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GCSE Maths · EdexcelChapter 23 · Probability
GCSE Maths · Probability · Foundation · Lesson 23.4

GCSE Maths: Listing outcomes and sample-space tables (Foundation)

Part of the GCSE Maths probability chapter (Foundation). Chapter overview · Higher version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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23.4

Listing outcomes and sample-space tables

Edexcel 1MA1: P6, P7
Grade 2-6

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 1HT
HHHHT
TTHTT
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.

+123456
1234567
2345678
3456789
45678910
567891011
6789101112
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.

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.

  1. 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
  2. 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
  3. Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
    P(one head and one tail) = 24
  4. 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.
+123456
1234567
2345678
3456789
45678910
567891011
6789101112

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.

  1. 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.
    +123456
    1234567
    2345678
    3456789
    45678910
    567891011
    6789101112
    Total number of outcomes = 6 × 6 = 36
  2. Highlight the cells where the total is 8. These are the favourable outcomes. Their number goes on the top of the fraction.
    +123456
    1234567
    2345678
    3456789
    45678910
    567891011
    6789101112
    Number of highlighted cells = 5
  3. 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.

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.

  1. 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
  2. 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
  3. Write the probability as a fraction. Probability = number of favourable outcomes ÷ total number of equally likely outcomes.
    P(an odd number and red) = 26
  4. 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.

  1. Split the problem into stages. We make one choice at a time, so we count the choices stage by stage.
    Sandwiches then Drinks
  2. 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
  3. 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
1
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
2
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
3
FTwo fair dice are rolled and the scores are added. Use the sample-space table to find the probability that the total is...
+123456
1234567
2345678
3456789
45678910
567891011
6789101112
(a)7
(b)12
(c)less than 5
(d)an even number
(e)a prime number
4
FTwo fair coins are tossed. Write down all the possible outcomes, using H for heads and T for tails. Then find the probability of two heads.
5
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
(b)at least 4
(c)1
6
FTwo fair dice are rolled and the scores are multiplied. Find the probability that the product is...
(a)12
(b)an odd number
(c)more than 20
7
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...
1231234
(a)5
(b)more than 4
(c)an even number
8
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
9
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.
10
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.
11
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
12
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.
13
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.
14
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.
15
RTwo fair dice are rolled and the scores are added. Show that a total of 7 is more likely than any other total.
16
RHow many different outcomes are there when...
(a)two fair dice
(b)three fair dice

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