Sample spaces - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Sample spaces for Edexcel GCSE Maths (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
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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.

Work out the probability that the total is 9.

Model answer

There are 6 × 6 = 36 equally likely outcomes.
Total of 9: (3, 6), (4, 5), (5, 4), (6, 3), so 4 outcomes
P(total is 9) = \(\frac{4}{36} = \frac{1}{9}\)

Explanation

A sample space grid of the two dice shows all 36 equally likely outcomes. Count the cells with a total of 9, remembering that (3, 6) and (6, 3) are different outcomes because the dice are different.

Mark scheme

1 mark
M1 for 36 outcomes (e.g. a sample space) or for identifying the 4 ways (3, 6), (4, 5), (5, 4), (6, 3)
correct answer with no working scores full marks
1 mark
A1 for \(\frac{4}{36}\) oe
accept \(\frac{1}{9}\) or 0.111...

Question 2

3 marks

There are five cards numbered 1, 2, 3, 4 and 5.
Leo takes two of the cards at random.

Work out the probability that the sum of the numbers on his two cards is odd.

Model answer

Possible pairs: 12, 13, 14, 15, 23, 24, 25, 34, 35, 45 (10 equally likely pairs)
Odd sum: 12, 14, 23, 25, 34, 45 (6 pairs)
P(sum is odd) = \(\frac{6}{10} = \frac{3}{5}\)

Explanation

List the pairs systematically so none is missed or repeated: all pairs starting with 1, then new pairs starting with 2, and so on. A sum is odd only when one card is odd and the other even, which gives a quick check: 3 odd cards × 2 even cards = 6 pairs.

Mark scheme

1 mark
M1 for a systematic list of the pairs, at least 7 correct with no repeats, or for 10 possible pairs (or 20 ordered pairs)
correct answer with no working scores full marks
1 mark
M1 for identifying the 6 pairs with an odd sum: 1 2, 1 4, 2 3, 2 5, 3 4, 4 5 (or 12 ordered pairs)
1 mark
A1 for \(\frac{6}{10}\) oe
accept \(\frac{12}{20}\), \(\frac{3}{5}\) or 0.6

Question 3

3 marks

Two fair four-sided spinners are each numbered 1, 2, 3 and 4.
Both spinners are spun.
The score is the larger of the two numbers. If the two numbers are the same, the score is that number.

Work out the probability that the score is 3.

Model answer

There are 4 × 4 = 16 equally likely outcomes.
Score 3 (larger number 3): (1, 3), (2, 3), (3, 3), (3, 1), (3, 2) - 5 outcomes
P(score is 3) = \(\frac{5}{16}\)

Explanation

Draw a 4 × 4 sample space and write the larger number in each cell. The score is 3 when one spinner shows 3 and the other shows 3 or less; (3, 3) is only one outcome, so be careful not to count it twice.

Mark scheme

1 mark
M1 for a sample space of 16 outcomes
correct answer with no working scores full marks
1 mark
M1 for identifying the 5 outcomes with score 3: (1, 3), (2, 3), (3, 3), (3, 1), (3, 2)
at least 4 correct, none wrong
1 mark
A1 for \(\frac{5}{16}\) oe
accept 0.3125

Question 4

5 marks

Ivy has an ordinary fair six-sided dice numbered 1 to 6.
She also has a fair six-sided dice with faces numbered 1, 2, 2, 3, n, n, where n is a whole number greater than 3.

Ivy rolls both dice and adds the two numbers.
The probability that the total is at least 10 is \(\frac{1}{6}\).

Work out the value of n.

Model answer

There are 6 × 6 = 36 equally likely outcomes.
P(total ≥ 10) = \(\frac{1}{6}\), so \(\frac{1}{6} \times 36 = 6\) outcomes must give a total of at least 10.
The faces 1, 2, 2, 3 cannot: the largest total is 3 + 6 = 9.
So the two faces showing n give 6 outcomes, 3 each.
With face n, the ordinary dice must show 4, 5 or 6 and not 3: n + 4 ≥ 10 and n + 3 < 10, so n = 6.
Check: n = 6 gives 6 + 4, 6 + 5, 6 + 6 for each n face, so 2 × 3 = 6 outcomes.

Explanation

Use the probability to find how many of the 36 outcomes must give a total of at least 10. Only the faces showing n can do this, and there are two of them, so each must combine with exactly 3 numbers on the ordinary dice. Trying values of n systematically (n = 5 gives 4 outcomes, n = 6 gives 6, n = 7 gives 8) also works.

Mark scheme

1 mark
P1 for 36 equally likely outcomes
correct answer with no working scores full marks
1 mark
P1 for recognising that faces 1, 2, 2 and 3 cannot give a total of at least 10 (the most is 3 + 6 = 9)
1 mark
P1 for 6 outcomes needed (\(\frac{1}{6} \times 36\)), so each face n gives 3 outcomes
1 mark
P1 for n + 4 ≥ 10 with n + 3 < 10, or a correct trial, e.g. n = 6: 6 + 4, 6 + 5, 6 + 6
or the number of outcomes 2(n − 3) = 6
1 mark
A1 for n = 6
cao

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