GCSE Maths · Probability

Sample space diagrams GCSE Questions and Worked Answers

A sample space lists all possible outcomes. For two events, use rows for the first result and columns for the second to keep every ordered pair. If all cells are equally likely, probability = favourable cells ÷ total cells.

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Start with the meaning

What you need to know about sample space diagrams

Toss one coin and then another. Record both results in order: HH, HT, TH or TT, where H means heads and T means tails. This complete list is the sample space. Listing systematically matters: 'one head' can happen in two different ways, HT and TH.

See the idea first

One row and one column identify one complete outcome

For two spinners, put the possible first results down the side and second results across the top. Each cell then records one ordered pair. In the table there are 2 × 3 = 6 pairs. If both spinners are fair with equal-sized labelled sectors and the spins are independent, all six cells are equally likely. You can then count favourable cells to calculate probability.

Independent fair spinner outcomes: first 1,2; second 1,2,3. Cells show ordered pairs, not sums.
First / second123
1(1,1)(1,2)(1,3)
2(2,1)(2,2)(2,3)
From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

List outcomes

What the problem asks: Toss two fair coins independently. List all outcomes.

How to solve it: Use HH, HT, TH, TT. Keep the order so no possible route disappears.

Count a condition

What the problem asks: For first spinner 1,2 and second spinner 1,2,3, both fair and independent, find the probability of a total of 3.

How to solve it: The favourable pairs are (1,2) and (2,1), two of the six equally likely cells. Probability = 2/6 = 1/3.

Check whether counting works

What the problem asks: Can you treat dice sums 2,3,…,12 as eleven equally likely outcomes?

How to solve it: No. Sum 2 has one ordered pair, (1,1); sum 7 has six. Count the 36 equally likely die pairs, not the eleven different sum labels.

A reliable routine

Use a sample-space table to calculate probability

Use this for a manageable number of combined outcomes. A systematic grid avoids missing or repeating ordered pairs. Favourable-count divided by total-count is valid only when those pairs are equally likely.

  1. List every possible first result and every possible second result.
  2. Fill one cell for each ordered pair; calculate a sum or product if requested.
  3. Check that fairness and independence justify equally likely cells, or use given weights instead.
  4. Count the cells meeting the condition and divide by the total number of equally likely cells.

Check: Repeated sum or product values do not collapse into one outcome. Each cell represents a different route and must retain its probability.

Fully worked

Sample space diagrams GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Two fair coins

2 marks
Question

Two fair coins are tossed independently. Find the probability of exactly one head.

The equally likely outcomes are HH, HT, TH and TT. Exactly one head occurs in HT and TH.

P=2/4=1/2P=2/4=1/2
Example 2

Coin and die

3 marks
Question

A fair coin and fair six-sided die are used independently. Find the probability of heads and a number greater than 4.

There are 2 × 6 = 12 equally likely ordered outcomes. The favourable outcomes are (H,5) and (H,6).

P=2/12=1/6P=2/12=1/6

Both conditions must hold.

Example 3

Two spinner sums

3 marks
Question

One fair spinner has sectors 1, 2. Another has sectors 1, 2, 3. They are spun independently. Find the probability their sum is at least 4.

The six pairs are (1,1), (1,2), (1,3), (2,1), (2,2), (2,3). Sums at least 4 occur in (1,3), (2,2), (2,3).

P=3/6=1/2P=3/6=1/2

'At least' includes 4.

Example 4

Two dice

3 marks
Question

Two fair six-sided dice are rolled independently. Find the probability of a total of 9.

There are 6 × 6 = 36 equally likely ordered pairs. Total 9: (3,6), (4,5), (5,4), (6,3).

P=4/36=1/9P=4/36=1/9

(3,6) and (6,3) are different cells.

Example 5

Products

3 marks
Question

Two independent fair spinners each have sectors 1, 2, 3. Find the probability the product is even.

An even product requires at least one 2, because the other possible values, 1 and 3, are odd and odd × odd is odd. The favourable pairs are (1,2), (2,1), (2,2), (2,3), (3,2): five cells out of nine.

P=5/9P=5/9

Count (2,2) only once.

Example 6

Unequal outcomes

3 marks
Question

Two independent coins each have P(heads) = 0.7. Explain why counting one HH cell out of four does not give its probability, and calculate P(HH).

The four outcome labels remain HH, HT, TH, TT, but are not equally likely. Independence lets us multiply the given probabilities:

P(HH)=0.7×0.7=0.49P(HH)=0.7\times0.7=0.49

It is not 1/4.

10 original questions · total 23 marks

Sample space diagrams GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 28 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed
1

Three coins

2 marks

Three fair coins are tossed independently. How many ordered outcomes are possible?

Show worked answer
2×2×2=82\times2\times2=8

They are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

2

Exactly two heads

2 marks

Three fair coins are tossed independently. Find the probability of exactly two heads.

Show worked answer

Favourable outcomes are HHT, HTH and THH, among eight equally likely outcomes.

P=3/8P=3/8
3

Coin and die

2 marks

A fair coin and fair die are used independently. Find P(tails and an even die result).

Show worked answer

Favourable outcomes: (T,2), (T,4), (T,6), among 12.

P=3/12=1/4P=3/12=1/4
4

Dice sum 4

2 marks

Two fair dice are rolled independently. Find P(total 4).

Show worked answer

Pairs (1,3), (2,2), (3,1) give 4.

P=3/36=1/12P=3/36=1/12
5

Dice sum 7

2 marks

Two fair dice are rolled independently. Find P(total 7).

Show worked answer

Pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

P=6/36=1/6P=6/36=1/6
6

Matching dice

2 marks

Two fair dice are rolled independently. Find the probability of matching numbers.

Show worked answer

The six doubles (1,1) through (6,6) are favourable.

P=6/36=1/6P=6/36=1/6
7

At least one six

3 marks

Two fair dice are rolled independently. Find P(at least one 6).

Show worked answer

Six cells have first die 6 and six have second die 6, but (6,6) is in both lists.

6+61=116+6-1=11 P=11/36P=11/36
8

Spinner sum

3 marks

Fair independent spinners have labels 1,2,3 and 2,4. Find P(sum greater than 5).

Show worked answer

The pairs and sums are (1,2) → 3, (1,4) → 5, (2,2) → 4, (2,4) → 6, (3,2) → 5, (3,4) → 7. Only (2,4) and (3,4) give sums greater than 5.

P=2/6=1/3P=2/6=1/3
9

Repeated labels

3 marks

A fair four-sector spinner has labels 1,1,2,3. Find P(1), and explain why counting distinct labels fails.

Show worked answer

Two of the four equal-sized sectors have label 1, so

P(1)=2/4=1/2P(1)=2/4=1/2

The three distinct labels are not equally likely. Preserve the two separate sectors labelled 1.

10

Check a claim

2 marks

A student says a total of 2 is as likely as a total of 6 with two fair independent dice, because each is one number. Explain.

Show worked answer

Total 2 has only (1,1), while total 6 has (1,5), (2,4), (3,3), (4,2), (5,1). Their probabilities are 1/36 and 5/36, so they are not equally likely.

Examiner-style feedback

Common sample space diagrams mistakes

Losing order

(2,5) and (5,2) are separate die outcomes.

Counting distinct totals

Equal likelihood belongs to the ordered cells, not usually to their sum labels.

Ignoring unequal chances

A biased spinner or coin needs probability weights; counting labels alone is not enough.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

  1. List every ordered pair.
  2. Preserve repeated results in different cells.
  3. Check equal likelihood.
  4. Count exactly the condition asked.
Quick answers

Sample space diagrams FAQ

Must a sample space be a table?

No. A list or a tree can also enumerate every outcome. A table is convenient for two sets of results.

Why do we need independence?

For fair devices used independently, every pair has the same product probability. Without independence, equally likely individual results do not guarantee equally likely pairs.

Build connected skills

What to revise next

Content standards

Curriculum and rights review

P7 sample spaces and theoretical probabilities across tiers. Counting cells is used only when their outcomes are equally likely. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references