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KS3 Maths · Year 8Chapter 12 · Probability of combined events
Year 8 · Chapter 12

Probability of combined events

By the end of this chapter you will be able to

  • List all the outcomes of two events systematically, and count how many are favourable
  • Use the fact that the probabilities of all outcomes add up to 1, to find the probability of the opposite event
  • Read probabilities from a two-way table or a Venn diagram
  • Draw a tree diagram for two events, with and without replacement, and multiply along the branches
  • Multiply probabilities of independent events, and find the expected number of outcomes in a number of trials

You should already know

  • Probability as a number from 0 to 1, and simple probability (Year 7)
  • Fractions, and simplifying fractions (Year 8, Chapter 3)
  • Reading tables and simple charts (Year 7 and Year 8)

Where this leads

  • Conditional probability and Venn diagrams with three sets (GCSE)
  • Expected values and simulations in real-life data (GCSE)
  • Probability with more than two events and dependent events in depth (Year 9)

National Curriculum for England, Key Stage 3 mathematics, Year 8. Content: List the outcomes of a combined event, and use a sample space, Use a two-way table and a Venn diagram to find probabilities, Find probabilities of combined events using tree diagrams and the rule that probabilities add up to 1, Find the probability of independent events, and the expected frequency of an outcome, Use probability to compare fairness and make predictions.

Exercise P  Do you remember?

Non-calculator
1
FHow many outcomes are there when you roll one fair six-sided die?
2
FWhat is the probability of getting a 3 on a fair six-sided die?
3
FWrite 0.25 as a fraction in its simplest form.
4
FWhat is 1 − 0.3?
5
FHow many different results are there when you toss two coins?
12.1

Sample spaces and listing outcomes

Year 8
List the outcomes of a combined event, and use a sample space
Video lesson: Sample spaces and listing outcomesComing soon
Tom rolls two dice and adds the scores. He wants to know how likely it is to get a total of 7, so he lists all the possible outcomes first, to make sure he does not miss any.

The sample space is the list of all possible outcomes. For two dice, there are 6 × 6 = 36 equally likely outcomes.

To list outcomes systematically, go through one event in order and pair it with every outcome of the other. This makes sure no outcome is missed.

Key ideas

  • List outcomes in a systematic way, so none are missed or repeated.
  • For two events with a and b outcomes, there are a × b outcomes altogether.
  • Probability = number of favourable outcomes ÷ total number of outcomes.
  • Each outcome in a fair sample space is equally likely.
Worked example 12.1.1
Two dice are rolled. Find the probability that the total is 7.

Method. Probability = favourable outcomes ÷ total outcomes. List or count the outcomes first.

  1. Count the total outcomes. Every outcome is equally likely.
    total = 36
  2. Count the favourable outcomes. Outcomes where the total is 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).
    favourable = 6
  3. Write the probability. Put the favourable outcomes over the total, then simplify.
    6/36 = 1/6
  4. Write the answer.
    1/6

Check. The probability must be between 0 and 1: 1/6 is.

Watch out. Check the total is the full list of outcomes. Each outcome should be counted once.

Worked example 12.1.2
Two coins are tossed. Find the probability of getting at least one head.

Method. List the sample space, then count the outcomes with at least one head.

  1. List the outcomes. The sample space is HH, HT, TH, TT.
    HH, HT, TH, TT
  2. Count the favourable. At least one head: HH, HT and TH (3 outcomes).
    3
  3. Write the probability. There are 4 outcomes altogether.
    3/4
  4. Write the answer.
    3/4

Check. Only TT has no head, so 1 − 1/4 = 3/4.

Worked example 12.1.3
A spinner has 1, 2, 3, 4 and a bag has a red and a blue token. Find the number of outcomes.

Method. Multiply the number of outcomes of each event.

  1. Count the spinner outcomes. The spinner has 4 outcomes.
    4
  2. Count the token outcomes. The token has 2 outcomes.
    2
  3. Multiply. The total number of outcomes is 4 × 2 = 8.
    8
  4. Write the answer.
    8 outcomes

Check. The 8 outcomes are 1R, 1B, 2R, 2B, 3R, 3B, 4R, 4B.

Watch out

Watch out: the outcome (1, 2) is different from (2, 1) when you roll two different dice. List both, unless the dice are the same.

Exam tip

Exam tip. Write a complete list or a grid before you count. Examiners check that you have counted each outcome once.

Exercise 12A  Sample spaces and listing outcomes

Non-calculator
1
FTwo fair dice are rolled and the scores are added. Find the probability that the total is 7.
2
FTwo fair dice are rolled and the scores are added. Find the probability that the total is 10.
3
FTwo fair dice are rolled and the scores are added. Find the probability that both dice show the same number.
4
FTwo fair dice are rolled and the scores are added. Find the probability that the total is more than 9.
5
FTwo fair dice are rolled and the scores are added. Find the probability that both numbers are even.
6
FTwo fair dice are rolled and the scores are added. Find the probability that the difference is 1.
7
FTwo fair dice are rolled and the scores are added. Find the probability that the total is less than 4.
8
FTwo fair dice are rolled and the scores are added. Find the probability that the total is 2.
9
FTwo coins are tossed. List the sample space and find the probability of getting two tails.
10
PSA spinner has the numbers 1, 2, 3 and 4. A coin is tossed. Find the probability of getting a 3 and a head.
11
FTwo dice are rolled. Write the number of outcomes where the first die shows 6.
12
FA bag has a red and a blue counter. Two counters are taken, one after the other, and put back each time. List the sample space.
13
RAli says that the sample space for two coins is HH and TT only. Is Ali right? Explain.
14
RExplain why the sample space for rolling two dice has 36 outcomes, not 21. Accept any answer that counts each ordered pair, such as (1, 2) and (2, 1).
15
PSA bag contains red and blue counters. Two are picked (with replacement). Write the probability of getting two reds, if there is one red and one blue counter.
12.2

The probability scale and the opposite event

Year 8
Use the fact that probabilities add up to 1, Find the probability of the opposite event
Video lesson: The probability scale and the opposite eventComing soon
A weather app says the probability of rain on Saturday is 2/5. Priya wants to know the probability that it does not rain, so she can plan a picnic.

Every probability is between 0 (impossible) and 1 (certain).

The probabilities of all the possible outcomes add up to 1. So if the probability of an event is P(A), the probability that it does not happen is 1 − P(A).

Key ideas

  • A probability is between 0 and 1.
  • The probabilities of all the outcomes add up to 1.
  • P(not A) = 1 − P(A).
  • Find a missing probability by subtracting the others from 1.
Worked example 12.2.1
The probability of rain is 2/5. Find the probability that it does not rain.

Method. The probabilities of all the outcomes add up to 1. So P(not A) = 1 − P(A).

  1. Write the rule. P(not A) = 1 − P(A).
    P(not A) = 1 − P(A)
  2. Put in the number. P(A) = 2/5.
    1 − 2/5
  3. Work it out. Subtract from 1.
    1 − 2/5 = 3/5
  4. Write the answer.
    3/5

Check. 2/5 + 3/5 = 1.

Worked example 12.2.2
A bag has red, blue and green counters. P(red) = 0.2 and P(blue) = 0.5. Find P(green).

Method. The three probabilities add up to 1, so the missing one is 1 minus the others.

  1. Add the known probabilities. 0.2 + 0.5 = 0.7.
    0.7
  2. Subtract from 1. 1 − 0.7 = 0.3.
    0.3
  3. Write the answer.
    0.3

Check. 0.2 + 0.5 + 0.3 = 1.

Worked example 12.2.3
A die is rolled. Find the probability of not getting a 6.

Method. The probabilities of all the outcomes add up to 1. So P(not A) = 1 − P(A).

  1. Write the rule. P(not A) = 1 − P(A).
    P(not A) = 1 − P(A)
  2. Put in the number. P(A) = 1/6.
    1 − 1/6
  3. Work it out. Subtract from 1.
    1 − 1/6 = 5/6
  4. Write the answer.
    5/6

Check. 1/6 + 5/6 = 1.

Watch out

Watch out: 1 − P(A) gives the probability that A does not happen, not P(A) subtracted from 0.

Exam tip

Exam tip. If you are asked for the probability of “not” something, try 1 minus the probability of the thing.

Exercise 12B  The opposite event and the sum to 1

Non-calculator
1
FThe probability that an event happens is 2/5. Find the probability that it does not happen.
2
FThe probability that an event happens is 1/6. Find the probability that it does not happen.
3
FThe probability that an event happens is 3/4. Find the probability that it does not happen.
4
FThe probability that an event happens is 1/3. Find the probability that it does not happen.
5
FThe probability that an event happens is 7/10. Find the probability that it does not happen.
6
FThe probability that an event happens is 5/8. Find the probability that it does not happen.
7
PSA bag has red, blue and green counters. P(red) = 0.2 and P(blue) = 0.5. Find P(green).
8
PSA spinner has sectors A, B and C. P(A) = 1/2 and P(B) = 1/3. Find P(C).
9
PSThe probability that a bus is late is 0.15. What is the probability that it is on time?
10
PSThe probability of getting a 6 on a fair die is 1/6. Find the probability of getting a number less than 6.
11
RA pack of cards: P(heart) = 1/4. Ali says the probability of not a heart is 3/4. Is Ali right? Explain.
12
RSam says that the probability of an event can be 1.2. Explain why this is not possible.
13
PSThree outcomes have probabilities 0.4, 0.35 and x. Find x.
12.3

Two-way tables and Venn diagrams

Year 8
Use a two-way table and a Venn diagram to find probabilities
Video lesson: Two-way tables and Venn diagramsComing soon
A school has 40 pupils in Year 8. The head of year records how they travel to school, and whether they are a boy or a girl. She puts the numbers in a two-way table, so that she can find probabilities.

A two-way table shows the counts for two events at once. Add the rows and columns to find the totals.

A Venn diagram shows which outcomes are in each set. The overlap is where both events happen.

Key ideas

  • A two-way table shows counts for two events: check that the total is the whole group.
  • Probability = count ÷ total for the whole group.
  • A Venn diagram region shows the number in that region only.
  • P(A or B) = the total in either set, counted once.
Worked example 12.3.1
Use the table: find the probability that a pupil is a boy who walks.

Method. Use the two-way table. Find the total for the whole group, then the count you want.

  1. Find the total. The total number of people is 40.
    total = 40
  2. Find the count in the table. The number with boy and walks is 12.
    12
  3. Write the probability. Put the count over the total, then simplify.
    12/40 = 3/10
  4. Write the answer.
    3/10

Check. The probability is the count divided by the total: 12 ÷ 40 = 0.300.

Worked example 12.3.2
A Venn diagram: 30 pupils, 18 like football, 12 like tennis, 7 like both. Find the probability that a pupil likes neither.

Method. Read the number in the correct region of the Venn diagram, then divide by the total number of people.

  1. Find the total. Add all the regions: 11 + 7 + 5 + 7 = 30.
    total = 30
  2. Find the region. The region for football only contains 11.
    11
  3. Write the probability. Divide the count by the total and simplify.
    11/30 = 11/30
  4. Write the answer.
    11/30

Check. The regions add up to 30, which is the whole group.

Worked example 12.3.3
Find P(a girl walks) from the table with 40 pupils.

Method. Use the two-way table. Find the total for the whole group, then the count you want.

  1. Find the total. The total number of people is 40.
    total = 40
  2. Find the count in the table. The number with girl and walks is 10.
    10
  3. Write the probability. Put the count over the total, then simplify.
    10/40 = 1/4
  4. Write the answer.
    1/4

Check. The probability is the count divided by the total: 10 ÷ 40 = 0.250.

Watch out

Watch out: the probability is out of the whole group, not out of the row or column you are looking at, unless the question asks about that group.

Exam tip

Exam tip. Check the totals of the table add up. A missing number can often be found by subtracting from the total.

Exercise 12C  Two-way tables and Venn diagrams

Non-calculator
1
FThe table shows how 40 pupils travel to school. Find the probability that a pupil is a boy who walks.
TravelWalkBusTotal
Boys12820
Girls101020
Total221840
2
FThe table shows how 40 pupils travel to school. Find the probability that a pupil is a girl.
TravelWalkBusTotal
Boys12820
Girls101020
Total221840
3
FThe table shows how 40 pupils travel to school. Find the probability that a pupil walks to school.
TravelWalkBusTotal
Boys12820
Girls101020
Total221840
4
FThe table shows how 40 pupils travel to school. Find the probability that a pupil is a girl who takes the bus.
TravelWalkBusTotal
Boys12820
Girls101020
Total221840
5
PSThe table shows how 40 pupils travel to school. Find the probability that a pupil takes the bus.
TravelWalkBusTotal
Boys12820
Girls101020
Total221840
6
FA Venn diagram shows 30 pupils: 18 like football, 12 like tennis, and 7 like both. Find the probability that a pupil likes football only.
7
FIn the same Venn diagram (30 pupils), find the probability that a pupil likes tennis only.
8
PSIn the same Venn diagram (30 pupils), find the probability that a pupil likes neither sport.
9
PSIn the same Venn diagram, find the probability that a pupil likes football or tennis.
10
RAli says the probability that a pupil likes both sports is 18/30, because 18 like football. Is Ali right? Explain.
11
RA two-way table has row totals 15 and 25, and column totals 18 and 22. Explain why the total must be 40.
12.4

Tree diagrams for combined events

Year 8
Draw a tree diagram for two events, with and without replacement, Multiply along the branches
Video lesson: Tree diagrams for combined eventsComing soon
A bag has 3 red and 5 blue counters. Mia takes two counters, one after the other, and tries to find the chance that both are red. She draws a tree diagram, so she can see every path.

A tree diagram shows each event as a set of branches. The probability of a path is found by multiplying the probabilities along its branches.

The probability of an outcome that can happen in more than one way is found by adding the paths.

With replacement the counter is put back, so the probabilities stay the same. Without replacement it is not put back, so the second probabilities change.

Key ideas

  • Multiply along the branches to find the probability of a path.
  • Add the probabilities of different paths that give the same result.
  • With replacement: the probabilities stay the same for each event.
  • Without replacement: use one fewer item in the second event.
Worked example 12.4.1
A bag has 3 red and 5 blue counters. A counter is taken and replaced, then another is taken. Find P(two reds).

Method. With replacement, the probabilities are the same for both events, so multiply them.

  1. Find the first probability. P(red) = 3/8, from 3 red out of 8.
    3/8
  2. Find the second probability. The counter is replaced, so P(red) is still 3/8.
    3/8
  3. Multiply along the branches. 3/8 × 3/8 = 9/64.
    9/64
  4. Write the answer.
    9/64

Check. 9/64 is about 0.14.

Worked example 12.4.2
The same bag: the first counter is red and NOT replaced. Find P(two reds).

Method. Without replacement, the second probability uses one fewer counter.

  1. Find the first probability. P(red) = 3/8.
    3/8
  2. Find the second probability. There are now 7 counters, 2 of them red, so P(red) = 2/7.
    2/7
  3. Multiply along the branches. 3/8 × 2/7 = 6/56 = 3/28.
    3/28
  4. Write the answer.
    3/28

Check. 3/28 is smaller than 9/64, because there are fewer reds left.

Worked example 12.4.3
Two coins are tossed. Find P(exactly one head).

Method. List the paths on the tree, and add the ones with exactly one head.

  1. Draw the tree. Each coin has H and T, each with probability 1/2.
    HH, HT, TH, TT
  2. Find the paths with one head. HT and TH.
    HT and TH
  3. Multiply and add. 1/2 × 1/2 = 1/4 for each path, and 1/4 + 1/4 = 1/2.
    1/4 + 1/4
  4. Write the answer.
    1/2

Check. There are 4 equally likely paths, and 2 of them have one head.

Watch out

Watch out: without replacement, the second probabilities change. Use one fewer counter, and one fewer of the colour you took first.

Exam tip

Exam tip. Write each branch probability on the branch. Then multiply along the path. Check that the branches from each point add up to 1.

Exercise 12D  Tree diagrams

Non-calculator
1
FA bag has 3 red and 5 blue counters. A counter is taken and replaced, then another is taken. Find P(two reds).
2
FThe same bag (3 red and 5 blue). A counter is taken and NOT replaced, then another is taken. Find P(two reds).
3
PSThe same bag, with replacement. Find P(one red and one blue), in either order.
4
PSThe same bag, with replacement. Find P(at least one red).
5
FA coin is tossed twice. Find the probability of getting two heads.
6
PSA spinner has red and blue sectors, with P(red) = 1/4. It is spun twice. Find P(red, then blue).
7
PSA bag has 4 white and 6 black counters. Two are taken without replacement. Find P(both white).
8
RExplain why P(two reds) without replacement is smaller than with replacement, for the same bag. Accept any answer that says there are fewer reds left, so the second probability is lower.
9
RA tree diagram for two coins has branches H and T at each stage. Ali says P(HT) is 1/2. Is Ali right? Explain.
12.5

Independent events, expected frequency and fairness

Year 8
Find the probability of independent events, Find the expected frequency of an outcome, Use probability to compare fairness
Video lesson: Independent events, expected frequency and fairnessComing soon
A game uses a fair die and a fair coin. Priya wants to know how often she would expect to get a 6 with a head in 120 rolls, and whether the game is fair for two players.

Two events are independent if the result of one does not affect the other. For independent events, P(A and B) = P(A) × P(B).

The expected frequency of an outcome is its probability multiplied by the number of trials: expected = P × number of trials.

A game is fair when each player has the same chance of winning.

Key ideas

  • For independent events, multiply the probabilities.
  • Expected frequency = probability × number of trials.
  • Expected values are predictions, not exact results.
  • A game is fair if each player has an equal chance of winning.
Worked example 12.5.1
A fair die and a fair coin are used. Find P(a 6 and a head).

Method. The events are independent, so multiply the probabilities.

  1. Find P(6). One face out of six.
    1/6
  2. Find P(head). One side out of two.
    1/2
  3. Multiply. 1/6 × 1/2 = 1/12.
    1/12
  4. Write the answer.
    1/12

Check. There are 12 equally likely outcomes, and one of them is 6H.

Worked example 12.5.2
A die is rolled 120 times. Find the expected number of 6s.

Method. Expected frequency = probability × number of trials.

  1. Find the probability. P(6) = 1/6.
    1/6
  2. Multiply by the number of trials. 1/6 × 120 = 20.
    20
  3. Write the answer.
    20

Check. 20 out of 120 is 1/6.

Worked example 12.5.3
A coin is tossed 50 times. Find the expected number of heads.

Method. Use expected frequency = probability × number of trials.

  1. Find the probability. P(head) = 1/2.
    1/2
  2. Multiply. 1/2 × 50 = 25.
    25
  3. Write the answer.
    25

Check. About half of 50 is 25.

Watch out

Watch out: expected frequency is a prediction. A coin tossed 50 times may not give exactly 25 heads.

Exam tip

Exam tip. When you find an expected number, say that it is an expected value, not a certain result.

Exercise 12E  Independent events, expected frequency and fairness

Non-calculator
1
FA fair coin is tossed and a fair die is rolled. Find P(head and 3).
2
FTwo coins are tossed. Find P(two heads).
3
FA die is rolled 120 times. Find the expected number of 6s.
4
FA coin is tossed 50 times. Find the expected number of tails.
5
PSA spinner has P(red) = 1/4. It is spun 200 times. How many times would you expect red?
6
PSA bag has 4 red and 6 blue counters. A counter is taken and replaced 50 times. How many reds would you expect?
7
PSA game: a player wins if a fair coin shows heads AND a fair die shows 6. Find the probability of winning.
8
RTwo players each have a chance of winning in a game. Player A wins if the spinner shows red (P = 1/2), and Player B wins if it shows blue. Is the game fair? Explain.
9
RPlayer A wins with a 1 or 2 on a die, and Player B wins with 3, 4, 5 or 6. Is the game fair? Explain.
10
RAli says that P(two heads) with two coins is 1/2, because there are 2 coins. Is Ali right? Explain.
11
PSA die is rolled 300 times. Estimate how many times a number bigger than 4 is rolled.

Chapter summary

SkillMethod
Sample spaceSystematic list of all equally likely outcomes; count favourable outcomes over the total
Opposite eventP(not A) = 1 − P(A)
Two-way table and VennUse the counts in the region, divided by the whole total
Tree diagramMultiply along each path; add paths with the same result
With and without replacementWith: same probabilities; without: one fewer item for the second event
Independent eventsP(A and B) = P(A) × P(B)
Expected frequencyProbability × number of trials

Can you…?

  • I can list the outcomes of two events and count the favourable ones
  • I can find the probability of the opposite event, and use the total of 1
  • I can read probabilities from a two-way table and a Venn diagram
  • I can draw a tree diagram, with and without replacement, and multiply along branches
  • I can find the probability of independent events
  • I can find an expected frequency and say whether a game is fair

Key terms

sample space
the list of all possible outcomes
outcome
one possible result of an event
event
one or more outcomes we are interested in
independent events
events where the result of one does not change the other
replacement
putting an item back before the next choice
expected frequency
the number of times an outcome should happen, on average, in a number of trials
fair game
a game where each player has an equal chance of winning

Exercise 12Z  Check-up

Non-calculator
1
Two dice are rolled. Find the probability that the total is 5.
2
The probability of rain is 0.35. Find the probability of no rain.
3
A bag has 2 red and 3 blue counters. A counter is taken and replaced, then another. Find P(two blues).
4
The same bag (2 red, 3 blue), without replacement. Find P(two blues).
5
A fair die is rolled 60 times. Find the expected number of 2s.
6
A coin and a die are both fair. Find P(tail and an odd number).
7
The probabilities of three outcomes are 0.2, 0.3 and x. Find x.
8
Two coins are tossed. Find P(at least one tail).
9
A tree diagram shows P(red) = 1/3 then P(red) = 1/3 (with replacement). Find P(two reds).

Total 29 marks. Allow about 29 minutes.

Exercise 12X  Exam-style questions

Non-calculator
1
FTwo fair dice are rolled and the scores are added. Find the probability that the total is 9.[2 marks]
2
FA bag has 4 red and 6 blue counters. A counter is taken and replaced, then another is taken. Find P(one red and one blue).[3 marks]
3
FA bag has 4 red and 6 blue counters. Two counters are taken without replacement. Find P(two blues).[3 marks]
4
FIn a survey of 40 pupils, 16 like cats, 14 like dogs, and 6 like both. Find the probability that a pupil likes neither.[4 marks]
5
FA die is rolled 180 times. Find the expected number of 5s.[2 marks]
6
FA fair coin is tossed and a fair spinner with 4 equal sectors is spun. Find P(head and sector 1).[2 marks]
7
FA game costs £1 to play. A player wins £5 with probability 1/10. Explain whether the game is fair to the player.[3 marks]
8
FAli says that the probability of rain on each of two days is 0.3, so the probability of rain on both days is 0.6. Is Ali right? Explain.[3 marks]
9
FA bag has 3 green and 2 yellow counters. Two are taken without replacement. Find the probability that the two counters are different colours.[4 marks]
10
FA spinner has 5 equal sectors numbered 1 to 5. It is spun 100 times. Estimate how many times the spinner will land on a number bigger than 3.[3 marks]
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