N5, P6, P7ProbabilityFoundation & HigherGrades 2–6

Sample Spaces & Listing

When two things happen together — two dice, a coin and a spinner, a starter and a main course — the possible outcomes multiply quickly. A sample space diagram or a systematic list captures every outcome exactly once, so probabilities become a matter of counting.

35 min AQA GCSE Maths

The big picture

The key word is $\textbf{systematic}$ . Listing outcomes in a fixed order (fix the first choice, run through every second choice, then move on) guarantees nothing is missed or repeated.

Counting also works without listing. If there are $m$ ways to do one thing and $n$ ways to do another, there are $m \times n$ ways to do both — the product rule. It is how AQA expects you to count PIN codes, outfits and menu choices.

What you'll learn

List all outcomes of combined events systematically
Draw and use sample space diagrams
Calculate probabilities from a sample space
Use the product rule for counting

Systematic listing

Fix one choice at a time. For a coin and a four-sided spinner: H1, H2, H3, H4, T1, T2, T3, T4 — eight outcomes.

Example 1A café offers 3 sandwiches (C, E, H) and 2 drinks (T, J). List all the possible meal deals.

1Fix sandwich C: CT, CJ. Then E: ET, EJ. Then H: HT, HJ.

Answer

CT, CJ, ET, EJ, HT, HJ — 6 combinations

Your turnSame method, new numbers — try it before moving on.

Three friends — Ali, Ben and Cara — line up in a queue. How many different orders are there?

Your answer

Show worked solution

Solution

1ABC, ACB, BAC, BCA, CAB, CBA.
2That is 6 orders (or $3 \times 2 \times 1$ ).

Answer: $6$

Sample space diagrams

For two events with several outcomes each, draw a grid: one event along the top, the other down the side, and the combined result in each cell.

Every cell is equally likely (if each event is fair), so probability $=$ number of cells that fit $\div$ total number of cells.

Totals of two fair dice: 36 equally likely cells. The shaded cells show a total of 6:

P(6) = \frac{5}{36}

Example 1Two fair dice are rolled and the scores added. Find the probability of a total of 9 or more.

1Count cells with 9, 10, 11 or 12:

4 + 3 + 2 + 1 = 10

2Total cells: 36.

Answer

\frac{10}{36} = \frac{5}{18}

Your turnSame method, new numbers — try it before moving on.

Two fair dice are rolled and the scores added. What is the probability of a total of 7?

Your answer

Show worked solution

Solution

1Cells totalling 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 cells.
2 $P = \frac{6}{36} = \frac{1}{6}$ .

Answer: $\frac{1}{6}$

Example 2A fair coin is flipped and a fair four-sided spinner (1–4) is spun. Find the probability of heads and an odd number.

1Outcomes:

2 \times 4 = 8

2Heads with odd: H1, H3 — 2 outcomes.

Answer

\frac{2}{8} = \frac{1}{4}

Your turnSame method, new numbers — try it before moving on.

Two fair dice are rolled and the scores multiplied. What is the probability the product is 12?

Your answer

Show hint

Find every pair that multiplies to 12, in both orders.

Show worked solution

Solution

1(2,6), (6,2), (3,4), (4,3) — 4 cells.
2 $P = \frac{4}{36} = \frac{1}{9}$ .

Answer: $\frac{1}{9}$

Tip — (2, 6) and (6, 2) are different outcomes — the first dice and the second dice are distinct.

The product rule for counting

If one choice can be made in $m$ ways and a second, independent choice in $n$ ways, the two can be made together in $m \times n$ ways. It extends to any number of choices.

If items cannot be reused (like people in a queue), the number of options drops by one each time: $5 \times 4 \times 3$ .

Example 1A code has 2 letters (A–Z) followed by 3 digits (0–9). How many codes are possible?

26 \times 26 \times 10 \times 10 \times 10

Answer

676{,}000

Your turnSame method, new numbers — try it before moving on.

A 4-digit PIN uses digits 0–9, and no digit can be repeated. How many PINs are possible?

Your answer

Show worked solution

Solution

1First digit: 10 choices, then 9, then 8, then 7.
2 $10 \times 9 \times 8 \times 7 = 5040$ .

Answer: $5040$

Grids versus multiplying

A sample space diagram is the product rule drawn out: 6 rows × 6 columns = 36 cells. For three dice, a grid would need 216 cells — which is exactly why counting by multiplication, or using tree diagrams, takes over.

Think like an examiner

In "list all the outcomes" questions, a systematic order earns the marks; a random list with a missing item loses them.
Draw a sample space grid when a question involves two dice or spinners — it prevents miscounting.
For the product rule, show the multiplication (e.g. $26 \times 26 \times 10^3$ ), not just the final number.

Counting

P = \dfrac{\text{cells that fit}}{\text{total cells}}

Using a sample space.

m \times n

Product rule: ways to make two choices.

6 \times 6 = 36

Outcomes for two dice.

Watch out for these

Counting (2, 5) and (5, 2) as one outcome.

They are different outcomes for two distinct dice.

Saying there are 11 equally likely totals for two dice (2 to 12).

There are 36 equally likely outcomes; the totals are not equally likely.

Adding choices: 3 sandwiches and 2 drinks give 5 meals.

Multiply:

3 \times 2 = 6

Allowing repeats when the question forbids them.

Reduce the options by one at each step.

Stretch yourself

Two fair dice are rolled. What is the probability that the two scores differ by at least 2?

Hint — Count the cells where the difference is 0 or 1, then subtract from 36.

Reveal solution

SolutionDifference 0: 6 cells. Difference 1: 10 cells ((1,2), (2,3), (3,4), (4,5), (5,6) and their reverses). So

6 + 10 = 16

cells differ by less than 2, and

36 - 16 = 20

differ by at least 2.

P = \frac{20}{36} = \frac{5}{9}

\blacksquare

Questions students ask

When should I use a sample space diagram rather than a tree diagram?

Grids are best for two events with several outcomes each (like two dice). Tree diagrams are better for sequences of events, especially with unequal probabilities or without replacement.

Does order matter in listing?

It depends on the context. For two dice, (1, 2) and (2, 1) differ. For choosing two people for a team, "Ali and Ben" is the same as "Ben and Ali".

What is the product rule called at A-level?

It becomes the basis of permutations and combinations — but at GCSE, the multiplication idea is all you need.

Key takeaways

List systematically: fix one choice, run through the others.
Sample space grids show every equally likely combined outcome.
Probability = favourable cells ÷ total cells.
Product rule: $m$ ways and $n$ ways give $m \times n$ ways.
Without repeats, reduce the options by one each time.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 33 marks

A fair spinner is numbered 1, 2, 3. A fair coin is flipped. The spinner is spun and the coin is flipped.

(a)List all the possible outcomes.[2]
Show mark scheme (a)
- B1At least 4 correct outcomes
- B1All 6: 1H, 1T, 2H, 2T, 3H, 3T with no repeats
(b)Work out the probability of getting tails and an even number.[1]
Show mark scheme (b)
- B1 $\frac{1}{6}$
Answer: $\frac{1}{6}$

Question 2Grade 53 marks

Two fair four-sided spinners, each numbered 1 to 4, are spun. The two scores are multiplied.

Work out the probability that the product is greater than 6.[3]
Show mark scheme
- M1A correct sample space of 16 products, or a systematic count
- M1Identifies the products greater than 6: 8, 8, 9, 12, 12, 16
- A1 $\frac{6}{16}$ or $\frac{3}{8}$
Answer: $\frac{3}{8}$

Question 3Grade 62 marks

A restaurant has 4 starters, 6 mains and 3 desserts. A "two-course deal" is a starter and main, or a main and dessert. How many different two-course deals are there?[2]
Show mark scheme
- M1 $4 \times 6 = 24$ or $6 \times 3 = 18$
- A1 $24 + 18 = 42$
Answer: $42$

Examiner note

Use multiplication within each type of deal, then add the two types — "or" between separate cases means add.

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

How many outcomes when two coins are flipped?
- $4$
- $2$
- $3$
- $8$
Show answer and explanation
Answer: $4$
HH, HT, TH, TT.
How many outcomes when two fair dice are rolled?
- $36$
- $12$
- $11$
- $6$
Show answer and explanation
Answer: $36$
$6 \times 6$ .
P(double six) with two fair dice?
- $\frac{1}{36}$
- $\frac{1}{6}$
- $\frac{2}{12}$
- $\frac{1}{12}$
Show answer and explanation
Answer: $\frac{1}{36}$
One cell of 36.
4 shirts and 5 pairs of trousers. How many outfits?
- $20$
- $9$
- $45$
- $10$
Show answer and explanation
Answer: $20$
$4 \times 5$ .
Two fair dice: P(total of 10)?
- $\frac{1}{12}$
- $\frac{1}{10}$
- $\frac{1}{36}$
- $\frac{1}{6}$
Show answer and explanation
Answer: $\frac{1}{12}$
(4,6), (5,5), (6,4): $\frac{3}{36}$ .
Two coins: P(at least one head)?
- $\frac{3}{4}$
- $\frac{1}{2}$
- $\frac{1}{4}$
- $\frac{2}{3}$
Show answer and explanation
Answer: $\frac{3}{4}$
HH, HT, TH out of 4.
Number of 3-letter arrangements of A, B, C, D, E with no repeats?
- $60$
- $125$
- $15$
- $120$
Show answer and explanation
Answer: $60$
$5 \times 4 \times 3$ .
Two fair dice: P(both scores even)?
- $\frac{1}{4}$
- $\frac{1}{2}$
- $\frac{1}{3}$
- $\frac{1}{6}$
Show answer and explanation
Answer: $\frac{1}{4}$
$3 \times 3 = 9$ cells of 36.

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