Sample spaces and frequency trees | AQA GCSE Maths Notes

Visual model

A sample space lists every outcome once

1

2

3

4

rows and columns prevent missed outcomes

Gold-standard guide

20 mins

What you will learn

List outcomes clearly and calculate probabilities.

Use a clear step-by-step method for sample spaces and frequency trees.

Check your answer and avoid the most common exam mistake.

Useful before you start

Core number skillsEarlier probability skillsShowing clear working

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Sample space: list all equally likely outcomes systematically — use a grid or table

Step 1

Draw a 6×6 sample space grid

Rows: die 1 (1–6)

Step 2

Identify cells where total = 8

(2,6), (3,5), (4,4), (5,3), (6,2) — that is 5 cells

Step 3

Calculate the probability

P(total = 8) = $\frac{5}{3}6$

Watch out

Students count outcomes for the same total multiple times (for example writing (3,5) and (5,3) as just one outcome)

Sample space

List every possible outcome once.

Frequency tree

Use branch totals to find missing frequencies before probabilities.

Worked example

Two fair dice are rolled. Find P(total = 8).

Draw a 6×6 sample space grid: Rows: die 1 (1–6). Columns: die 2 (1–6). Total outcomes = 36.

Identify cells where total = 8: (2,6), (3,5), (4,4), (5,3), (6,2) — that is 5 cells.

Calculate the probability: P(total = 8) = $\frac{5}{3}6$ .

Final answer

$\frac{5}{3}6$

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

Two fair dice are rolled. Find P(total = 8).

3 marks4 minssample-spaces-and-frequency-trees-worked

Show solution

Worked solution

1.Draw a 6×6 sample space grid: Rows: die 1 (1–6). Columns: die 2 (1–6). Total outcomes = 36.
2.Identify cells where total = 8: (2,6), (3,5), (4,4), (5,3), (6,2) — that is 5 cells.
3.Calculate the probability: P(total = 8) = $\frac{5}{3}6$ .

Final answer

$\frac{5}{3}6$

Mark points

M1: draw a 6×6 sample space grid
M1: identify cells where total = 8
M1: calculate the probability
A1: $\frac{5}{3}6$

Watch out

g. writing (3,5) and (5,3) as just one outcome). A full grid prevents this.Always count each ordered pair as a distinct outcome.

Diagnosticrecall

P(total = 7) when two dice are rolled.

1 mark2 minssample-spaces-and-frequency-trees-q1

Show solution

Worked solution

1.Spot the skill: Sample space: list all equally likely outcomes systematically — use a grid or table.
2.Use the draw a 6×6 sample space grid stage first, then identify cells where total = 8.
3.Keep the final answer visible: $\frac{6}{3}6$ = $\frac{1}{6}$ .

Final answer

$\frac{6}{3}6$ = $\frac{1}{6}$

Mark points

M1: use the correct sample space: list all equally likely outcomes systematically — use a grid or table.frequency trees: branch probabilities multiply along a path, all branches from one node sum to 1.p(event) = number of favourable cells / total cells.
A1: $\frac{6}{3}6$ = $\frac{1}{6}$

Watch out

g. writing (3,5) and (5,3) as just one outcome). A full grid prevents this.Always count each ordered pair as a distinct outcome.

Easyprocedure

P(at least one 6) when two dice are rolled.

2 marks3 minssample-spaces-and-frequency-trees-q2

Show solution

Worked solution

1.Spot the skill: Sample space: list all equally likely outcomes systematically — use a grid or table.
2.Use the identify cells where total = 8 stage first, then calculate the probability.
3.Keep the final answer visible: $1\frac{1}{3}6$ .

Final answer

$1\frac{1}{3}6$

Mark points

M1: use the correct sample space: list all equally likely outcomes systematically — use a grid or table.frequency trees: branch probabilities multiply along a path, all branches from one node sum to 1.p(event) = number of favourable cells / total cells.
A1: $1\frac{1}{3}6$

Watch out

g. writing (3,5) and (5,3) as just one outcome). A full grid prevents this.Always count each ordered pair as a distinct outcome.

Mediumreasoning

A bag has 4 red, 2 blue balls. Two are drawn with replacement. Find P(both red).

3 marks4 minssample-spaces-and-frequency-trees-q3

Show solution

Worked solution

1.Spot the skill: Sample space: list all equally likely outcomes systematically — use a grid or table.
2.Use the calculate the probability stage first, then draw a 6×6 sample space grid.
3.Keep the final answer visible: $\frac{4}{6}$ × $\frac{4}{6}$ = $\frac{4}{9}$ .

Final answer

$\frac{4}{6}$ × $\frac{4}{6}$ = $\frac{4}{9}$

Mark points

M1: use the correct sample space: list all equally likely outcomes systematically — use a grid or table.frequency trees: branch probabilities multiply along a path, all branches from one node sum to 1.p(event) = number of favourable cells / total cells.
A1: $\frac{4}{6}$ × $\frac{4}{6}$ = $\frac{4}{9}$

Watch out

g. writing (3,5) and (5,3) as just one outcome). A full grid prevents this.Always count each ordered pair as a distinct outcome.

Hardproblem solving

P(different colours) when drawing twice with replacement from the bag above.

3 marks5 minssample-spaces-and-frequency-trees-q4

Show solution

Worked solution

1.Spot the skill: Sample space: list all equally likely outcomes systematically — use a grid or table.
2.Use the draw a 6×6 sample space grid stage first, then identify cells where total = 8.
3.Keep the final answer visible: 2 × ( $\frac{4}{6}$ × $\frac{2}{6}$ ) = $\frac{8}{1}8$ = $\frac{4}{9}$ .

Final answer

2 × ( $\frac{4}{6}$ × $\frac{2}{6}$ ) = $\frac{8}{1}8$ = $\frac{4}{9}$

Mark points

M1: use the correct sample space: list all equally likely outcomes systematically — use a grid or table.frequency trees: branch probabilities multiply along a path, all branches from one node sum to 1.p(event) = number of favourable cells / total cells.
A1: 2 × ( $\frac{4}{6}$ × $\frac{2}{6}$ ) = $\frac{8}{1}8$ = $\frac{4}{9}$

Watch out

g. writing (3,5) and (5,3) as just one outcome). A full grid prevents this.Always count each ordered pair as a distinct outcome.

Exam-stylemulti-step

In a frequency tree: P(late) = 0.1 on a bus day. Bus day P = 0.6. Find P(late and bus day).

4 marks6 minssample-spaces-and-frequency-trees-q5

Show solution

Worked solution

1.Spot the skill: Sample space: list all equally likely outcomes systematically — use a grid or table.
2.Use the identify cells where total = 8 stage first, then calculate the probability.
3.Keep the final answer visible: 0.06.

Final answer

0.06

Mark points

M1: use the correct sample space: list all equally likely outcomes systematically — use a grid or table.frequency trees: branch probabilities multiply along a path, all branches from one node sum to 1.p(event) = number of favourable cells / total cells.
A1: 0.06

Watch out

g. writing (3,5) and (5,3) as just one outcome). A full grid prevents this.Always count each ordered pair as a distinct outcome.

Grade 9 stretchproblem solving

Spinner A shows 1, 2 or 3. Spinner B shows X or Y. All outcomes are equally likely. Find P(even number and Y).

4 marks7 minssample-space-g9

Show solution

Worked solution

1.List the six possible ordered outcomes.
2.Identify the one matching both conditions.

Final answer

$\frac{1}{6}$

Mark points

M1: identify six outcomes
A1: $\frac{1}{6}$

Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Timed checkpoint

12 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Sample spaces and frequency trees - 2 marksP(total = 7) when two dice are rolled.Mark answer

Answer

$\frac{6}{3}6$ = $\frac{1}{6}$

2Probability basics - 2 marksP(A) = 0.35. Find P(not A).Mark answer

Answer

0.65

3Relative frequency - 2 marksExpected frequency of an outcome with probability 0.4 in 250 trials.Mark answer

Answer

100

4Venn diagrams and set notation - 3 marksIn a Venn diagram, the 'only A' region has 9, 'only B' has 7, 'both' has 4. Find P(A).Mark answer

Answer

P(A) = $1\frac{3}{2}0$

Mastery check

I can explain the method for sample spaces and frequency trees.
I can show clear working without skipping key steps.
g. writing (3,5) and (5,3) as just one outcome). A full grid prevents this.Always count each ordered pair as a distinct outcome.

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