Sample Space Diagrams

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How to list every outcome of two events in a sample space diagram, and how to use the grid to work out probabilities, including when the two numbers are added or multiplied.

What the exam asks

A sample space diagram shows every possible outcome when two things happen together, such as throwing a coin and spinning a spinner. Some exam papers call it a possibility space diagram.

In the exam you are often asked to list all the possible combinations, for 2 marks. You may also be asked to complete a sample space diagram, for 1 mark, and then use it to work out a probability, for 1 or 2 marks.

Harder questions ask which of two outcomes is more likely. Then you must work out both probabilities and compare them.

Listing outcomes in a grid

To list every combination, keep the first choice fixed and go through every second choice in order. Then move on to the next first choice and do the same again. This stops you missing or repeating a combination.

A grid does this for you. Put one event down the side and the other across the top. Each cell inside the grid is one outcome: one result from the side together with one result from the top.

Every row has the same number of cells, so you can check your count:

\text{number of outcomes} = \text{first choices} \times \text{second choices}

Worked example 1: Listing combinations

A café sells a lunch deal. You choose one wrap, bean (B) or cheese (C), and one drink, water (W), juice (J) or milk (M). List all the possible combinations.

  1. There are 2 wraps and 3 drinks. Each wrap can go with each drink.
  2. Your turn. How many combinations should there be?

    2 \times 3 = 6 combinations

  3. Keep bean fixed and go through the drinks: BW, BJ, BM. Then keep cheese fixed: CW, CJ, CM. That makes 6, as expected.
  4. Answer: BW, BJ, BM, CW, CJ and CM.

Explore the size of a sample space

Try it: the slider labelled First-stage choices sets how many numbers the first spinner has (the rows). The slider labelled Second-stage choices sets the second spinner (the columns). In each cell, the first number is the first spinner and the second number is the second spinner. Count the cells each time: it is always rows times columns. Set both sliders to 6 to see the 36 outcomes for two ordinary dice.

Probabilities from the grid

When every cell is equally likely, count the cells that fit and divide by the total number of cells:

\text{probability} = \frac{\text{number of cells that fit}}{\text{total number of cells}}

The cells are equally likely when the coins, dice and spinners are fair and every sector of a spinner is the same size. If a coin is biased, or a spinner has one big sector, the cells are not equally likely and this method does not work.

If a fair spinner has four equal sectors labelled 1, 2, 2 and 3, give each sector its own row. The number 2 then appears in two rows, because it is twice as likely as 1 or 3.

Worked example 2: A coin and a spinner

Mia throws a fair coin and spins a fair four-sided spinner numbered 1, 2, 3 and 4. Work out the probability that she gets tails and an even number.

  1. The coin goes down the side and the spinner across the top. There are 2 \times 4 = 8 outcomes, and they are all equally likely.
  2. Your turn. How many outcomes have tails and an even number?

    T2 and T4: 2 outcomes

  3. Answer: \frac{2}{8} = \frac{1}{4}

Grids of totals and products

Often the two numbers are added or multiplied to make a score. Each cell then shows the side number plus the top number, or the side number times the top number. A + or \times in the corner of the grid reminds you which.

Different cells can show the same score. Spinner 1 with dice 3, and spinner 3 with dice 1, both give a total of 4, but they are different outcomes. Count every cell, not just the different scores.

Worked example 3: Completing a grid of totals

Jo spins a fair three-sided spinner numbered 1, 2 and 3, and rolls a fair ordinary dice. She adds the two numbers to get her score. Complete the sample space diagram, then work out the probability that her score is 7.

  1. Each cell is the spinner number plus the dice number. One cell is still empty.
  2. Your turn. What score goes in the empty cell, for spinner 2 and dice 6?

    2 + 6 = 8

  3. There are 3 \times 6 = 18 cells, all equally likely.
  4. A score of 7 is in 3 cells: 1 + 6, 2 + 5 and 3 + 4
  5. Answer: \frac{3}{18} = \frac{1}{6}

Worked example 4: Totals are not equally likely

Two fair ordinary dice are rolled and the two numbers are added. Leo says, “There are 11 possible totals, from 2 to 12, so the probability of a total of 4 is \frac{1}{11}.” Is Leo correct? Give a reason for your answer.

  1. Draw the grid of totals. There are 6 \times 6 = 36 cells, all equally likely.
  2. Your turn. How many cells show a total of 4?

    3 cells: 1 + 3, 2 + 2 and 3 + 1

  3. So the probability of a total of 4 is \frac{3}{36} = \frac{1}{12}
  4. Answer: Leo is not correct. The 11 totals are not equally likely, and the probability of a total of 4 is \frac{1}{12}

Worked example 5: A grid of products

Bag A has four cards numbered 1, 2, 3 and 4. Bag B has three cards numbered 1, 2 and 3. Amir takes a card at random from each bag and multiplies the two numbers. Work out the probability that the product is more than 6.

  1. Each cell is the bag A number times the bag B number. There are 4 \times 3 = 12 cells, all equally likely.
  2. “More than 6” does not include 6 itself, so the two cells showing 6 do not count.
  3. Your turn. How many cells show a product more than 6?

    3 cells: 8, 9 and 12

  4. Answer: \frac{3}{12} = \frac{1}{4}

Worked example 6: Which is more likely?

Spinner A is fair, with three equal sectors numbered 1, 2 and 3. Spinner B is fair, with five equal sectors numbered 1, 2, 3, 4 and 5. Priya spins both spinners and adds the two numbers. Is Priya more likely to get an even total or an odd total?

  1. Draw the grid of totals. There are 3 \times 5 = 15 cells, all equally likely.
  2. Your turn. How many cells show an even total?

    8 cells: 3 in the first row, 2 in the second row and 3 in the third row

  3. The other 15 - 8 = 7 cells show an odd total.
  4. P(\text{even}) = \frac{8}{15} and P(\text{odd}) = \frac{7}{15}
  5. Answer: an even total is more likely, because \frac{8}{15} is greater than \frac{7}{15}

Common mistakes

  • Repeating or missing combinations. Keep one choice fixed, go through every other choice in order, and check the count is first choices times second choices.
  • Writing combinations that run together. Write each outcome as its own pair, such as BW or (2, 5), with a clear gap between pairs.
  • Dividing by the number of different totals. Two dice give 36 equally likely cells, so divide by 36, not by the 11 different totals.
  • Assuming every outcome is equally likely. Count cells only when the coins, dice or spinners are fair and the sectors are equal.
  • Comparing counts instead of probabilities. Write each probability as a fraction, with its own denominator, before you compare.
  • Misreading “more than” and “or more”. “More than 6” leaves out 6, but “6 or more” includes it.

Practice questions