The probability scale and probability of single eventsComing soon

We use chance every day: “It will probably rain”, “I might win the raffle”, “There is no way I will finish that today.” In maths we measure chance with a number called a probability. The first words to learn are these. An experiment is anything we do where the result is not certain, such as rolling a die. Each possible result is an outcome. An event is the outcome (or group of outcomes) we are interested in, such as “the die shows an even number”. If every outcome has the same chance, we say the outcomes are equally likely, and a fair experiment, where nothing is rigged, is called random. A coin that is not fair is called biased.

The probability scale. Every probability is a number from 0 to 1. An event that cannot happen has probability 0 (impossible). An event that must happen has probability 1 (certain). An event that is as likely to happen as not has probability 1/2 (an even chance). Everything else lies between: the nearer to 1, the more likely. Probabilities can be written as fractions, decimals or percentages, so 1/2 = 0.5 = 50%. They are never written as ratios, and they are never bigger than 1 (or 100%) or smaller than 0.

The probability scale. A is at 1/4 (unlikely) and B is at 3/4 (likely).

Theoretical probability. When all the outcomes are equally likely we can calculate the probability without doing the experiment. Picture a bag with 3 red counters and 2 blue counters. There are 5 counters and each one is equally likely to be picked. Three of them are red, so the chance of red is 3 out of 5.

P(red) = 3 out of 5 = 3/5 = 0.6 = 60%.

The rule is: probability = number of favourable outcomes ÷ total number of outcomes. “Favourable” just means “the ones we are looking for”. We write P(red) for “the probability of red”. To use the rule: (1) list or count all the possible outcomes, this is the bottom of the fraction; (2) count the outcomes that match the event, this is the top; (3) simplify the fraction. A single die has six equally likely outcomes, 1, 2, 3, 4, 5, 6. Three of them are even (2, 4, 6), so P(even) = 3/6 = 1/2.

A fair spinner with 6 equal sections. P(red) = 3/6 = 1/2, P(green) = 1/6.

Key ideas

Probability is a number from 0 (impossible) to 1 (certain). 1/2 is an even chance.
Write probabilities as fractions, decimals or percentages, never as a ratio.
For equally likely outcomes: probability = favourable outcomes ÷ total outcomes.
The bottom of the fraction is all the outcomes. The top is only the ones we want.
Simplify fractions: 3/6 = 1/2. Probability can never be more than 1 or less than 0.

Worked example 23.1.1

A fair six-sided die is rolled. Find the probability that the score is a prime number.

Key words. An outcome is one possible result of an experiment. An event is the outcome (or group of outcomes) we are interested in. Equally likely outcomes have the same chance. Probability = number of favourable outcomes ÷ total number of equally likely outcomes, and it is always between 0 and 1.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

List all the possible outcomes. The total number of equally likely outcomes goes on the bottom of the probability fraction.
1, 2, 3, 4, 5, 6
Total number of outcomes = 6
Pick out the outcomes where the score is a prime number. These are the favourable outcomes. Their number goes on the top.
2, 3, 5
Number of favourable outcomes = 3
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the score is a prime number) = 36
Simplify the fraction. Divide the top and the bottom by their highest common factor, 3. The value does not change but the numbers are smaller.
3 ÷ 3 = 1 and 6 ÷ 3 = 2
36 = 12

Check. The other 3 outcomes do not fit the event, so P(the event does not happen) = 12. Then 36 + 36 = 66 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Worked example 23.1.2

A bag contains 5 red counters, 3 blue counters and 2 green counters. One counter is picked at random. Find the probability that it is red or green.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Find the total number of counters. Every counter is equally likely to be picked, so add up all the counters. This is the bottom of the fraction.
5 + 3 + 2 = 10
Count the counters that are red and green. These are the favourable outcomes. This number goes on the top.
5 + 2 = 7
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the counter is red or green) = 710

Check. The other 3 outcomes do not fit the event, so P(the event does not happen) = 310. Then 710 + 310 = 1010 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Worked example 23.1.3

A fair spinner has 8 equal sections numbered 1, 2, 3, 3, 4, 5, 5, 5. Find the probability that it lands on 5.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

List all the possible outcomes. The total number of equally likely outcomes goes on the bottom of the probability fraction. Count every section, even when a number appears more than once.
1, 2, 3, 3, 4, 5, 5, 5
Total number of sections = 8
Pick out the outcomes where the spinner lands on 5. These are the favourable outcomes. Their number goes on the top.
5, 5, 5
Number of favourable sections = 3
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the spinner lands on 5) = 38

Check. The other 5 outcomes do not fit the event, so P(the event does not happen) = 58. Then 38 + 58 = 88 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Worked example 23.1.4

The probability that a bus is on time is 7/20. Write this probability as a decimal and as a percentage.

Method. Probabilities can be written as fractions, decimals or percentages. Change the fraction by making the denominator 100 or by dividing.

Start with the fraction. A fraction, a decimal and a percentage are three ways of writing the same number.
720
Make the denominator 100. A percentage is a number out of 100. Multiply the top and the bottom by the same number.
20 × 5 = 100, so 7 × 5 = 35
720 = 35100
Write it as a decimal and a percentage. 35 out of 100 is 35%, and 35 hundredths is 0.35.
35100 = 0.35 = 35%

Check. Multiply the decimal by the denominator: 0.35 × 20 = 7, which is the numerator 7. ✓

Watch out. A probability as a percentage can be written 25%, as a decimal 0.25 and as a fraction 14. They are all the same number.

Watch out

The bottom of the fraction is all the outcomes, not the number of outcomes you do not want. In a bag with 3 red and 2 blue counters, P(red) is 3/5, not 3/2 and not 3 : 2. And a number that appears twice on a spinner counts twice.

23.2

Probabilities add up to 1: the probability of “not A”

Edexcel 1MA1: P4, P3
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Probabilities add up to 1: the probability of “not A”Coming soon

Look at a single roll of a die. Exactly one of the numbers 1, 2, 3, 4, 5 or 6 must come up. These six outcomes cover every possibility, so we call them exhaustive. They cannot happen at the same time, so we also call them mutually exclusive (the die cannot show 2 and 5 together). Because exactly one of them must happen, the probabilities of all the outcomes add up to 1: 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1. This is a very useful fact, because if we know all the probabilities but one, we can work out the missing one by taking the rest away from 1.

An event A either happens or it does not. The two probabilities fill the whole bar: P(A) + P(not A) = 1.

The complement. For any event A, the event “A does not happen” is called not A (or the complement of A, written A′). Since A either happens or does not happen, and these two cannot both happen, P(not A) = 1 − P(A). For example, if the probability that it rains is 0.3, then the probability that it does not rain is 1 − 0.3 = 0.7. It is usually quicker to work out “not A” this way than to count all the outcomes that are not A. Use it whenever you see the words not, fails, misses or at least one (you will see the last one in section 23.5).

Fractions, decimals and percentages. The rule works in every form. With a fraction, write 1 as a fraction with the same denominator: 1 − 3/8 = 8/8 − 3/8 = 5/8. With a decimal, 1 − 0.35 = 0.65. With a percentage, 100% − 35% = 65%.

Missing probabilities. If several outcomes are listed in a table and one probability is missing, add the known ones and take the total from 1. Adding for “or”. If two events cannot happen together (mutually exclusive), the probability of one or the other is found by adding their probabilities. For example, if P(red) = 0.2 and P(blue) = 0.35 for a spinner, then P(red or blue) = 0.2 + 0.35 = 0.55. We use this idea a lot in sections 23.5 and 23.6.

Key ideas

The probabilities of all the possible outcomes of an experiment add up to 1.
P(not A) = 1 − P(A), in any form: fractions, decimals or percentages.
Mutually exclusive events cannot happen together. Exhaustive outcomes cover every possibility.
To find a missing probability: add the known ones and subtract the total from 1.
For mutually exclusive events, P(A or B) = P(A) + P(B).

Worked example 23.2.1

The probability that Priya’s bus is late is 0.35. Work out the probability that her bus is not late.

Key words. The complement of an event is “the event does not happen”, written not A or A′. Outcomes are exhaustive if together they cover every possibility, so their probabilities add up to 1.

Method. The probabilities of an event happening and not happening add up to 1 (or 100%), so subtract the given probability from 1.

Spot the event and its opposite. Either “late” happens or “not late” happens, never both, so these two events are opposites that cover every possibility.
P(late) = 0.35
Write the rule. An event either happens or it does not, so the two probabilities add up to 1. Therefore P(not A) = 1 − P(A).
P(not late) = 1 − P(late)
Subtract from 1. Line up the decimal points and take the probability away from 1.
1 − 0.35 = 0.65

Check. Add the two probabilities: 0.35 + 0.65 = 1. ✓

Watch out. The answer must be between 0 and 1. If you get a negative number or a number bigger than 1 you have subtracted the wrong way round.

Worked example 23.2.2

The probability that Josh wins a game is 38. Work out the probability that he does not win.

Method. The probabilities of an event happening and not happening add up to 1 (or 100%), so subtract the given probability from 1.

Spot the event and its opposite. Either “win” happens or “not win” happens, never both, so these two events are opposites that cover every possibility.
P(win) = 38
Write the rule. An event either happens or it does not, so the two probabilities add up to 1. Therefore P(not A) = 1 − P(A).
P(not win) = 1 − P(win)
Write 1 as a fraction with the same denominator. We can only subtract fractions with the same denominator. 1 whole is 8 out of 8.
1 = 88
Subtract the numerators. The denominator stays the same.
88 − 38 = 58

Check. Add the two probabilities: 38 + 58 = 1. ✓

Watch out. The answer must be between 0 and 1. If you get a negative number or a number bigger than 1 you have subtracted the wrong way round.

Colour	Red	Blue	Green	Yellow
Probability	0.3	0.25	0.15	x

Worked example 23.2.3

A biased spinner can land on red, blue, green or yellow. The table shows the probabilities. Work out the value of x, the probability of yellow.

Method. The probabilities of all the possible outcomes add up to 1. Add the ones we know, then take the total away from 1.

Write down what we know. The outcomes listed cover every possibility, so their probabilities add up to 1. One of them is missing.
P(red) = 0.3
P(blue) = 0.25
P(green) = 0.15
P(yellow) = ?
Add the known probabilities. Line up the decimal points carefully.
0.3 + 0.25 + 0.15 = 0.7
Subtract the total from 1. All the probabilities add up to 1, so the missing one is 1 minus the total of the others.
1 − 0.7 = 0.3

Check. Add all the probabilities, including the one we found: 0.3 + 0.25 + 0.15 + 0.3 = 1. ✓

Watch out. Do not forget that the missing probability makes the total exactly 1 (not more, not less).

Worked example 23.2.4

In a game a player can win, draw or lose. P(win) = 12 and P(draw) = 16. Work out P(lose).

Method. The probabilities of all the possible outcomes add up to 1. Add the ones we know, then take the total away from 1.

Write down what we know. The outcomes listed cover every possibility, so their probabilities add up to 1. One of them is missing.
P(win) = 12
P(draw) = 16
P(lose) = ?
Add the known probabilities. Use the common denominator 6 so that we can add them.
36 + 16 = 46 = 23
Subtract the total from 1. All the probabilities add up to 1, so the missing one is 1 minus the total of the others.
66 − 46 = 26 = 13

Check. Add all the probabilities, including the one we found: 12 + 16 + 13 = 1. ✓

Watch out. Do not forget that the missing probability makes the total exactly 1 (not more, not less).

Worked example 23.2.5

Use the spinner table above. Work out the probability that the spinner lands on red or blue.

Method. Red and blue cannot both happen on one spin, so the events are mutually exclusive. For mutually exclusive events we add the probabilities.

Decide whether the events can happen together. The spinner lands on one colour only, so red and blue cannot both happen.
Mutually exclusive
Write the rule. For mutually exclusive events, P(A or B) = P(A) + P(B).
P(red or blue) = P(red) + P(blue)
Add the probabilities. Line up the decimal points.
0.3 + 0.25 = 0.55

Check. The opposite event is green or yellow: 0.15 + 0.3 = 0.45 (since x = 0.3), and 0.55 + 0.45 = 1. ✓

Watch out. Add for “or” only when the events cannot happen together.

Watch out

Do not stop after finding the total of the known probabilities. The missing probability is 1 minus that total. And check that every probability is between 0 and 1 and that all of them add up to exactly 1. Also do not say P(not A) = P(A) − 1: that is negative.

Exercise 23B Adding to 1 and the probability of “not”

Non-calculator

FThe probability of event A is given. Work out the probability that A does not happen.

(a)P(A) = 27

(b)P(A) = 59

(c)P(A) = 310

(d)P(A) = 712

(e)P(A) = 120

Done

FWork out the probability that each event does not happen.

(a)P(rain) = 0.7

(b)P(win) = 0.45

(c)P(late) = 0.08

(d)P(pass) = 85%

(e)P(sunny) = 62%

Done

FA biased spinner lands on red, blue, green or yellow. The table shows the probabilities. Work out the probability that it lands on yellow.

Colour	Red	Blue	Green	Yellow
Probability	0.2	0.35	0.15	?

Done

FA bag has red, blue and green counters only. P(red) = 310 and P(blue) = 15. Work out P(green).

Done

PSThe table shows the probabilities of the scores on a biased die.

Score	1	2	3	4	5	6
Probability	0.1	0.15	0.2	0.2	0.15	?

(a)Work out the probability of scoring 6

(b)Work out the probability of scoring an even number

(c)Work out the probability of not scoring 6

Done

RThe probability that a train arrives early is 0.1, the probability that it is on time is 0.6 and the probability that it is late is 0.2. Explain why there must be another possible outcome, and find its probability.

Done

RAli says: “The probabilities of A, B and C are 0.5, 0.3 and 0.4, and A, B and C are the only possible outcomes.” Explain why Ali cannot be correct.

Done

FThe probability that Ravi passes a driving test is 0.58. The probability that Aisha passes is 0.72. Work out the probability that...

(a)Ravi does not pass

(b)Aisha does not pass

Done

PSThe spinner has 12 equal sections. Each section is red, blue or green. P(red) = 13 and P(blue) = 14. How many sections are green?

Done

PSA bag has red, blue and yellow counters only. P(red) = 0.4 and P(blue) = 0.35. Work out P(yellow), and then P(red or yellow).

(a)P(yellow)

(b)P(red or yellow)

Done

RP(A) = 0.7 and P(B) = 0.5. Explain why A and B cannot be mutually exclusive.

Done

PSThe probabilities that a spinner lands on A, B, C or D are x, 2x, 3x and 0.4. Work out the value of x and the probability that it lands on C.

(a)The value of x

(b)The probability of C

Done

23.3

Relative frequency, expected outcomes and fairness

Edexcel 1MA1: P1, P2, P3, P5
Grade 2-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Relative frequency, expected outcomes and fairnessComing soon

Sometimes we cannot calculate a probability because the outcomes are not equally likely. Think of a drawing pin: it can land point up or point down, but there is no reason for each to be 1/2. Then we find out by trying it. Each go is a trial, and the number of times the event happens is its frequency. A table of these results is a frequency table. The relative frequency is the fraction of the trials in which the event happened:

relative frequency = frequency ÷ total number of trials

If a pin lands point up 18 times in 50 trials, the relative frequency of “point up” is 18/50 = 9/25 = 0.36. We use the relative frequency as our estimate of the probability.

More trials give better estimates. A small number of trials can be misleading. The graph below shows the relative frequency of heads when a coin was tossed 500 times (the results come from a computer simulation of a fair coin). After 10 tosses the relative frequency was 0.3. After 100 tosses it was 0.55, and after 500 tosses it was 0.478. As the number of trials grows, the relative frequency settles down towards the theoretical probability 0.5. This is why scientists, insurers and doctors collect as much data as they can. But a fair coin will not usually give exactly half heads: there is always some random variation.

Relative frequency of heads (blue) settles near the theoretical probability 0.5 (dashed line) as the number of tosses increases.

Expected outcomes. If we know the probability of an event, we can predict roughly how many times it will happen in a number of trials: expected number = probability × number of trials. If P(6) = 1/6 and a die is rolled 120 times, we expect 1/6 × 120 = 20 sixes. We do not promise exactly 20: it is an average over many sets of trials.

Fair or biased? A fair die has all six outcomes equally likely. To test a die, roll it many times and compare each relative frequency with 1/6. If they are all close, there is no evidence of bias. If one is far away, and the number of trials is large, then the die may be biased. We say “may be” because we can never be completely sure from experiment alone. Throughout this chapter we assume experiments are random (nothing is rigged) and that each trial does not affect the next.

Key ideas

Relative frequency = frequency ÷ total number of trials. It estimates the probability.
More trials usually give a better estimate. Results from few trials can be very different from the true probability.
Expected number = probability × number of trials. It is an average, not a promise.
Compare relative frequencies with the theoretical probability to judge fairness. A fair die does not give exactly equal results.
Relative frequencies from all the possible outcomes of one experiment add up to 1.

Worked example 23.3.1

A spinner is spun 80 times. It lands on red 28 times. Work out the relative frequency of red, in its simplest form.

Key words. Frequency is how many times something happened. Relative frequency = frequency ÷ total number of trials. It is an estimate of the probability. An expected number is probability × number of trials.

Method. Relative frequency is the number of times the event happened divided by the number of trials.

Write the rule. Relative frequency = frequency ÷ total number of trials. It estimates the probability.
relative frequency = frequencytrials
Put the numbers in. The event happened 28 times in 80 trials.
2880
Simplify the fraction. Divide the top and the bottom by their highest common factor, 4. The value does not change but the numbers are smaller.
28 ÷ 4 = 7 and 80 ÷ 4 = 20
2880 = 720

Check. Multiply back: 720 × 80 = 28, which is the frequency 28. ✓

Watch out. The bottom of the fraction is the total number of trials, not the number of times the event did not happen.

Worked example 23.3.2

A fair die is rolled 120 times. How many times would you expect it to land on a 5?

Method. The expected number of times something happens is the probability multiplied by the number of trials.

Write down the rule. We multiply how likely the event is on one trial by the number of trials.
expected number = probability × number of trials
Write down the probability and the number of trials. We need both numbers.
P(5) = 16, number of trials = 120
Find the expected number. Expected number = probability × number of trials. Take 16 of 120: divide by 6, then multiply by 1.
120 ÷ 6 = 20
20 × 1 = 20

Check. Work backwards: 20 ÷ 120 = 0.17, which is the probability we started with. ✓

Watch out. An expected number is an average over many trials. It does not promise exactly that many will happen.

Worked example 23.3.3

A factory checks 500 bulbs. 12 are faulty. (a) Estimate the probability that a bulb is faulty. (b) The factory makes 8000 bulbs in a week. Estimate how many are faulty.

Method. Use the relative frequency of faulty bulbs as the probability, then multiply by the number of bulbs.

Work out the relative frequency. Faulty bulbs divided by the number checked.
relative frequency = 12500 = 3125
Use it as the estimated probability. We use the experiment to estimate the probability.
P(faulty) ≈ 3125 = 0.024
Multiply by the number of bulbs made. Expected number = probability × number of trials.
3125 × 8000 = 8000 ÷ 125 × 3 = 64 × 3 = 192

Check. 12 faulty in 500 means 24 faulty in 1000, so in 8000 we expect 24 × 8 = 192. ✓

Watch out. The answer is an estimate, so we write “about 192” and not “exactly 192”.

Worked example 23.3.4

Dan tosses a coin 200 times and gets 124 heads. Is there evidence that the coin is biased?

Method. Work out the relative frequency of heads and compare it with 1/2, which is the probability for a fair coin.

Work out the relative frequency of heads. Frequency of heads divided by the number of tosses.
124200 = 3150 = 0.62
Find the expected number of heads for a fair coin. A fair coin gives P(heads) = 1/2.
12 × 200 = 100 heads
Compare. 124 heads is quite a lot more than 100. With 200 trials this is a fairly big difference.
0.62 compared with 0.5
Write a conclusion. Because the gap is large for so many tosses, it suggests the coin may be biased towards heads. We cannot be certain.
There is some evidence the coin is biased.

Check. If the coin were fair, 100 heads is the average. 124 is 24 more, which is unusual but not impossible. More tosses would give a stronger conclusion. ✓

Watch out. Never say a coin is “definitely” biased from one experiment.

Watch out

A relative frequency from a few trials can be far from the real probability. Tom tossing a coin 10 times and getting 7 heads does not prove that P(heads) = 0.7. Also, the expected number is an average: expecting 20 sixes in 120 rolls does not mean you will get exactly 20.

Exercise 23C Relative frequency

Non-calculator

FWrite down the relative frequency of the event. Give each answer as a fraction in its simplest form.

(a)The event happened 9 times in 30 trials

(b)The event happened 24 times in 80 trials

(c)The event happened 15 times in 60 trials

(d)The event happened 35 times in 140 trials

Done

FWork out the relative frequency of each event as a decimal.

(a)6 out of 20 trials

(b)27 out of 60 trials

(d)7 out of 8 trials

Done

FTom spins a spinner 80 times. The table shows his results. Find the relative frequency of each colour, as a fraction in its simplest form.

Colour	Red	Blue	Green
Frequency	24	31	25

(a)Red

(b)Blue

(c)Green

(d)Add the three relative frequencies

Done

FA coin is tossed several times. The table shows the number of heads after different numbers of tosses. Work out the relative frequency of heads each time, as a decimal.

Number of tosses	20	50	100	200
Number of heads	12	27	52	98

(a)After 20 tosses

(b)After 50 tosses

(c)After 100 tosses

(d)After 200 tosses

(e)Which result is the best estimate of P(heads)? Explain.

Done

PSIn a survey 200 students were asked how they travel to school. 38 walk, 62 take the bus, 70 come by car and the rest cycle. A student is chosen at random from the 200. Estimate the probability that the student...

(a)walks

(b)cycles

Done

RTom tosses a coin 10 times and gets 7 heads. He says: “P(heads) must be 0.7.” Explain why Tom may be wrong.

Done

RElla spun a spinner 20 times and landed on red 6 times. Raj spun the same spinner 200 times and landed on red 58 times. Who has the better estimate of P(red)? Give a reason, and write both estimates as decimals.

Done

RA fair coin was tossed 400 times by a computer. The table gives the number of heads after 20, 50, 100, 200 and 400 tosses and the graph shows the relative frequency of heads.

Number of tosses	20	50	100	200	400
Heads	12	28	50	102	184

(a)Work out the relative frequency of heads after 50 tosses, as a decimal.

(b)Work out the relative frequency of heads after 400 tosses, as a decimal.

(c)The theoretical probability of heads is 0.5. Which is closer to 0.5: the result after 20 tosses or after 400 tosses?

Done

PSA bag contains 20 counters that are red or blue. Sam picks a counter, records its colour and puts it back. He does this 150 times and gets red 93 times. Use his results to estimate the number of red counters in the bag.

Done

Exercise 23D Expected outcomes, fair and biased

Calculator allowed

FA fair six-sided die is rolled. How many times would you expect to roll...

(a)a 3 in 60 rolls

(b)an even number in 120 rolls

(c)a number bigger than 4 in 300 rolls

(d)a 6 in 450 rolls

Done

FWork out the expected number of times that each event happens.

(a)A fair coin is tossed 500 times: heads

(b)A spinner has P(red) = 3/8 and is spun 200 times: red

(c)The probability of rain on a day is 0.2: rainy days in 365 days

(d)P(a part is faulty) = 2%: faulty parts in 5000

Done

PSA drawing pin is dropped 60 times and lands point up 24 times. Use his results to answer the questions.

(a)Relative frequency of point up

(b)Estimate for 150 drops

Done

PSAt a fair, a game costs £2 to play. The probability of winning is 112 and a win pays £10 prize money. In one day the game is played 300 times. Work out the expected profit for the fair.

Done

PSMaria rolls a die 120 times. The table shows her results.

Score	1	2	3	4	5	6
Frequency	17	23	19	25	16	20

(a)How many times would you expect each score if the die is fair?

(b)Which score is furthest from the expected number?

(c)Is there evidence that the die is biased? Give a reason.

Done

RMaya tosses a fair coin and gets heads 5 times in a row. She says: “Tails is more likely next time, to even things out.” Is Maya correct? Explain.

Done

RDie A is rolled 30 times and shows a 6 on 5 rolls. Die B is rolled 300 times and shows a 6 on 70 rolls. Which die is more likely to be biased? Show your working.

Done

PSThe probability that Ben wins a game is 0.15. Work out the least number of games Ben must play to expect to win at least 12 games.

Done

RExplain why an expected number of wins does not have to be a whole number, and why it does not tell us exactly how many wins there will be.

Done

23.4

Listing outcomes and sample-space tables

Edexcel 1MA1: P6, P7
Grade 2-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Listing outcomes and sample-space tablesComing soon

To use the rule probability = favourable ÷ total we must know all the outcomes. When there are two or more events, such as tossing two coins, it is easy to miss one. The cure is to list the outcomes systematically: in a fixed order, so that nothing is missed or repeated.

Listing. For two coins, write what the first coin shows, then every possibility for the second: first coin heads (H): HH, HT. First coin tails (T): TH, TT. That gives 4 outcomes: HH, HT, TH, TT. Notice that HT and TH are different outcomes (the first coin and the second coin are different), and each of the four is equally likely. So P(two heads) = 1/4, and P(one head and one tail) = 2/4 = 1/2, not 1/3.

coin 2 \ coin 1	H	T
H	HH	HT
T	TH	TT

Two coins: the four equally likely outcomes. The two highlighted cells give “one head and one tail”.

Sample-space tables. For two events with several outcomes each, a table (a sample space diagram) is neat. Put the outcomes of one event along the top and the other down the side. Each cell gives one combined outcome, such as the total of two dice. Count the cells that match the question, and count all the cells. For two dice there are 6 × 6 = 36 equally likely outcomes.

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

The total of two dice. The highlighted cells show a total of 8: five of the 36 cells, so P(total 8) = 5/36.

Counting without listing. Sometimes we only need to know how many outcomes there are. If there are 3 starters and 2 main courses, each starter can go with each main course, so there are 3 × 2 = 6 different two-course meals. This is the product rule for counting: multiply the number of choices at each stage. It also tells us the size of the sample space: two dice give 6 × 6 = 36, three coins give 2 × 2 × 2 = 8.

Key ideas

List outcomes in a fixed order so none is missed or repeated. HT and TH are different outcomes.
A sample-space table shows every combined outcome of two events as a cell.
Probability = number of matching cells ÷ total number of cells.
Two dice have 36 outcomes. Two coins have 4. Three coins have 8.
Product rule: number of outcomes = (choices at stage 1) × (choices at stage 2) × ...

Worked example 23.4.1

Two fair coins are tossed. List all the possible outcomes and find the probability of getting one head and one tail.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

List all the possible outcomes. The total number of equally likely outcomes goes on the bottom of the probability fraction. Write the first coin first, then the second.
HH, HT, TH, TT
Total number of outcomes = 4
Pick out the outcomes where one head and one tail. These are the favourable outcomes. Their number goes on the top.
HT, TH
Number of favourable outcomes = 2
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(one head and one tail) = 24
Simplify the fraction. Divide the top and the bottom by their highest common factor, 2. The value does not change but the numbers are smaller.
2 ÷ 2 = 1 and 4 ÷ 2 = 2
24 = 12

Check. The other 2 outcomes do not fit the event, so P(the event does not happen) = 12. Then 24 + 24 = 44 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Worked example 23.4.2

Two fair dice are rolled and their scores are added. Find the probability that the total is 8.

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

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely. Each cell shows the total of the two scores.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the total is 8. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Number of highlighted cells = 5
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is 8) = 536

Check. The other 31 outcomes do not fit the event, so P(the event does not happen) = 3136. Then 536 + 3136 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Worked example 23.4.3

Spinner A has the numbers 1, 2, 3 and spinner B has the colours red (R) and blue (B). Each spinner is spun once. List all the outcomes and find the probability of getting an odd number and red.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

List all the possible outcomes. The total number of equally likely outcomes goes on the bottom of the probability fraction. List spinner A first, then spinner B: for each number there are two colours.
1R, 1B, 2R, 2B, 3R, 3B
Total number of outcomes = 6
Pick out the outcomes where an odd number and red. These are the favourable outcomes. Their number goes on the top.
1R, 3R
Number of favourable outcomes = 2
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(an odd number and red) = 26
Simplify the fraction. Divide the top and the bottom by their highest common factor, 2. The value does not change but the numbers are smaller.
2 ÷ 2 = 1 and 6 ÷ 2 = 3
26 = 13

Check. The other 4 outcomes do not fit the event, so P(the event does not happen) = 23. Then 26 + 46 = 66 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Worked example 23.4.4

A café sells 3 sandwiches (cheese, ham, egg) and 4 drinks. How many different sandwich and drink combinations are there?

Method. To count how many outcomes there are, multiply the number of choices at each stage.

Split the problem into stages. We make one choice at a time, so we count the choices stage by stage.
Sandwiches then Drinks
Write down the number of choices at each stage. For each stage we count how many choices there are.
Sandwiches: 3 choices
Drinks: 4 choices
Multiply the numbers of choices. Each choice at one stage can be paired with every choice at the next stage, so the numbers multiply (this is the product rule for counting).
3 × 4 = 12

Check. For two stages, picture a table with 3 rows and 4 columns: it has 3 × 4 = 12 cells. ✓

Watch out. Multiply the numbers of choices. Do not add them.

Watch out

Do not forget that HT and TH are different outcomes. If you count “one head and one tail” as a single outcome you get 3 outcomes and the wrong answer 1/3. Also be careful to count all the cells, not just the highlighted ones, for the bottom of the fraction.

Exercise 23E Listing outcomes and sample-space tables

Non-calculator

FA fair coin is tossed and a fair die is rolled. Write the outcomes as, for example, H3 (heads and a 3). Find...

(a)the number of possible outcomes

(b)the probability of heads and a 6

(c)the probability of tails and an odd number

Done

FThree fair coins are tossed. List all the possible outcomes systematically. Then find the probability of getting...

(a)exactly two heads

(b)at least one tail

(c)all three coins the same

Done

FTwo fair dice are rolled and the scores are added. Use the sample-space table to find the probability that the total is...

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

(a)7

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the total is 7. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Number of highlighted cells = 6
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is 7) = 636
Simplify the fraction. Divide the top and the bottom by their highest common factor, 6. The value does not change but the numbers are smaller.
6 ÷ 6 = 1 and 36 ÷ 6 = 6
636 = 16

Check. The other 30 outcomes do not fit the event, so P(the event does not happen) = 56. Then 636 + 3036 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

(b)12

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the total is 12. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Number of highlighted cells = 1
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is 12) = 136

Check. The other 35 outcomes do not fit the event, so P(the event does not happen) = 3536. Then 136 + 3536 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

(c)less than 5

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the total is less than 5. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Number of highlighted cells = 6
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is less than 5) = 636
Simplify the fraction. Divide the top and the bottom by their highest common factor, 6. The value does not change but the numbers are smaller.
6 ÷ 6 = 1 and 36 ÷ 6 = 6
636 = 16

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

(d)an even number

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the total is even. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Number of highlighted cells = 18
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is even) = 1836
Simplify the fraction. Divide the top and the bottom by their highest common factor, 18. The value does not change but the numbers are smaller.
18 ÷ 18 = 1 and 36 ÷ 18 = 2
1836 = 12

Check. The other 18 outcomes do not fit the event, so P(the event does not happen) = 12. Then 1836 + 1836 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

(e)a prime number

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the total is a prime number. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Number of highlighted cells = 15
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is a prime number) = 1536
Simplify the fraction. Divide the top and the bottom by their highest common factor, 3. The value does not change but the numbers are smaller.
15 ÷ 3 = 5 and 36 ÷ 3 = 12
1536 = 512

Check. The other 21 outcomes do not fit the event, so P(the event does not happen) = 712. Then 1536 + 2136 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Done

FTwo fair dice are rolled. The difference is the bigger score take away the smaller score (it is 0 if the scores are equal). Find the probability that the difference is...

(a)0

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely. Each cell shows the difference (bigger − smaller).
− 1 2 3 4 5 6
1 0 1 2 3 4 5
2 1 0 1 2 3 4
3 2 1 0 1 2 3
4 3 2 1 0 1 2
5 4 3 2 1 0 1
6 5 4 3 2 1 0
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the difference is 0. These are the favourable outcomes. Their number goes on the top of the fraction.
− 1 2 3 4 5 6
1 0 1 2 3 4 5
2 1 0 1 2 3 4
3 2 1 0 1 2 3
4 3 2 1 0 1 2
5 4 3 2 1 0 1
6 5 4 3 2 1 0
Number of highlighted cells = 6
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the difference is 0) = 636
Simplify the fraction. Divide the top and the bottom by their highest common factor, 6. The value does not change but the numbers are smaller.
6 ÷ 6 = 1 and 36 ÷ 6 = 6
636 = 16

−	1	2	3	4	5	6
1	0	1	2	3	4	5
2	1	0	1	2	3	4
3	2	1	0	1	2	3
4	3	2	1	0	1	2
5	4	3	2	1	0	1
6	5	4	3	2	1	0

−	1	2	3	4	5	6
1	0	1	2	3	4	5
2	1	0	1	2	3	4
3	2	1	0	1	2	3
4	3	2	1	0	1	2
5	4	3	2	1	0	1
6	5	4	3	2	1	0

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

(b)at least 4

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely. Each cell shows the difference (bigger − smaller).
− 1 2 3 4 5 6
1 0 1 2 3 4 5
2 1 0 1 2 3 4
3 2 1 0 1 2 3
4 3 2 1 0 1 2
5 4 3 2 1 0 1
6 5 4 3 2 1 0
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the difference is at least 4. These are the favourable outcomes. Their number goes on the top of the fraction.
− 1 2 3 4 5 6
1 0 1 2 3 4 5
2 1 0 1 2 3 4
3 2 1 0 1 2 3
4 3 2 1 0 1 2
5 4 3 2 1 0 1
6 5 4 3 2 1 0
Number of highlighted cells = 6
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the difference is at least 4) = 636
Simplify the fraction. Divide the top and the bottom by their highest common factor, 6. The value does not change but the numbers are smaller.
6 ÷ 6 = 1 and 36 ÷ 6 = 6
636 = 16

−	1	2	3	4	5	6
1	0	1	2	3	4	5
2	1	0	1	2	3	4
3	2	1	0	1	2	3
4	3	2	1	0	1	2
5	4	3	2	1	0	1
6	5	4	3	2	1	0

−	1	2	3	4	5	6
1	0	1	2	3	4	5
2	1	0	1	2	3	4
3	2	1	0	1	2	3
4	3	2	1	0	1	2
5	4	3	2	1	0	1
6	5	4	3	2	1	0

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely. Each cell shows the difference (bigger − smaller).
− 1 2 3 4 5 6
1 0 1 2 3 4 5
2 1 0 1 2 3 4
3 2 1 0 1 2 3
4 3 2 1 0 1 2
5 4 3 2 1 0 1
6 5 4 3 2 1 0
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the difference is 1. These are the favourable outcomes. Their number goes on the top of the fraction.
− 1 2 3 4 5 6
1 0 1 2 3 4 5
2 1 0 1 2 3 4
3 2 1 0 1 2 3
4 3 2 1 0 1 2
5 4 3 2 1 0 1
6 5 4 3 2 1 0
Number of highlighted cells = 10
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the difference is 1) = 1036
Simplify the fraction. Divide the top and the bottom by their highest common factor, 2. The value does not change but the numbers are smaller.
10 ÷ 2 = 5 and 36 ÷ 2 = 18
1036 = 518

−	1	2	3	4	5	6
1	0	1	2	3	4	5
2	1	0	1	2	3	4
3	2	1	0	1	2	3
4	3	2	1	0	1	2
5	4	3	2	1	0	1
6	5	4	3	2	1	0

−	1	2	3	4	5	6
1	0	1	2	3	4	5
2	1	0	1	2	3	4
3	2	1	0	1	2	3
4	3	2	1	0	1	2
5	4	3	2	1	0	1
6	5	4	3	2	1	0

Check. The other 26 outcomes do not fit the event, so P(the event does not happen) = 1318. Then 1036 + 2636 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Done

FTwo fair dice are rolled and the scores are multiplied. Find the probability that the product is...

(a)12

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely. Each cell shows the product.
× 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the product is 12. These are the favourable outcomes. Their number goes on the top of the fraction.
× 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Number of highlighted cells = 4
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the product is 12) = 436
Simplify the fraction. Divide the top and the bottom by their highest common factor, 4. The value does not change but the numbers are smaller.
4 ÷ 4 = 1 and 36 ÷ 4 = 9
436 = 19

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

Check. The other 32 outcomes do not fit the event, so P(the event does not happen) = 89. Then 436 + 3236 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

(b)an odd number

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely. Each cell shows the product.
× 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the product is odd. These are the favourable outcomes. Their number goes on the top of the fraction.
× 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Number of highlighted cells = 9
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the product is odd) = 936
Simplify the fraction. Divide the top and the bottom by their highest common factor, 9. The value does not change but the numbers are smaller.
9 ÷ 9 = 1 and 36 ÷ 9 = 4
936 = 14

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

Check. The other 27 outcomes do not fit the event, so P(the event does not happen) = 34. Then 936 + 2736 = 3636 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

(c)more than 20

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely. Each cell shows the product.
× 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the product is more than 20. These are the favourable outcomes. Their number goes on the top of the fraction.
× 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Number of highlighted cells = 6
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the product is more than 20) = 636
Simplify the fraction. Divide the top and the bottom by their highest common factor, 6. The value does not change but the numbers are smaller.
6 ÷ 6 = 1 and 36 ÷ 6 = 6
636 = 16

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Done

FSpinner A has the numbers 1, 2, 3 and spinner B has the numbers 1, 2, 3, 4. Both are fair and are spun once. The two numbers are added. Draw a sample-space table and find the probability that the total is...

(a)5

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the spinner A scores along the top and the spinner B scores down the side. Every cell is one outcome, and all the cells are equally likely.
+ 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
Total number of outcomes = 4 × 3 = 12
Highlight the cells where the total is 5. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
Number of highlighted cells = 3
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is 5) = 312
Simplify the fraction. Divide the top and the bottom by their highest common factor, 3. The value does not change but the numbers are smaller.
3 ÷ 3 = 1 and 12 ÷ 3 = 4
312 = 14

+	1	2	3
1	2	3	4
2	3	4	5
3	4	5	6
4	5	6	7

+	1	2	3
1	2	3	4
2	3	4	5
3	4	5	6
4	5	6	7

Check. The other 9 outcomes do not fit the event, so P(the event does not happen) = 34. Then 312 + 912 = 1212 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

(b)more than 4

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the spinner A scores along the top and the spinner B scores down the side. Every cell is one outcome, and all the cells are equally likely.
+ 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
Total number of outcomes = 4 × 3 = 12
Highlight the cells where the total is more than 4. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
Number of highlighted cells = 6
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is more than 4) = 612
Simplify the fraction. Divide the top and the bottom by their highest common factor, 6. The value does not change but the numbers are smaller.
6 ÷ 6 = 1 and 12 ÷ 6 = 2
612 = 12

+	1	2	3
1	2	3	4
2	3	4	5
3	4	5	6
4	5	6	7

+	1	2	3
1	2	3	4
2	3	4	5
3	4	5	6
4	5	6	7

Check. The other 6 outcomes do not fit the event, so P(the event does not happen) = 12. Then 612 + 612 = 1212 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

(c)an even number

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the spinner A scores along the top and the spinner B scores down the side. Every cell is one outcome, and all the cells are equally likely.
+ 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
Total number of outcomes = 4 × 3 = 12
Highlight the cells where the total is even. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
Number of highlighted cells = 6
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is even) = 612
Simplify the fraction. Divide the top and the bottom by their highest common factor, 6. The value does not change but the numbers are smaller.
6 ÷ 6 = 1 and 12 ÷ 6 = 2
612 = 12

+	1	2	3
1	2	3	4
2	3	4	5
3	4	5	6
4	5	6	7

+	1	2	3
1	2	3	4
2	3	4	5
3	4	5	6
4	5	6	7

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Done

PSA two-digit number is made by choosing the tens digit from 1, 2, 3 and the units digit from 1, 2, 3 (digits can be repeated). List all the possible numbers, then find the probability that a number chosen at random is...

(a)even

(b)bigger than 25

Done

FThe letters A, B and C are put in a row in a random order, using each letter once. List all the possible orders. Find the probability that A is first.

Done

PSA café offers 3 sandwiches (cheese, ham, egg), 2 soups (tomato, leek) and 4 drinks.

(a)How many different sandwich and soup combinations are there?

(b)How many different meals of sandwich, soup and drink are there?

(c)A customer picks a sandwich and a soup at random. Find the probability of ham and tomato.

Done

PSA code has one letter chosen from A, B, C, followed by one digit chosen from 1, 2, 3, 4. A code is chosen at random. Find the probability that it is...

(a)A followed by an even digit

(b)B3 or C3

Done

RAli and Beth play a game. Two fair dice are rolled and the scores are added. Ali wins if the total is 8 or more. Beth wins if the total is less than 8. Is the game fair? Show your working.

Done

PSTwo fair dice, one red and one blue, are rolled. Find the probability that the red die shows a bigger score than the blue die.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the red die scores along the top and the blue die scores down the side. Every cell is one outcome, and all the cells are equally likely. Write each cell as (red, blue).
red \ blue 1 2 3 4 5 6
1 1,1 2,1 3,1 4,1 5,1 6,1
2 1,2 2,2 3,2 4,2 5,2 6,2
3 1,3 2,3 3,3 4,3 5,3 6,3
4 1,4 2,4 3,4 4,4 5,4 6,4
5 1,5 2,5 3,5 4,5 5,5 6,5
6 1,6 2,6 3,6 4,6 5,6 6,6
Total number of outcomes = 6 × 6 = 36
Highlight the cells where red is bigger than blue. These are the favourable outcomes. Their number goes on the top of the fraction.
red \ blue 1 2 3 4 5 6
1 1,1 2,1 3,1 4,1 5,1 6,1
2 1,2 2,2 3,2 4,2 5,2 6,2
3 1,3 2,3 3,3 4,3 5,3 6,3
4 1,4 2,4 3,4 4,4 5,4 6,4
5 1,5 2,5 3,5 4,5 5,5 6,5
6 1,6 2,6 3,6 4,6 5,6 6,6
Number of highlighted cells = 15
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(red is bigger than blue) = 1536
Simplify the fraction. Divide the top and the bottom by their highest common factor, 3. The value does not change but the numbers are smaller.
15 ÷ 3 = 5 and 36 ÷ 3 = 12
1536 = 512

red \ blue	1	2	3	4	5	6
1	1,1	2,1	3,1	4,1	5,1	6,1
2	1,2	2,2	3,2	4,2	5,2	6,2
3	1,3	2,3	3,3	4,3	5,3	6,3
4	1,4	2,4	3,4	4,4	5,4	6,4
5	1,5	2,5	3,5	4,5	5,5	6,5
6	1,6	2,6	3,6	4,6	5,6	6,6

red \ blue	1	2	3	4	5	6
1	1,1	2,1	3,1	4,1	5,1	6,1
2	1,2	2,2	3,2	4,2	5,2	6,2
3	1,3	2,3	3,3	4,3	5,3	6,3
4	1,4	2,4	3,4	4,4	5,4	6,4
5	1,5	2,5	3,5	4,5	5,5	6,5
6	1,6	2,6	3,6	4,6	5,6	6,6

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Done

RTom says: “When I toss two coins there are three outcomes: two heads, one head, no heads. So P(two heads) = 1/3.” Explain why Tom is wrong and give the correct probability.

Done

RTwo fair dice are rolled and the scores are added. Show that a total of 7 is more likely than any other total.

Method. Count how many cells of the sample-space table give each total, then compare.

Draw the sample-space table of totals. Each cell is one of 36 equally likely outcomes.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Count the cells for each total. Look along each diagonal of equal totals.
2: 1, 3: 2, 4: 3, 5: 4, 6: 5, 7: 6, 8: 5, 9: 4, 10: 3, 11: 2, 12: 1
Compare. The largest count is for 7.
7 occurs 6 times, and every other total occurs 5 times or fewer.
Write the probability. 6 out of 36.
P(7) = 636 = 16

Check. The counts add up: 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36. ✓

Watch out. A table makes it easy to see why 7 is the most common total.

Done

RHow many different outcomes are there when...

(a)two fair dice

(b)three fair dice

Done

PSTwo fair 4-sided dice, numbered 1 to 4, are rolled and the scores are added. Find the probability that the total is a multiple of 3.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely. Totals that are multiples of 3 are 3 and 6.
+ 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
Total number of outcomes = 4 × 4 = 16
Highlight the cells where the total is a multiple of 3. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
Number of highlighted cells = 5
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is a multiple of 3) = 516

+	1	2	3	4
1	2	3	4	5
2	3	4	5	6
3	4	5	6	7
4	5	6	7	8

+	1	2	3	4
1	2	3	4	5
2	3	4	5	6
3	4	5	6	7
4	5	6	7	8

Check. The other 11 outcomes do not fit the event, so P(the event does not happen) = 1116. Then 516 + 1116 = 1616 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Done

23.5

Independent events: the AND rule and the OR rule

Edexcel 1MA1: P8, P6
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Independent events: the AND rule and the OR ruleComing soon

Two events are independent if the result of one has no effect on the result of the other. Tossing a coin and rolling a die are independent: whatever the coin shows, the die is still fair. Rain on two different, far-apart days is usually treated as independent too. In real life we assume independence only when it is reasonable.

The AND rule. To find the probability that both of two independent events happen, multiply the probabilities: P(A and B) = P(A) × P(B). Why? Look at the table of the 12 outcomes when a coin is tossed and a die is rolled. Only one cell, H6, is “heads and a 6”, so the probability is 1/12. And 1/2 × 1/6 = 1/12. Half of all the outcomes have heads, and one sixth of those have a 6, so we take one sixth of one half, which is multiplying.

coin \ die	1	2	3	4	5	6
H	H1	H2	H3	H4	H5	H6
T	T1	T2	T3	T4	T5	T6

Coin and die: 12 equally likely outcomes. Only H6 gives “heads and a 6”, so P = 1/12 = 1/2 × 1/6.

The OR rule. Two events are mutually exclusive if they cannot happen at the same time (a die cannot show 2 and 5 on the same roll). For such events, the probability that either happens is found by adding: P(A or B) = P(A) + P(B). For example P(roll 2 or 5) = 1/6 + 1/6 = 2/6 = 1/3.

Do not mix them up. “And” (both things, one after the other) means multiply, and the answer is smaller. “Or” (one thing or the other, in the same trial) means add, and the answer is bigger. The AND rule needs independent events. The OR rule needs mutually exclusive events.

At least one. “At least one 6 in two rolls” means one 6, or two sixes. That is lots of cases, so use the opposite instead: P(at least one) = 1 − P(none). P(no 6 on one roll) = 5/6, so P(no 6 in two rolls) = 5/6 × 5/6 = 25/36, and P(at least one 6) = 1 − 25/36 = 11/36. Notice that this is not 1/6 + 1/6 = 1/3: that adds events that are not mutually exclusive (both rolls could be sixes).

Key ideas

Independent events do not affect each other. The AND rule: P(A and B) = P(A) × P(B).
Mutually exclusive events cannot happen together. The OR rule: P(A or B) = P(A) + P(B).
“And” means multiply (smaller answer). “Or” means add (bigger answer).
P(at least one) = 1 − P(none).
We should say what we assume: for example, that the coin is fair and the events are independent.

Worked example 23.5.1

A fair coin is tossed and a fair die is rolled. Find the probability of getting heads and a 6.

Key words. Two events are independent if one happening does not change the probability of the other. Mutually exclusive events cannot happen at the same time. For independent events, P(A and B) = P(A) × P(B).

Method. For independent events, “A and B” means we multiply their probabilities: P(A and B) = P(A) × P(B).

Check that the events are independent. The outcome of one does not change the chances for the other. So we can multiply their probabilities.
Independent events
Write down the probability of each event. We need each probability on its own.
P(heads) = 12
P(a 6) = 16
Multiply the probabilities. For independent events, P(A and B) = P(A) × P(B). Multiply the tops together and the bottoms together.
12 × 16 = 112

Check. Both things have to happen, so the answer must be smaller than each single probability: 112 is smaller than 16. ✓

Watch out. Multiply for “and”. Do not add the probabilities.

Worked example 23.5.2

The probability of rain on Saturday is 0.3 and the probability of rain on Sunday is 0.4. Assume the two days are independent. Find the probability that it rains on both days.

Method. For independent events, “A and B” means we multiply their probabilities: P(A and B) = P(A) × P(B).

Check that the events are independent. Rain on one day does not change the chance of rain on the other (we assume this). So we can multiply their probabilities.
Independent events
Write down the probability of each event. We need each probability on its own.
P(rain on Saturday) = 0.3
P(rain on Sunday) = 0.4
Multiply the probabilities. For independent events, P(A and B) = P(A) × P(B). Multiply as if there were no decimal points, then count the decimal places.
0.3 × 0.4
3 × 4 = 12, and there are 2 decimal places altogether
= 0.12

Check. Both things have to happen, so the answer must be smaller than each single probability: 0.12 is smaller than 0.3. ✓

Watch out. Multiply for “and”. Do not add the probabilities.

Worked example 23.5.3

A bag has 10 counters: 3 red, 2 blue and 5 green. One counter is picked at random. Find the probability that it is red or blue.

Method. For mutually exclusive events, “A or B” means we add their probabilities: P(A or B) = P(A) + P(B).

Check that the events are mutually exclusive. One counter cannot be both red and blue. So we can add their probabilities.
Mutually exclusive events
Write down the probability of each event. We need each probability on its own.
P(red) = 310
P(blue) = 15
Add the probabilities. For mutually exclusive events, P(A or B) = P(A) + P(B).
310 + 15 = 310 + 210 = 510 = 12

Check. The opposite event has probability 1 − 12 = 12, and the two add to 1. ✓

Watch out. Add for “or”, but only when the events cannot both happen.

Worked example 23.5.4

A fair die is rolled twice. Find the probability of getting at least one 6.

Method. P(at least one) = 1 − P(none). Find the probability of none by multiplying, then subtract from 1.

Use the opposite event. “At least one 6” is the opposite of “no 6 every time”. It is easier to find the opposite and subtract from 1.
P(at least one) = 1 − P(none)
Find the probability of the event not happening on one trial. Use P(not A) = 1 − P(A).
P(not 6) = 1 − 16 = 56
Multiply for all 2 trials. The trials are independent, so multiply the 2 probabilities.
P(none) = 56 × 56 = 2536
Subtract from 1. P(at least one) = 1 − P(none).
3636 − 2536 = 1136

Check. P(none) + P(at least one) = 2536 + 1136 = 1. ✓

Watch out. Do not add the probabilities of each trial. For “at least one” always think “1 minus none”.

Worked example 23.5.5

Three independent tests are passed with probabilities 0.8, 0.9 and 0.75. Find (a) the probability that a student passes all three tests, (b) the probability that a student fails at least one test.

Method. For “all three” multiply the three probabilities. “Fails at least one” is the opposite of “passes all three”.

Check the events are independent. The tests do not affect each other, so we can multiply.
Independent
Multiply the three probabilities. P(all three) = P(1st) × P(2nd) × P(3rd).
0.8 × 0.9 = 0.72
0.72 × 0.75 = 0.54
Find the opposite event. Fails at least one means NOT passing all three.
P(fails at least one) = 1 − 0.54 = 0.46

Check. Failing the first test alone has probability 0.2, which is less than 0.46, and 0.46 is bigger as it includes more ways to fail. ✓

Watch out. “At least one” is almost always quicker as 1 minus the opposite.

Watch out

Do not add when you should multiply. The probability of heads and a 6 is 1/2 × 1/6 = 1/12, not 1/2 + 1/6. And do not use the OR rule on events that can both happen: “a 6 on the first roll or a 6 on the second roll” is not 1/6 + 1/6, because both can be sixes.

23.6

Tree diagrams: with and without replacement

Edexcel 1MA1: P8, P6
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Tree diagrams: with and without replacementComing soon

A tree diagram is a picture of what can happen in two or more events, one after the other. Each stage has its own set of branches. On each branch we write the probability of that result, and at the end of each complete route we write the combined outcome. Here is a bag with 3 red and 2 blue counters. One counter is picked, its colour is noted and it is put back. Then a second counter is picked.

Tree diagram for two picks with replacement from 3 red (R) and 2 blue (B) counters. The probabilities on the second branches are the same as on the first.

Rules for tree diagrams. (1) The probabilities on the branches leaving any one point add up to 1, because one of those things must happen. (2) To find the probability of a complete route, multiply along the branches (the “and” rule: first thing and second thing). (3) To find the probability of an event that can happen by more than one route, add the probabilities of those routes (the “or” rule: this route or that route). (4) The probabilities on all the complete routes add up to 1, which is a very good check.

Without replacement. What if the first counter is not put back? Then there are only 4 counters left for the second pick, and the numbers on the second branches change. If the first counter was red, 2 red and 2 blue are left, so the second branches are 2/4 and 2/4. If the first was blue, 3 red and 1 blue are left, so they are 3/4 and 1/4. The two picks are now dependent: the first result changes the probabilities for the second. This is the key difference: with replacement the branches are the same each time; without replacement the bottom number falls by one and the top number for the colour just taken falls by one.

The same bag, but the first counter is not replaced. The second-stage fractions have changed.

Not just bags. Trees work for any events: wet or dry days, a free throw that is scored or missed, a test that is passed or failed. If the probabilities on the second stage depend on the first result, draw different second branches for each first branch. And for “at least one”, remember the shortcut: find the probability of the opposite (none) and subtract from 1.

Key ideas

Branch probabilities from one point add up to 1.
Multiply along the branches (and). Add the routes you want (or).
The probabilities of all the complete routes add up to 1.
With replacement: the same branches at each stage (independent). Without replacement: the numbers change (dependent).
For “at least one”, use 1 − P(none).

Worked example 23.6.1

A bag has 3 red and 2 blue counters. A counter is picked, replaced, and then a second counter is picked. Find the probability that both counters are red.

Key words. A tree diagram shows every outcome of two or more events in order. The probabilities on the branches from one point always add to 1. Multiply along a route (“and”) and add the routes you want (“or”).

Method. Find the routes through the tree that give the event, multiply the probabilities along each route, then add the routes together.

Draw the tree diagram. Write each probability on its branch. The branches from one point must add up to 1. The ends of the branches show every possible outcome.
Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.
red then red
Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
red then red: 35 × 35 = 925

Check. The routes not used add up to 1625, and 925 + 1625 = 1. ✓

Watch out. Multiply along the branches (and), then add between the routes that give the event (or). Do not add along a route.

Worked example 23.6.2

Use the same bag and the same experiment (counter replaced). Find the probability that the two counters are different colours.

Method. Find the routes through the tree that give the event, multiply the probabilities along each route, then add the routes together.

Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.
red then blue
blue then red
Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
red then blue: 35 × 25 = 625
blue then red: 25 × 35 = 625
Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.
625 + 625 = 1225

Check. The routes not used add up to 1325, and 1225 + 1325 = 1. ✓

Watch out. Multiply along the branches (and), then add between the routes that give the event (or). Do not add along a route.

Worked example 23.6.3

A bag has 4 red and 6 blue counters. Two counters are picked at random without replacement. Find the probability that they are the same colour.

Method. Find the routes through the tree that give the event, multiply the probabilities along each route, then add the routes together.

Draw the tree diagram. Write each probability on its branch. The branches from one point must add up to 1. The ends of the branches show every possible outcome.
Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.
red then red
blue then blue
Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
red then red: 410 × 39 = 1290 = 215
blue then blue: 610 × 59 = 3090 = 13
Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.
215 + 13 = 215 + 515 = 715

Check. The routes not used add up to 815, and 715 + 815 = 1. ✓

Watch out. Multiply along the branches (and), then add between the routes that give the event (or). Do not add along a route.

Worked example 23.6.4

The probability that it rains on any day is 0.3. Two days are independent. Find the probability that it rains on exactly one of the two days.

Method. Find the routes through the tree that give the event, multiply the probabilities along each route, then add the routes together.

Draw the tree diagram. Write each probability on its branch. The branches from one point must add up to 1. The ends of the branches show every possible outcome.
Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.
rain then dry
dry then rain
Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
rain then dry: 0.3 × 0.7 = 0.21
dry then rain: 0.7 × 0.3 = 0.21
Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.
0.21 + 0.21 = 0.42

Check. The routes not used add up to 0.58, and 0.42 + 0.58 = 1. ✓

Watch out. Multiply along the branches (and), then add between the routes that give the event (or). Do not add along a route.

Worked example 23.6.5

A bag has 5 red and 3 blue counters. Two counters are picked without replacement. Find the probability that at least one is blue.

Method. Use P(event) = 1 − P(opposite event). Find the routes of the opposite event on the tree, multiply along them, add them, and subtract from 1.

Think about the opposite event. It is easier to find the probability that the event does not happen, then subtract from 1. The routes that do not give "at least one is blue" are:
red then red
Multiply along each of these routes. Each route is "first thing and second thing", so we multiply the branch probabilities.
red then red: 58 × 47 = 2056 = 514
Subtract from 1. The event and its opposite together make probability 1.
1 − 514 = 914

Check. Directly: the routes that give the event add up to 914, which agrees. ✓

Watch out. Multiply along the branches (and), then add between the routes that give the event (or). Do not add along a route.

Worked example 23.6.6

A bag has 6 red and 4 blue counters. Three counters are picked without replacement. Find the probability that exactly two are red.

Method. Find the routes through the tree that give the event, multiply the probabilities along each route, then add the routes together.

Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.
red then red then blue
red then blue then red
blue then red then red
Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
red then red then blue: 610 × 59 × 48 = 120720 = 16
red then blue then red: 610 × 49 × 58 = 120720 = 16
blue then red then red: 410 × 69 × 58 = 120720 = 16
Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.
16 + 16 + 16 = 36 = 12

Check. The routes not used add up to 12, and 12 + 12 = 1. ✓

Watch out. Multiply along the branches (and), then add between the routes that give the event (or). Do not add along a route.

Worked example 23.6.7

The probability that Kim is late on Monday is 0.1. If she is late on Monday, the probability she is late on Tuesday is 0.4. If she is on time on Monday, the probability she is late on Tuesday is 0.05. Find the probability that she is late on exactly one of the two days.

Method. Find the routes through the tree that give the event, multiply the probabilities along each route, then add the routes together.

Draw the tree diagram. Write each probability on its branch. The branches from one point must add up to 1. The ends of the branches show every possible outcome.
Find the routes through the tree that give the event. Each complete route is one possible outcome. Pick out the ones that match.
late then on time
on time then late
Multiply along each of these routes. Each route means “first thing and then second thing”, so we multiply the branch probabilities.
late then on time: 0.1 × 0.6 = 0.06
on time then late: 0.9 × 0.05 = 0.045
Add the routes. The routes are different ways for the event to happen (this or that), and they cannot happen together, so we add.
0.06 + 0.045 = 0.105

Check. The routes not used add up to 0.895, and 0.105 + 0.895 = 1. ✓

Watch out. Multiply along the branches (and), then add between the routes that give the event (or). Do not add along a route.

Watch out

With replacement the second-stage probabilities stay the same. Without replacement they change: the denominator goes down by one. Also check that your final route probabilities add up to 1, and do not add along a branch or multiply between different routes.

23.7

Sets and Venn diagrams

Edexcel 1MA1: P6, P8
Grade 3-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Sets and Venn diagramsComing soon

A set is a collection of items called elements. We write a set in curly brackets: A = {2, 4, 6, 8}. The symbol ∈ means “is an element of”, so 4 ∈ A, and ∉ means “is not an element of”, so 5 ∉ A. The universal set, written ξ, is everything we are talking about, for example all the whole numbers from 1 to 10. The empty set {} or ∅ has no elements. n(A) means the number of elements of A, so n(A) = 4 above. If every element of B is also in A we write B ⊂ A (B is a subset of A).

Combining sets. Let A = {even numbers} and B = {multiples of 3} for ξ = {1, 2, ..., 12}, so A = {2, 4, 6, 8, 10, 12} and B = {3, 6, 9, 12}. A ∩ B (“A intersection B”) contains the elements in both A and B: {6, 12}. A ∪ B (“A union B”) contains the elements in A or B or both: {2, 3, 4, 6, 8, 9, 10, 12}. A′ (“A complement” or “not A”) contains everything in ξ that is not in A: {1, 3, 5, 7, 9, 11}. A handy way to remember: ∩ looks like the letter A for “And”, and ∪ is like a cup for “or”.

A Venn diagram for the sets above. The overlap holds A ∩ B = {6, 12}. The rectangle is ξ.

Venn diagrams. A Venn diagram shows sets as circles inside a rectangle that stands for ξ. The overlap of two circles is A ∩ B. Anything inside the rectangle but outside every circle is in none of the sets. We can write either the elements or the number of elements in each region. When counting, fill in the middle first and then work outwards. Be careful: a number given for a circle (such as “18 play football”) includes the overlap.

Probability from a Venn diagram. If one item is chosen at random from everything in ξ, then all the items are equally likely, so P(event) = n(event) ÷ n(ξ). For example, with 30 students in total and 13 of them in A only, P(A only) = 13/30. The total on the bottom is everything in ξ, including the numbers outside the circles.

Higher: the addition formula. The numbers in A and in B both count the overlap, so P(A) + P(B) counts the overlap twice. Subtract it once: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Also P(A′) = 1 − P(A), and P(neither) = 1 − P(A ∪ B).

Key ideas

ξ is the universal set. ∈ means “is in”. n(A) is the number of elements in A.
A ∩ B = in both A and B. A ∪ B = in A or B or both. A′ = not in A.
Fill in a Venn diagram from the middle outwards. Subtract the overlap from each circle total.
P(event) = n(event) ÷ n(ξ). The bottom number is everything in ξ.
Higher: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Worked example 23.7.1

ξ = {1, 2, 3, ..., 12}, A = {even numbers} and B = {multiples of 3}. List the elements of (a) A ∩ B, (b) A ∪ B, (c) A′.

Key words. A set is a collection of items. ξ is the universal set (everything we are considering). A ∩ B (“A intersection B”) is what is in both A and B. A ∪ B (“A union B”) is what is in A or B or both. A′ is what is not in A. n(A) is the number of items in A.

Method. Write out both sets, then use the meaning of ∩ (both), ∪ (either) and ′ (not in) to pick out the elements.

Write out the sets. List every element.
ξ = {1, 2, ..., 12}
A = {2, 4, 6, 8, 10, 12}
B = {3, 6, 9, 12}
A ∩ B: in both sets. Find the numbers that appear in both lists.
A ∩ B = {6, 12}
A ∪ B: in A or B (or both). Combine the lists, writing each number once.
A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12}
A′: not in A. Everything in ξ that is not even, so the odd numbers.
A′ = {1, 3, 5, 7, 9, 11}

Check. n(A) + n(A′) = 6 + 6 = 12 = n(ξ), as it should. ✓

Watch out. Do not write a repeated element twice in A ∪ B.

Worked example 23.7.2

30 students were asked about sport. 18 play football (F), 12 swim (S) and 5 do both. Draw a Venn diagram to show this information.

Method. Fill in the overlap first, then work outwards: subtract the overlap from each circle total, and finally find the number outside both circles from the total.

Start with the overlap. The people in both sets go in the middle where the circles overlap. This is F ∩ S.
Both football and swimming: 5
Work out F only. The whole of circle F has 18 people, but 5 of them are also in S. They are already in the middle, so take them away.
18 − 5 = 13
Work out S only. In the same way, circle S has 12 people, 5 of whom are in the middle.
12 − 5 = 7
Work out the number in neither set. The numbers in the circles add up to 13 + 5 + 7 = 25. The rest of the 30 are outside both circles.
30 − 25 = 5
Draw the Venn diagram. Write each number in its region. The numbers must add up to the total.

Check. Total of all regions: 13 + 5 + 7 + 5 = 30. Circle F: 13 + 5 = 18. Circle S: 7 + 5 = 12. ✓

Watch out. The number given for a circle includes the overlap, so you must subtract the overlap to find “only”.

Worked example 23.7.3

The Venn diagram shows the number of students who have a pet cat (A) and a pet dog (B). Find n(A ∪ B) and n(A′).

Method. Work out which regions belong to each set, then add the numbers in those regions.

Find the regions of A ∪ B. ∪ means “or”: the items in either set, or both.
in A only: 8
in B only: 7
in both A and B: 5
Add them. Count each region once.
8 + 5 + 7 = 20, so n(A ∪ B) = 20
Find the regions of A′. The dash ′ means “not in”: everything outside that set, including outside the circles.
in none of the sets (outside the circles): 4
in B only: 7
Add them. A′ is everything outside circle A.
7 + 4 = 11, so n(A′) = 11

Check. n(A) = 13, and n(A) + n(A′) = 13 + 11 = 24 = n(ξ). ✓

Watch out. A′ includes the number outside both circles.

Worked example 23.7.4

Use the Venn diagram above. A student is chosen at random from the 24 students. Find the probability that the student has a dog but not a cat, that is P(A′ ∩ B).

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Find the total number of items. Add the numbers in every region, including the ones outside the circles. This goes on the bottom.
4 + 8 + 7 + 5 = 24
Find the number of items in the event. ∩ means “and”: only the items in both sets. The dash ′ means “not in”: everything outside that set, including outside the circles.
in B only: 7
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(dog but not cat) = 724

Check. The other 17 outcomes do not fit the event, so P(the event does not happen) = 1724. Then 724 + 1724 = 2424 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. The total on the bottom is everything in the universal set ξ, not just the numbers inside the circles.

Worked example 23.7.5

40 students were asked which of three clubs they belong to: art (A), band (B) and chess (C). 20 are in A, 18 in B and 14 in C. 8 are in both A and B, 6 in both A and C, 5 in both B and C, and 3 are in all three. Draw a Venn diagram.

Method. With three sets, work from the middle outwards: the triple overlap first, then the pairs, then the single circles, and finally the outside.

Start in the very middle. The people in all three sets go where all three circles overlap.
A ∩ B ∩ C: 3
Work out the regions where exactly two circles overlap. Each pair total includes the middle region, so subtract the middle number.
A and B only: 8 − 3 = 5
A and C only: 6 − 3 = 3
B and C only: 5 − 3 = 2
Work out the regions that are in one circle only. Each circle total includes the middle and the two-way overlaps, so subtract all of them.
A only: 20 − 5 − 3 − 3 = 9
B only: 18 − 5 − 2 − 3 = 8
C only: 14 − 3 − 2 − 3 = 6
Work out the number in none of the sets. Add all seven numbers inside the circles and subtract from the total.
9 + 8 + 6 + 5 + 3 + 2 + 3 = 36, so 40 − 36 = 4
Draw the Venn diagram. Write each number in its region.

Check. All the regions add up to 40, the total. Circle A: 9 + 5 + 3 + 3 = 20. ✓

Watch out. Always start from the middle. Subtracting in the wrong order gives negative numbers.

Worked example 23.7.6

P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2. Find P(A ∪ B) and the probability that neither A nor B happens.

Method. Use the addition formula P(A ∪ B) = P(A) + P(B) − P(A ∩ B), then use the opposite event for “neither”.

Write the formula. Adding P(A) and P(B) counts the overlap twice, so subtract it once.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Put in the numbers. Add first, then subtract.
0.5 + 0.4 − 0.2 = 0.9 − 0.2 = 0.7
Find “neither”. Neither is the opposite of A or B (or both).
P(neither) = 1 − 0.7 = 0.3

Check. A only is 0.5 − 0.2 = 0.3, B only is 0.4 − 0.2 = 0.2, both is 0.2, neither is 0.3. Total: 0.3 + 0.2 + 0.2 + 0.3 = 1. ✓

Watch out. If you forget to subtract P(A ∩ B) your total can exceed what is possible.

Watch out

In a Venn diagram the number given for a circle includes the overlap. If 18 play football and 5 play football and swim, only 13 play football only. Also remember that A′ includes the people outside both circles, and for a probability the bottom number is everyone in ξ.

Exercise 23J Probability from Venn diagrams

Non-calculator

FThe Venn diagram shows 24 students. A student is chosen at random. Find the probability that the student is in...

(a)A

(b)B

(c)A ∩ B

(d)A ∪ B

(e)A′

(f)A ∩ B′

Done

PS30 students were asked about sport. 18 play football (F), 12 swim (S) and 5 do both. A student is chosen at random. Find the probability that the student...

(a)plays football but does not swim

(b)plays neither sport

(c)plays at least one of the sports

(d)plays exactly one of the sports

Done

PS40 people were asked about pets. 25 own a cat (C), 17 own a dog (D) and 6 own both. A person is chosen at random. Find the probability that the person owns...

(a)a cat only

(b)a dog

(c)no cat

(d)neither pet

Done

FThe numbers of students in a year group who walk (W) and cycle (C) to school are shown. Find the probability that a student chosen at random walks or cycles.

Done

PSIn a Venn diagram, n(A) = 12, n(B) = 9, n(A ∩ B) = 4 and n(ξ) = 30. Find the probability that a randomly chosen element is in neither A nor B.

Done

RIn a group of 24 students, 13 are in set A and 12 are in set B, and 5 are in both. Ali says: “P(A) = 1324 and P(B) = 1224, so P(A or B) = 2524.” Explain why Ali must be wrong.

Done

PSThe Venn diagram shows how 40 students use three types of transport to school: bus (A), bike (B) and car (C). A student is chosen at random. Find the probability that the student uses...

(a)all three types of transport

(b)bus and bike only

(c)exactly one type of transport

(d)no transport from these three

Done

PSP(A) = 0.6, P(B) = 0.3 and P(A ∩ B) = 0.1. Work out...

(a)P(A ∪ B)

(b)P(neither)

(c)P(A ∩ B′)

Done

RP(A) = 0.45, P(B) = 0.35 and P(A ∪ B) = 0.65. Find P(A ∩ B). Are A and B mutually exclusive? Give a reason.

Done

23.8

Two-way tables and frequency trees

Edexcel 1MA1: P1, P9, P6
Grade 2-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Two-way tables and frequency treesComing soon

When people (or items) are sorted by two things at once, such as year group and how they travel, a two-way table keeps the information tidy. One category goes down the side and the other goes across the top. Each cell shows how many are in both categories, and the totals are written at the ends of the rows and the bottom of the columns.

	Walk	Bus	Total
Boys	12	6	18
Girls	9	8	17
Total	21	14	35

A two-way table for 35 students. 12 boys walk. 6 boys take the bus. The 18 boys in total are the row total.

Completing a table. Every row adds up to its row total and every column adds up to its column total. So look for a row or column with only one number missing and work it out by adding (if the total is missing) or by subtracting the known numbers from the total (if a cell is missing). Then repeat. Reading probabilities. If one person is chosen at random from the whole table, P(event) = number in the event ÷ grand total. For example P(girl) = 17/35 and P(boy who walks) = 12/35.

Frequency trees. A frequency tree looks like a tree diagram but shows numbers of people instead of probabilities. The first box holds the total. Each branch splits the number into groups, and each group splits again. The numbers coming out of a box must add up to the number in the box. If you are given a fraction or a percentage, multiply the number in the box by it. A frequency tree turns probabilities into expected numbers, so it is a friendly way to see what a tree diagram means: the probability of a route is its number divided by the grand total.

A frequency tree for 35 students: first girls or boys, then left-handed or right-handed.

Key ideas

A two-way table has row totals and column totals. Every row and column adds up to its total.
To complete it, find a row or column with only one missing number.
P(event) from a table = number in the event ÷ total in the group chosen from.
A frequency tree shows numbers, not probabilities. Numbers leaving a box add up to the number in it.
Take a fraction (or percentage) of the number in the previous box, not of the original total.

Worked example 23.8.1

35 students were asked how they travel to school. Some of the information is shown. Complete the two-way table.

	Walk	Bus	Total
Boys	12	?	18
Girls	?	8	?
Total	21	?	35

Method. In a two-way table every row and every column adds up to its total. Find a row or column with only one missing number, work it out, then repeat.

Find the missing number in the Boys row, Bus column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
18 − 12 = 6
Find the missing number in the Total row, Bus column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
35 − 21 = 14
Find the missing number in the Girls row, Walk column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
21 − 12 = 9
Find the missing number in the Girls row, Total column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
35 − 18 = 17
Write the completed table. Check that every row and every column adds up to its total.
Walk Bus Total
Boys 12 6 18
Girls 9 8 17
Total 21 14 35

Check. Check the grand total: the row totals add to 35 and the column totals add to 35, and both equal 35. ✓

Watch out. Start with a row or column that has only one missing number.

Worked example 23.8.2

Use the completed table above. One student is chosen at random. Find the probability that the student is a girl who walks, and the probability that the student takes the bus.

Method. Read the numbers from the completed table. The bottom of each fraction is the grand total, 35.

Find the grand total. It is the bottom right number.
35 students
Girls who walk. Where the Girls row meets the Walk column.
9, so P(girl who walks) = 935
Students who take the bus. Use the Bus column total.
14, so P(bus) = 1435 = 25

Check. P(walk) + P(bus) = 21/35 + 14/35 = 1. ✓

Watch out. Use the column total, not just one cell, for “the student takes the bus”.

Worked example 23.8.3

In a survey of 80 students, 35 are girls. 18 of the girls and 14 of the boys are left-handed. Draw a frequency tree.

Method. Work out each number on the tree by finding a fraction of the number in the box before it.

Start with the total. The first box of a frequency tree holds everyone (or everything) in the survey.
Students: 80
Work out the first branches. Multiply the total by the fraction (or use the number given). The numbers on the first branches must add up to the total.
Girls: 35 of 80 = 80 ÷ 5 × 3 = 48
Boys: 25 of 80 = 80 ÷ 5 × 2 = 32
Check: 48 + 32 = 80
Work out the second branches. Each second branch is a fraction of the number on the branch before it, not of the total.
Girls, Left: 18 of 48 = 6
Girls, Right: 78 of 48 = 42
Boys, Left: 14 of 32 = 8
Boys, Right: 34 of 32 = 24
Draw the frequency tree. Write each number in its box. The numbers coming out of a box must add up to the number in it.

Check. The numbers from each box add up to the number in that box. ✓

Watch out. Take the fraction of the number on the previous branch, not of the original total.

Worked example 23.8.4

Use the frequency tree in the previous example. A student is chosen at random from the 80. Find the probability that the student is a left-handed girl, and the probability that the student is left-handed.

Method. Read the numbers off the frequency tree. The bottom is 80.

Left-handed girls. Follow the Girls branch then Left.
6 students, so P = 680 = 340
All left-handed students. Add the left-handed girls and the left-handed boys.
6 + 8 = 14
Write the probability. Left-handed students over all students.
1480 = 740

Check. P(right-handed) = 3340, and the two add to 1. ✓

Watch out. “Left-handed” includes both the girls and the boys.

Watch out

In a two-way table, take care which total goes on the bottom. For “a girl who walks” use the grand total (35). In a frequency tree, a second-stage fraction is a fraction of the number on its own branch, not of the original total: 18 of the 48 girls is 6, not 18 of 80.

Exercise 23K Two-way tables

Non-calculator

FThe two-way table shows how 40 students voted for a school trip.

	Zoo	Museum	Total
Boys	14	6	20
Girls	9	11	20
Total	23	17	40

(a)How many girls voted for the zoo?

(b)How many students voted for the museum?

(c)How many boys are there?

Done

FWork out the missing numbers in each two-way table.

(a)A club has 50 members. Some of the information is shown. Complete the table.

	Swim	Run	Total
Adults	?	11	30
Children	8	?	?
Total	?	23	50

Method. In a two-way table every row and every column adds up to its total. Find a row or column with only one missing number, work it out, then repeat.

Find the missing number in the Adults row, Swim column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
30 − 11 = 19
Find the missing number in the Total row, Swim column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
50 − 23 = 27
Find the missing number in the Children row, Run column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
23 − 11 = 12
Find the missing number in the Children row, Total column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
50 − 30 = 20
Write the completed table. Check that every row and every column adds up to its total.
Swim Run Total
Adults 19 11 30
Children 8 12 20
Total 27 23 50

Check. Check the grand total: the row totals add to 50 and the column totals add to 50, and both equal 50. ✓

Watch out. Start with a row or column that has only one missing number.

(b)60 people were asked whether they prefer tea or coffee. Complete the table.

	Tea	Coffee	Total
Men	16	?	35
Women	?	11	?
Total	?	?	60

Method. In a two-way table every row and every column adds up to its total. Find a row or column with only one missing number, work it out, then repeat.

Find the missing number in the Men row, Coffee column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
35 − 16 = 19
Find the missing number in the Total row, Coffee column. The total at the bottom of a column is the sum of the numbers in the column.
19 + 11 = 30
Find the missing number in the Women row, Total column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
60 − 35 = 25
Find the missing number in the Women row, Tea column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
25 − 11 = 14
Find the missing number in the Total row, Tea column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
60 − 30 = 30
Write the completed table. Check that every row and every column adds up to its total.
Tea Coffee Total
Men 16 19 35
Women 14 11 25
Total 30 30 60

Check. Check the grand total: the row totals add to 60 and the column totals add to 60, and both equal 60. ✓

Watch out. Start with a row or column that has only one missing number.

(c)A shop sells shirts in three sizes in two colours. Complete the table.

	Blue	White	Total
Small	7	?	18
Medium	?	15	27
Large	9	?	?
Total	?	?	70

Method. In a two-way table every row and every column adds up to its total. Find a row or column with only one missing number, work it out, then repeat.

Find the missing number in the Small row, White column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
18 − 7 = 11
Find the missing number in the Medium row, Blue column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
27 − 15 = 12
Find the missing number in the Total row, Blue column. The total at the bottom of a column is the sum of the numbers in the column.
7 + 12 + 9 = 28
Find the missing number in the Large row, Total column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
70 − 18 − 27 = 25
Find the missing number in the Large row, White column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
25 − 9 = 16
Find the missing number in the Total row, White column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
70 − 28 = 42
Write the completed table. Check that every row and every column adds up to its total.
Blue White Total
Small 7 11 18
Medium 12 15 27
Large 9 16 25
Total 28 42 70

Check. Check the grand total: the row totals add to 70 and the column totals add to 70, and both equal 70. ✓

Watch out. Start with a row or column that has only one missing number.

Done

FThe two-way table shows the type of ticket sold for a concert. One ticket is chosen at random from the 50. Find the probability that it is...

	Saturday	Sunday	Total
Adult	15	10	25
Child	5	20	25
Total	20	30	50

(a)an adult ticket

(b)for Saturday

(c)a child ticket for Sunday

(d)not an adult ticket

Done

RIn a year group of 45 students, some are in the school band and some are not. The two-way table shows the information. Phil says: “Boys are more likely to be in the band than girls.” Is Phil right? Show your working.

	In band	Not in band	Total
Boys	8	12	20
Girls	6	19	25
Total	14	31	45

Done

PS120 customers at a cafe were asked whether they had a hot drink or a cold drink and whether they had a snack. The table shows the results. A customer is chosen at random. Find the probability that the customer...

	Snack	No snack	Total
Hot drink	27	18	45
Cold drink	21	54	75
Total	48	72	120

(a)had a hot drink and a snack

(b)had a snack

(c)had a cold drink but no snack

Done

PSThe table shows 600 tickets sold for a school play. The tickets were Adult or Child, for Friday or Saturday. Complete the table and then estimate how many of the next 150 tickets sold are Child tickets for Saturday.

	Friday	Saturday	Total
Adult	150	?	330
Child	?	120	?
Total	?	?	600

Done

RIn a table the numbers in the cells are 6, 9, 14 and 11 and the table has no totals written in. Explain how to find the grand total and what the probability of choosing the second cell is.

Done

PSA doctor records whether 50 patients are Under 40 or 40 and over, and whether they have had a flu jab. A patient is chosen at random. The table shows the results. Write the answers as fractions in their simplest forms.

	Jab	No jab	Total
Under 40	3	22	25
40 and over	11	14	25
Total	14	36	50

(a)Find the probability that the patient is under 40 and had a jab

(b)Find the probability that the patient has not had a jab

Done

PSA two-way table shows 80 students. The table has 4 cells and totals. The row totals are 50 and 30 and the column totals are 45 and 35. The number in the top-left cell is 28. Complete the table.

Method. In a two-way table every row and every column adds up to its total. Find a row or column with only one missing number, work it out, then repeat.

Find the missing number in the Row 1 row, Column 2 column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
50 − 28 = 22
Find the missing number in the Row 2 row, Column 1 column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
45 − 28 = 17
Find the missing number in the Row 2 row, Column 2 column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
35 − 22 = 13
Write the completed table. Check that every row and every column adds up to its total.
Column 1 Column 2 Total
Row 1 28 22 50
Row 2 17 13 30
Total 45 35 80

Check. Check the grand total: the row totals add to 80 and the column totals add to 80, and both equal 80. ✓

Watch out. Start with a row or column that has only one missing number.

Done

Exercise 23L Frequency trees

Calculator allowed

PSAt a festival 200 people were asked about their diet. 35 of them are adults and the rest are children. 14 of the adults and 12 of the children are vegetarian. Draw a frequency tree and find how many people are vegetarian.

Done

PSIn a survey of 150 people, 60% own a smartphone. Of the people who own a smartphone, 40% use a banking app. Draw a frequency tree and find how many of the 150 use a banking app.

Done

PSA factory makes 5000 items. 3% are faulty. A test finds 90% of faulty items and wrongly flags 2% of the good items. Complete a frequency tree and find how many items the test flags in total.

Done

23.9

Conditional probability and probability with algebra

Edexcel 1MA1: P9, P8
Grade 6-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Conditional probability and probability with algebraComing soon

Sometimes we are told something has already happened, and we are asked how likely something else is given that. This is conditional probability. It is written P(A | B), read “the probability of A given B”. For example, “P(girl | walks)” means: choose a student who walks, and ask how likely it is that this student is a girl.

The word given shrinks the sample space. We are no longer choosing from everyone. We are choosing only from the group we are told about. So the bottom of the probability fraction is the size of that group, not the grand total.

	Walk	Bus	Total
Boys	12	6	18
Girls	9	8	17
Total	21	14	35

P(girl | walks): look only at the Walk column. 9 of the 21 students who walk are girls, so P = 9/21 = 3/7.

From a two-way table: restrict to the row or column named after “given”. From a Venn diagram: P(A | B) = n(A ∩ B) ÷ n(B). From a tree diagram: use the rule P(A | B) = P(A and B) ÷ P(B), where P(A and B) is a route probability and P(B) is found by adding the routes where B happens. On a tree with dependent events the branch for the second stage is already a conditional probability: for example, the branch probability 3/9 on the second stage after a red is P(second red | first red).

Order matters. P(A | B) is usually not the same as P(B | A). The probability that a person who owns a pet is a child is different from the probability that a child owns a pet. Independence. Events A and B are independent exactly when knowing B does not change the probability of A, that is P(A | B) = P(A), which is the same as P(A and B) = P(A) × P(B).

Algebra and probability. Probabilities can contain unknowns. If a bag has x red counters and 5 blue counters, then P(red) = x/(x + 5). If we are told P(red) = 3/5, we form an equation and solve it. We use two facts again and again: all the probabilities of the outcomes add up to 1, and probabilities of picks without replacement multiply with the numbers going down by one.

Key ideas

P(A | B) means the probability of A given that B has happened. “Given” shrinks the sample space to the group B.
Table or Venn: P(A | B) = n(A ∩ B) ÷ n(B). Tree: P(A | B) = P(A and B) ÷ P(B).
P(A | B) is not usually the same as P(B | A).
A and B are independent when P(A | B) = P(A), that is P(A and B) = P(A) × P(B).
With algebra: write each probability as a fraction containing the unknown, make an equation, and solve it. All probabilities add up to 1.

Worked example 23.9.1

The table shows how 35 students travel to school. A student who walks is chosen at random. Find the probability that the student is a girl.

	Walk	Bus	Total
Boys	12	6	18
Girls	9	8	17
Total	21	14	35

Key words. Conditional probability is the probability of an event given that another event has happened. The word “given” shrinks the sample space to just the group we are told about. P(A | B) = n(A ∩ B) ÷ n(B).

Method. For a conditional probability, shrink the sample space to the group we are told about (the “given” group), then work out probability = favourable ÷ size of that group.

Narrow down to the group we are told about: the student walks. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.
Walk column total = 12 + 9 = 21 students
Count how many in that group also have the event: the student is a girl. This is the number we want. It goes on the top.
9 of them are girls, so 9
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the student is a girl | the student walks) = 921
Simplify the fraction. Divide the top and the bottom by their highest common factor, 3. The value does not change but the numbers are smaller.
9 ÷ 3 = 3 and 21 ÷ 3 = 7
921 = 37

Check. The other 12 outcomes do not fit the event, so P(the event does not happen) = 47. Then 921 + 1221 = 2121 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. The bottom of a conditional probability is the size of the group after “given”, not the total of everyone.

Worked example 23.9.2

The Venn diagram shows 24 students: A is the set who study art and B is the set who study music. A student who studies music is chosen at random. Find P(A | B).

Method. For a conditional probability, shrink the sample space to the group we are told about (the “given” group), then work out probability = favourable ÷ size of that group.

Narrow down to the group we are told about: B. The word “given” tells us the choice has already been made from this group only. The size of this group goes on the bottom.
n(B) = 5 + 7 = 12 students study music
Count how many in that group also have the event: A. This is the number we want. It goes on the top.
n(A ∩ B) = 5 students are in both
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(A | B) = 512

Check. The other 7 outcomes do not fit the event, so P(the event does not happen) = 712. Then 512 + 712 = 1212 = 1, as the probabilities of an event happening and not happening must add up to 1. ✓

Watch out. The bottom of a conditional probability is the size of the group after “given”, not the total of everyone.

Worked example 23.9.3

A bag has 4 red and 6 blue counters. Two counters are taken without replacement. Given that the second counter is red, find the probability that the first counter was also red.

Method. P(A | B) = P(A and B) ÷ P(B). Use the tree to find P(B) and P(A and B), then divide.

Write the rule. Conditional probability: the probability of the event given the condition is the probability of both, divided by the probability of the condition.
P(first red | second red) = P(first red and second red)P(second red)
Find P(second red). Find every route on the tree where the condition happens, multiply along each, and add.
red then red: 410 × 39 = 1290 = 215
blue then red: 610 × 49 = 2490 = 415
215 + 415 = 615 = 25
P(second red) = 25
Find P(first red and second red). Of those routes, pick the ones where the event also happens.
red then red: 410 × 39 = 1290 = 215
P(first red and second red) = 215
Divide. Write both probabilities over a common denominator, so we can just divide the numerators.
215 ÷ 615 = 2 ÷ 6 = 13

Check. Think of 6 equally likely cases where the condition holds, of which 2 also have the event: 13. ✓

Watch out. The bottom is the probability of the given condition, not 1 and not the total.

Worked example 23.9.4

A medical test is given to 1000 people. 2% have the condition. The test is positive for 90% of people with the condition and for 5% of people without it. A person has a positive test. Find the probability that they have the condition.

Method. Use expected numbers (a frequency tree). The “given positive” group is everyone with a positive test, so find its size.

Find the number with the condition. 2% of 1000.
0.02 × 1000 = 20 people have the condition, 980 do not.
Find the positive tests in each group. 90% of 20 and 5% of 980.
0.9 × 20 = 18
0.05 × 980 = 49
Find the total number of positive tests. This is the size of the group we are told about (“given positive”).
18 + 49 = 67
Find how many of these have the condition. Positive tests among those with the condition.
18 people
Write the probability. Number with the condition and positive, over all positives.
1867
Show the frequency tree. This summarises all the numbers.

Check. 18 ÷ 67 = 0.269: only about 27%, even though the test looks accurate, because the condition is rare. ✓

Watch out. The bottom is all the positive tests, not all 1000 people and not just the 20 who have the condition.

Worked example 23.9.5

P(A) = 0.4, P(B) = 0.5 and P(A and B) = 0.2. Find P(A | B) and say whether A and B are independent.

Method. Use P(A | B) = P(A and B) ÷ P(B), then compare with P(A).

Write the rule. Conditional probability formula.
P(A | B) = P(A and B) ÷ P(B)
Substitute. Use the given values.
0.2 ÷ 0.5 = 2 ÷ 5 = 0.4
Compare with P(A). If P(A | B) = P(A), knowing B does not change the chance of A.
P(A | B) = 0.4 = P(A)
Conclude.
A and B are independent.

Check. P(A) × P(B) = 0.4 × 0.5 = 0.2 = P(A and B). ✓

Watch out. Independence is tested by P(A | B) = P(A).

Worked example 23.9.6

A bag has 3 red counters and x blue counters. One counter is picked at random. P(blue) = 58. Form an equation and find the value of x.

Method. Write P(blue) as a fraction containing x, put it equal to 5/8, and solve.

Count all the counters. The total is red plus blue.
Total = 3 + x
Write P(blue). Blue counters over all counters.
P(blue) = x3 + x
Form the equation. We are told P(blue) = 5/8.
x3 + x = 58
Cross-multiply. Multiply both sides by 8 and by (3 + x) to clear the fractions.
8x = 5(3 + x)
Expand and solve. Multiply out the bracket, subtract 5x from both sides, then divide by 3.
8x = 15 + 5x
3x = 15
x = 5

Check. With 5 blue and 3 red there are 8 counters and P(blue) = 5/8. ✓

Watch out. Remember that the number of blue counters is also part of the total on the bottom.

Worked example 23.9.7

A bag has 5 red counters and x blue counters. Two counters are picked at random without replacement. The probability that both are red is 29. Show that x² + 9x − 70 = 0 and hence find x.

Method. Write P(both red) as a product of two fractions in terms of x, equate to 2/9, rearrange into a quadratic and solve by factorising.

Write the two probabilities. After the first red counter is taken, there is one fewer red and one fewer counter in total.
P(1st red) = 55 + x, P(2nd red) = 44 + x
Multiply and set equal to 2/9. Both red means first and second.
55 + x × 44 + x = 20(5 + x)(4 + x) = 29
Cross-multiply. Multiply both sides by 9 and by the bottom, then divide by 2.
20 × 9 = 2(5 + x)(4 + x)
90 = (5 + x)(4 + x)
Expand and rearrange. (5 + x)(4 + x) = 20 + 9x + x². Subtract 90 from both sides.
20 + 9x + x² = 90
x² + 9x − 70 = 0
Factorise and solve. Two numbers that multiply to −70 and add to 9 are 14 and −5.
(x + 14)(x − 5) = 0
x = −14 or x = 5
Choose the sensible answer. The number of counters cannot be negative.
x = 5

Check. 5 red and 5 blue: P(both red) = 5/10 × 4/9 = 20/90 = 2/9. ✓

Watch out. Reject the negative solution in context.

Watch out

P(A | B) and P(B | A) are different. And the bottom of a conditional probability is the size of the given group: 9/21 for “a girl, given walks”, not 9/35. With algebra, do not forget that a counter taken without replacement also reduces the total: both the top and the bottom change.

Exercise 23M Conditional probability

Non-calculator

FThe table shows how 35 students travel to school. Find the probability that...

	Walk	Bus	Total
Boys	12	6	18
Girls	9	8	17
Total	21	14	35

(a)the student is a boy, given that the student takes the bus

(b)the student walks to school, given that the student is a girl

(c)the student takes the bus, given that the student is a boy

Done

FThe Venn diagram shows 22 students: A studies Spanish and B studies French. A student is chosen at random. Find...

(a)P(A | B)

(b)P(B | A)

(c)P(A | B′)

Done

FA fair die is rolled. Given that the score is even, find the probability that the score is greater than 3.

Done

FA card is chosen from a standard pack of 52 cards. Given that the card is a heart, find the probability that it is a king.

Done

FA bag has 4 red and 6 blue counters. Two counters are taken at random without replacement. Find...

(a)the probability that the second counter is red, given that the first was red

(b)the probability that the first counter was red, given that the second is red

(c)the probability that the counters are different colours, given that the second is blue

Done

PSThe probability of rain on Monday is 0.3. If it rains on Monday, the probability of rain on Tuesday is 0.6. If it is dry on Monday, the probability of rain on Tuesday is 0.2. Find the probability that...

(a)it rains on both days

(b)it rains on Tuesday

(c)it rained on Monday, given that it rains on Tuesday

Done

PSA test is given to 2000 people. 1% have a disease. The test is positive for 95% of people with the disease and for 4% of people without it. Use a frequency tree to find the probability that a person with a positive test has the disease. Give your answer to 2 decimal places.

Done

RZara says: “P(A | B) is always the same as P(B | A).” In a Venn diagram n(A) = 13, n(B) = 10 and n(A ∩ B) = 4. Show that Zara is wrong.

Done

RP(A) = 0.5, P(B) = 0.4 and P(A and B) = 0.1. Show that A and B are not independent.

Done

RThe table shows 80 students. Find the two probabilities below, and say what they tell you about playing an instrument and being in Year 10.

	Instrument	None	Total
Year 10	16	14	30
Year 11	24	26	50
Total	40	40	80

(a)P(Year 10 | plays an instrument)

(b)P(Year 10)

(c)What the answers tell you

Done

PSA bag contains 3 red and 5 blue counters. A counter is picked at random and put aside. A second counter is then picked. Given that the second counter is blue, find the probability that the first counter was red.

Done

PSThe Venn diagram shows 40 students who travel by bus (A), bike (B) or car (C). Find...

(a)P(A | B)

(b)P(C | A′)

(c)P(A ∩ B | C)

Done

Chapter summary

Skill	Method
Probability of an event	favourable outcomes ÷ total outcomes, for equally likely outcomes. Always between 0 and 1.
Probability of not A	P(not A) = 1 − P(A). All the probabilities of the outcomes add up to 1.
Relative frequency	frequency ÷ number of trials. It estimates the probability. More trials give a better estimate.
Expected number	probability × number of trials
Fair or biased?	Compare relative frequencies with the theoretical probability, using plenty of trials. Fair does not mean exactly equal.
Listing outcomes	List systematically or use a sample-space table. Probability = matching cells ÷ all cells (two dice: 36).
Counting outcomes	Product rule: multiply the number of choices at each stage.
Independent events	P(A and B) = P(A) × P(B). “And” means multiply.
Mutually exclusive events	P(A or B) = P(A) + P(B). “Or” means add.
At least one	P(at least one) = 1 − P(none)
Tree diagrams	Multiply along the branches, add the routes. Without replacement the numbers on the second branches change.
Sets and Venn diagrams	∩ both, ∪ either, A′ not in A, ξ everything. Fill in from the middle outwards. P = n(event) ÷ n(ξ).
Two-way tables	Rows and columns add up to their totals. Probability = number ÷ grand total.
Frequency trees	Numbers, not probabilities. Take a fraction of the number in the box before it.
Higher: P(A or B)	P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Higher: conditional probability	P(A \| B) = P(A and B) ÷ P(B) = n(A ∩ B) ÷ n(B). “Given” shrinks the sample space.
Higher: probability and algebra	Write each probability with the unknown, form an equation (probabilities add to 1, or a route probability equals a given value) and solve.

Can you…?

I can use the words impossible, unlikely, even chance, likely and certain, and the probability scale from 0 to 1.
I can work out the probability of a single event when outcomes are equally likely.
I can use P(not A) = 1 − P(A) and find a missing probability.
I can work out a relative frequency and use it to estimate a probability.
I can find an expected number of outcomes.
I can say whether an experiment suggests a die or coin is fair or biased, and explain why more trials are better.
I can list outcomes systematically and use a sample-space table.
I can use the product rule to count outcomes.
I can use the AND rule for independent events and the OR rule for mutually exclusive events.
I can draw and use tree diagrams with and without replacement.
I can use set notation and Venn diagrams to count and to find probabilities.
I can complete and use two-way tables and frequency trees.
Higher: I can find conditional probabilities from tables, Venn diagrams and tree diagrams.
Higher: I can form and solve equations from probability information.

Key terms

Probability: A number from 0 to 1 that measures how likely an event is.
Outcome: One possible result of an experiment.
Event: An outcome or group of outcomes we are interested in.
Equally likely: Outcomes that each have the same chance of happening.
Random: Not predictable or rigged: each outcome has its fair chance.
Biased: Not fair: some outcomes are more likely than they should be.
Theoretical probability: Probability worked out from equally likely outcomes: favourable ÷ total.
Trial: One go of an experiment.
Frequency: The number of times an event happens.
Relative frequency: Frequency divided by the number of trials: an estimate of probability.
Expected number: Probability multiplied by the number of trials.
Complement: The event “A does not happen”, written not A or A′.
Exhaustive: Covering every possible outcome, so the probabilities add up to 1.
Mutually exclusive: Events that cannot happen at the same time.
Independent: Events where one has no effect on the probability of the other.
Dependent: Events where the result of one changes the probability of the other.
Sample space: The list or table of all possible outcomes.
Product rule for counting: Number of outcomes = (choices at stage 1) × (choices at stage 2) × ...
Tree diagram: A diagram of branches showing the outcomes of events in order, with their probabilities.
With replacement: An item is put back after it is picked, so probabilities stay the same.
Without replacement: An item is not put back, so later probabilities change.
Set: A collection of items called elements, written in curly brackets.
Universal set (ξ): The set of everything being considered.
Intersection (A ∩ B): The elements in both A and B.
Union (A ∪ B): The elements in A or B or both.
Venn diagram: A diagram that uses overlapping circles inside a rectangle to show sets.
Two-way table: A table showing two categories at once, with row and column totals.
Frequency tree: A tree showing numbers of people or items instead of probabilities.
Conditional probability: The probability of an event given that another event has happened: P(A | B).

Exercise 23Z Check-up (non-calculator)

Non-calculator

A bag contains 5 red, 3 blue and 2 green counters. One counter is picked at random. Find the probability that it is blue.

Done

The probability that a bus is late is 0.35. Work out the probability that the bus is not late.

Done

A spinner can land on red, blue or green. P(red) = 12 and P(blue) = 16. Work out P(green).

Done

A drawing pin is dropped 50 times and lands point up 18 times. Write down the relative frequency of point up as a fraction in its simplest form.

Done

The probability that a spinner lands on red is 15. The spinner is spun 150 times. How many times would you expect it to land on red?

Done

Two fair coins are tossed. Find the probability that both land on heads.

Done

Two fair dice are rolled and their scores are added. Find the probability that the total is 9.

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the total is 9. These are the favourable outcomes. Their number goes on the top of the fraction.
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Number of highlighted cells = 4
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the total is 9) = 436
Simplify the fraction. Divide the top and the bottom by their highest common factor, 4. The value does not change but the numbers are smaller.
4 ÷ 4 = 1 and 36 ÷ 4 = 9
436 = 19

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Done

A bag has 3 red and 2 blue counters. Two counters are taken at random without replacement. Find the probability that both are red.

Done

The table shows how 40 students voted for a school trip. A girl is chosen at random. Find the probability that she voted for the zoo.

	Zoo	Museum	Total
Boys	14	6	20
Girls	9	11	20
Total	23	17	40

Done

A bag has x red counters and 4 blue counters. P(red) = 23. Find the value of x.

Done

Total 58 marks. Allow about 58 minutes.

Exercise 23X Exam-style questions

Non-calculator

FA bag contains 4 red counters, 5 blue counters and 1 green counter. One counter is picked at random. Find the probability that the counter is...[3 marks]

(a)green

(b)not blue

(c)red or green

Done

FThe table shows the probability that a biased spinner lands on each colour.[4 marks]

Colour	Red	Blue	Green	Yellow
Probability	0.3	0.25	0.35	?

(a)Work out the probability that it lands on yellow.

(b)The spinner is spun 200 times. How many times would you expect it to land on blue?

Done

FPriya rolled a die 60 times. The results are in the table.[3 marks]

Score	1	2	3	4	5	6
Frequency	8	12	9	11	10	10

(a)Work out the relative frequency of a 6.

(b)Priya says the die is biased. Is she right? Give a reason.

Done

FTwo fair dice are rolled and the scores are multiplied. Find the probability that the product is 12.[3 marks]

Method. Probability = favourable outcomes ÷ total outcomes, when all the outcomes are equally likely. Count the total first, then count the outcomes we want.

Draw the sample space table. Write the first die scores along the top and the second die scores down the side. Every cell is one outcome, and all the cells are equally likely.
× 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Total number of outcomes = 6 × 6 = 36
Highlight the cells where the product is 12. These are the favourable outcomes. Their number goes on the top of the fraction.
× 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
Number of highlighted cells = 4
Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
P(the product is 12) = 436
Simplify the fraction. Divide the top and the bottom by their highest common factor, 4. The value does not change but the numbers are smaller.
4 ÷ 4 = 1 and 36 ÷ 4 = 9
436 = 19

Watch out. Write the answer as a fraction, decimal or percentage, never as a ratio such as 1 : 6. A fraction is best left in its simplest form.

Done

FA fair coin is tossed and a fair spinner with the numbers 1, 2, 3, 4, 5 is spun. Find the probability of heads and an even number.[2 marks]

Done

FA bag has 3 red counters and 2 blue counters. A counter is picked at random, replaced, and a second counter is picked.[4 marks]

(a)Complete the tree diagram.

(b)Find the probability that the two counters are the same colour.

Done

FA bag has 4 red and 6 blue counters. Two counters are taken at random without replacement. Find the probability that the counters are different colours.[4 marks]

Done

F40 students were asked if they play chess (A) and draughts (B). 22 play chess, 18 play draughts and 7 play both.[4 marks]

(a)Draw a Venn diagram.

(b)A student is chosen at random. Find the probability that the student plays neither game.

Done

F70 members of a club were asked if they prefer swimming or running. Some of the results are shown.[4 marks]

	Swimming	Running	Total
Men	20	?	35
Women	?	25	?
Total	30	?	70

(a)Complete the two-way table.

Method. In a two-way table every row and every column adds up to its total. Find a row or column with only one missing number, work it out, then repeat.

Find the missing number in the Men row, Running column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
35 − 20 = 15
Find the missing number in the Total row, Running column. The numbers in a row add up to the row total, so subtract the known numbers from the total.
70 − 30 = 40
Find the missing number in the Women row, Swimming column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
30 − 20 = 10
Find the missing number in the Women row, Total column. The numbers in a column add up to the column total, so subtract the known numbers from the total.
70 − 35 = 35
Write the completed table. Check that every row and every column adds up to its total.
Swimming Running Total
Men 20 15 35
Women 10 25 35
Total 30 40 70

Check. Check the grand total: the row totals add to 70 and the column totals add to 70, and both equal 70. ✓

Watch out. Start with a row or column that has only one missing number.

(b)A member is chosen at random. Find the probability that it is a woman who prefers swimming.

Done

F120 people were asked about their diet. 23 of them are adults. 14 of the adults and 12 of the children are vegetarian. Complete a frequency tree and find how many of the 120 people are vegetarian.[4 marks]

Done

FThe probability that a train is late is 0.1 on any day, independently. Find the probability that it is late on at least one of 3 days. Give your answer as a decimal.[3 marks]

Done

FSam says: “When I toss two coins there are three outcomes, so the probability of getting two heads is 1/3.” Explain what is wrong with Sam’s statement.[2 marks]

Done

FA game costs £2 to play. The probability of winning is 18 and the prize is £12. 400 people play. Work out the expected profit.[4 marks]

Done

FA bag has 6 red and 4 blue counters. Two counters are taken at random without replacement. Given that the second counter is blue, find the probability that the first counter was red.[4 marks]

Done

FA bag has 4 red counters and x blue counters. Two counters are taken at random without replacement. The probability that both are red is 215. Show that x² + 7x − 78 = 0 and hence find x.[5 marks]

Done

FP(A) = 0.5 and P(B) = 0.4 and P(A and B) = 0.2.[5 marks]

(a)Work out P(A or B).

(b)Work out the probability that neither A nor B happens.

(c)Are A and B independent? Give a reason.

Done

← Chapter 22: Constructions, loci, bearings and vectors Chapter 24: Statistics →

×	1	2	3	4	5	6
1	1	2	3	4	5	6
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3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

red \ blue	1	2	3	4	5	6
1	1,1	2,1	3,1	4,1	5,1	6,1
2	1,2	2,2	3,2	4,2	5,2	6,2
3	1,3	2,3	3,3	4,3	5,3	6,3
4	1,4	2,4	3,4	4,4	5,4	6,4
5	1,5	2,5	3,5	4,5	5,5	6,5
6	1,6	2,6	3,6	4,6	5,6	6,6

red \ blue	1	2	3	4	5	6
1	1,1	2,1	3,1	4,1	5,1	6,1
2	1,2	2,2	3,2	4,2	5,2	6,2
3	1,3	2,3	3,3	4,3	5,3	6,3
4	1,4	2,4	3,4	4,4	5,4	6,4
5	1,5	2,5	3,5	4,5	5,5	6,5
6	1,6	2,6	3,6	4,6	5,6	6,6

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

red \ blue	1	2	3	4	5	6
1	1,1	2,1	3,1	4,1	5,1	6,1
2	1,2	2,2	3,2	4,2	5,2	6,2
3	1,3	2,3	3,3	4,3	5,3	6,3
4	1,4	2,4	3,4	4,4	5,4	6,4
5	1,5	2,5	3,5	4,5	5,5	6,5
6	1,6	2,6	3,6	4,6	5,6	6,6

red \ blue	1	2	3	4	5	6
1	1,1	2,1	3,1	4,1	5,1	6,1
2	1,2	2,2	3,2	4,2	5,2	6,2
3	1,3	2,3	3,3	4,3	5,3	6,3
4	1,4	2,4	3,4	4,4	5,4	6,4
5	1,5	2,5	3,5	4,5	5,5	6,5
6	1,6	2,6	3,6	4,6	5,6	6,6

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

GCSE Maths Probability (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

The probability scale and probability of single events

Key ideas

Watch out

Exercise 23A The probability scale and single events

Probabilities add up to 1: the probability of “not A”

Key ideas

Watch out

Exercise 23B Adding to 1 and the probability of “not”

Relative frequency, expected outcomes and fairness

Key ideas

Watch out

Exercise 23C Relative frequency

Exercise 23D Expected outcomes, fair and biased

Listing outcomes and sample-space tables

Key ideas

Watch out

Exercise 23E Listing outcomes and sample-space tables

Independent events: the AND rule and the OR rule

Key ideas

Watch out

Exercise 23F Independent events: AND and OR

Tree diagrams: with and without replacement

Key ideas

Watch out

Exercise 23G Tree diagrams: independent events and replacement

Exercise 23H Tree diagrams: dependent events and no replacement

Sets and Venn diagrams

Key ideas

Watch out

Exercise 23I Sets and Venn diagrams

Exercise 23J Probability from Venn diagrams

Two-way tables and frequency trees

Key ideas

Watch out

Exercise 23K Two-way tables

Exercise 23L Frequency trees

Conditional probability and probability with algebra

Key ideas

Watch out

Exercise 23M Conditional probability

Exercise 23N Probability and algebra

Chapter summary

Can you…?

Key terms

Exercise 23Z Check-up (non-calculator)

Exercise 23X Exam-style questions

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

red \ blue	1	2	3	4	5	6
1	1,1	2,1	3,1	4,1	5,1	6,1
2	1,2	2,2	3,2	4,2	5,2	6,2
3	1,3	2,3	3,3	4,3	5,3	6,3
4	1,4	2,4	3,4	4,4	5,4	6,4
5	1,5	2,5	3,5	4,5	5,5	6,5
6	1,6	2,6	3,6	4,6	5,6	6,6

red \ blue	1	2	3	4	5	6
1	1,1	2,1	3,1	4,1	5,1	6,1
2	1,2	2,2	3,2	4,2	5,2	6,2
3	1,3	2,3	3,3	4,3	5,3	6,3
4	1,4	2,4	3,4	4,4	5,4	6,4
5	1,5	2,5	3,5	4,5	5,5	6,5
6	1,6	2,6	3,6	4,6	5,6	6,6

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36