Sample spaces - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Sample spaces for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
Instant marking
No sign-up needed

20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Two fair six-sided dice are rolled and the two scores are added together.
The table shows all the possible totals.

Work out the probability that the total is 8.

Table with columns "Dice 1 / Dice 2", "1", "2", "3", "4", "5", "6". Dice 1 / Dice 2: 1; 1: 2; 2: 3; 3: 4; 4: 5; 5: 6; 6: 7 | Dice 1 / Dice 2: 2; 1: 3; 2: 4; 3: 5; 4: 6; 5: 7; 6: 8 | Dice 1 / Dice 2: 3; 1: 4; 2: 5; 3: 6; 4: 7; 5: 8; 6: 9 | Dice 1 / Dice 2: 4; 1: 5; 2: 6; 3: 7; 4: 8; 5: 9; 6: 10 | Dice 1 / Dice 2: 5; 1: 6; 2: 7; 3: 8; 4: 9; 5: 10; 6: 11 | Dice 1 / Dice 2: 6; 1: 7; 2: 8; 3: 9; 4: 10; 5: 11; 6: 12.

Model answer

There are 6 × 6 = 36 equally likely outcomes.
A total of 8 appears 5 times: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
P(total = 8) = \(\dfrac{5}{36}\)

Explanation

Each cell of the sample space is one equally likely outcome. Count the cells containing 8 and divide by the 36 cells; the totals themselves (2 to 12) are not equally likely.

Mark scheme

1 mark
M1 for counting 5 outcomes with a total of 8, or a fraction with denominator 36
correct answer with no working scores full marks
1 mark
A1 for \(\dfrac{5}{36}\)
cao

Question 2

2 marks

Two fair six-sided dice are rolled.

Work out the probability that at least one of the dice shows a 6.

Model answer

In the 6 × 6 sample space, the outcomes with a 6 are the 6 in the last row and the 6 in the last column, but (6, 6) is in both.
Number of outcomes = 6 + 6 − 1 = 11
P(at least one 6) = \(\dfrac{11}{36}\)

Explanation

Count from the sample space carefully: (6, 6) must only be counted once. Alternatively, there are 5 × 5 = 25 outcomes with no 6, so the answer is \(1 - \dfrac{25}{36}\).

Mark scheme

1 mark
M1 for 11 outcomes out of 36, or \(1 - \dfrac{25}{36}\)
correct answer with no working scores full marks
1 mark
A1 for \(\dfrac{11}{36}\)
cao; \(\dfrac{12}{36}\) (counting (6, 6) twice) scores 0

Question 3

3 marks

Two fair spinners each have four equal sections numbered 1, 2, 3 and 4.
Both spinners are spun. The score is the higher of the two numbers (if the numbers are the same, the score is that number).
The table shows some of the possible scores.

Work out the probability that the score is 4.

Model answer

Completed rows: first spinner 3: 3, 3, 3, 4; first spinner 4: 4, 4, 4, 4
There are 4 × 4 = 16 equally likely outcomes.
Score 4: one 4 in each of the first three rows and all four in the last row = 7
P(score = 4) = \(\dfrac{7}{16}\)

Explanation

The score is 4 whenever at least one spinner shows 4. Completing the grid shows this happens 4 + 4 − 1 = 7 times, because (4, 4) must only be counted once.

Mark scheme

1 mark
M1 for completing the rows: 3, 3, 3, 4 and 4, 4, 4, 4
or a full list of the 16 outcomes; correct answer with no working scores full marks
1 mark
M1 for counting 7 outcomes with a score of 4
1 mark
A1 for \(\dfrac{7}{16}\)
cao

Question 4

4 marks

Spinner A has 5 equal sections numbered 1 to 5.
Spinner B has \(n\) equal sections numbered 1 to \(n\), where \(n\) is at least 5.
Both spinners are spun.
The probability that both spinners land on the same number is \(\dfrac{1}{6}\).

Work out the value of \(n\).

Model answer

There are \(5 \times n = 5n\) equally likely outcomes.
The numbers are the same for (1, 1), (2, 2), (3, 3), (4, 4), (5, 5): 5 outcomes (since \(n \ge 5\)).
\(\dfrac{5}{5n} = \dfrac{1}{6}\)
\(\dfrac{1}{n} = \dfrac{1}{6}\), so \(n = 6\)

Explanation

Think of the sample space as a grid with 5 rows and \(n\) columns. Matching numbers lie on the 'diagonal', and there are only 5 of them because spinner A stops at 5. Setting up an equation for the probability gives \(n\).

Mark scheme

1 mark
P1 for the number of outcomes in the sample space, \(5n\)
correct answer with no working scores full marks
1 mark
P1 for 5 outcomes where the numbers are the same, (1, 1) to (5, 5)
1 mark
P1 for \(\dfrac{5}{5n} = \dfrac{1}{6}\) oe
e.g. \(\dfrac{1}{n} = \dfrac{1}{6}\)
1 mark
A1 for \(n = 6\)
cao

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

AI Tutor Hints first, answers after you try

Hi! Stuck? Ask me for a hint and I’ll help you work it out. Once you’ve checked your answer I can explain exactly where the marks are.