Relative frequency and the probability scale - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Relative frequency and the probability scale 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
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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

Here are seven numbers.
\(0.7 \qquad 1.3 \qquad -0.2 \qquad \dfrac{3}{2} \qquad \dfrac{2}{3} \qquad 0 \qquad 105\%\)

Write down all the numbers that cannot be probabilities.

Model answer

A probability must be between 0 and 1 inclusive.
1.3 > 1, −0.2 < 0, \(\dfrac{3}{2} = 1.5 > 1\), 105% = 1.05 > 1
So 1.3, −0.2, \(\dfrac{3}{2}\) and 105% cannot be probabilities.

Explanation

Every probability lies on the scale from 0 (impossible) to 1 (certain). Convert fractions and percentages to decimals to check them; 0 is allowed because it is the probability of an impossible event.

Mark scheme

2 marks
B2 for 1.3, −0.2, \(\dfrac{3}{2}\) and 105% and no others
B1 for two or three of these with no incorrect numbers, or all four with one incorrect number

Question 2

2 marks

In a survey of 80 workers in a town, 28 said that they cycle to work.
There are 12 000 workers in the town.

Work out an estimate for the number of workers in the town who cycle to work.

Model answer

Relative frequency = \(\dfrac{28}{80} = 0.35\)
Estimate = 0.35 × 12 000 = 4200

Explanation

The survey result gives a relative frequency, which estimates the probability that a worker chosen at random cycles. Multiply by the number of workers in the town. It is only an estimate, and it relies on the sample being representative of the town.

Mark scheme

1 mark
M1 for \(\dfrac{28}{80} \times 12\,000\) oe
relative frequency 0.35 seen implies M1 if used with 12 000
1 mark
A1 for 4200
cao

Question 3

3 marks

Sami rolls a biased dice.
After 40 rolls, the relative frequency of a six is 0.3
He rolls the dice 20 more times.
After all 60 rolls, the relative frequency of a six is 0.35

How many sixes did Sami roll in the last 20 rolls?

Model answer

Sixes in the first 40 rolls: 0.3 × 40 = 12
Sixes in all 60 rolls: 0.35 × 60 = 21
Sixes in the last 20 rolls: 21 − 12 = 9

Explanation

Relative frequency × number of trials gives the frequency, so turn both relative frequencies back into numbers of sixes. The difference is the number of sixes in the extra rolls. You cannot just average 0.3 and 0.35 because the groups are different sizes.

Mark scheme

1 mark
P1 for the number of sixes in the first 40 rolls, 0.3 × 40 = 12
correct answer with no working scores full marks
1 mark
P1 for the number of sixes in 60 rolls, 0.35 × 60 = 21
1 mark
A1 for 9
cao

Question 4

4 marks

A bag contains \(n\) counters. 6 of the counters are red.
Ana takes a counter at random, records its colour and puts it back. After a large number of trials, the relative frequency of taking a red counter is 0.2

Ana then puts 5 more red counters into the bag.
Work out an estimate for the probability that a counter taken at random from the bag is now red.

Model answer

\(\dfrac{6}{n} \approx 0.2\), so \(n \approx 6 \div 0.2 = 30\)
After adding 5 red counters: 6 + 5 = 11 red counters out of 30 + 5 = 35
Estimated P(red) = \(\dfrac{11}{35}\)

Explanation

The relative frequency after many trials is a good estimate of the probability, which is the fraction of the counters that are red. That lets you estimate the total. Adding red counters increases both the number of reds and the total number of counters.

Mark scheme

1 mark
P1 for an estimate of the number of counters, 6 ÷ 0.2 = 30
oe, e.g. \(\dfrac{6}{n} = 0.2\); correct answer with no working scores full marks
1 mark
P1 for 11 red counters
1 mark
P1 for 35 counters in total
ft their 30
1 mark
A1 for \(\dfrac{11}{35}\)
oe

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