Practise 28 exam-style questions on Frequency trees and probability experiments 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
Asha thinks a coin is biased. She throws the coin 400 times and it lands on heads 232 times.
She is going to throw the coin another 750 times. Estimate the number of times the coin will land on heads.
Model answer
Relative frequency of heads \(= \frac{232}{400} = 0.58\) Estimated number of heads \(= 0.58 \times 750 = 435\)
Explanation
Use the relative frequency from the experiment as an estimate of the probability, then multiply by the number of new trials to get the expected number. The coin may be biased, so do not use 0.5.
Mark scheme
1 mark
M1 for \(\frac{232}{400} \times 750\)
oe, e.g. 0.58 × 750
1 mark
A1 for 435
cao; correct answer with no working scores full marks
Question 2
2 marks
Jamal throws a coin 5000 times. It lands on heads 2780 times.
Jamal says, "The coin is fair." Is Jamal correct? You must give a reason for your answer.
Model answer
Relative frequency of heads \(= \frac{2780}{5000} = 0.556\) For a fair coin we would expect about 0.5, i.e. about 2500 heads. 5000 is a large number of throws, so the relative frequency should be very close to the probability. 0.556 is too far from 0.5 for that, so the coin is probably biased towards heads. Jamal is probably not correct.
Explanation
With a large number of trials, the relative frequency should be very close to the true probability. 280 more heads than expected in 5000 throws is a big difference, so the coin is likely to be biased. With only a few throws you could not draw this conclusion.
Mark scheme
1 mark
M1 for the relative frequency of heads, \(\frac{2780}{5000}\) = 0.556, or for the expected number of heads for a fair coin, 2500
1 mark
C1 for No (the coin is probably biased), because 0.556 is well above 0.5 (or 2780 is well above 2500) and 5000 is a large number of throws
must refer to the large number of throws as well as the comparison
Question 3
3 marks
Holly rolls a dice 50 times. The relative frequency of getting a six is 0.24.
She rolls the dice another 50 times. The relative frequency of getting a six in all 100 rolls is 0.19.
How many sixes did Holly get in the second 50 rolls?
Model answer
Sixes in the first 50 rolls = 0.24 × 50 = 12 Sixes in all 100 rolls = 0.19 × 100 = 19 Sixes in the second 50 rolls = 19 − 12 = 7
Explanation
Turn each relative frequency back into a number of sixes by multiplying by the number of rolls it is based on. The second relative frequency is for all 100 rolls, not for the second 50, so subtract to find the sixes in the second set. Averaging the relative frequencies does not work.
Mark scheme
1 mark
P1 for the number of sixes in the first 50 rolls: 0.24 × 50 = 12
correct answer with no working scores full marks
1 mark
P1 for the number of sixes in all 100 rolls: 0.19 × 100 = 19
1 mark
A1 for 7
cao
Question 4
4 marks
150 people were asked if they wear glasses. 70 of the people were women. The probability that a person chosen at random from the 150 people is a woman who wears glasses is 0.22. The probability that a man chosen at random from the men wears glasses is 0.35.
One of the people who wear glasses is chosen at random. Work out the probability that this person is a woman.
Model answer
Women who wear glasses = 0.22 × 150 = 33 Men = 150 − 70 = 80. Men who wear glasses = 0.35 × 80 = 28 Total who wear glasses = 33 + 28 = 61 P(woman, given the person wears glasses) \(= \frac{33}{61}\) (= 0.540...)
Explanation
Read each probability carefully: 0.22 is a fraction of all 150 people, but 0.35 is a fraction of the 80 men only. Change both into numbers of people, as on a frequency tree. The person is chosen from those who wear glasses, so divide by 61.
Mark scheme
1 mark
P1 for the number of women who wear glasses: 0.22 × 150 = 33
correct answer with no working scores full marks
1 mark
P1 for the number of men who wear glasses: 0.35 × 80 = 28
using 150 − 70 = 80 men
1 mark
P1 for the total number who wear glasses: 33 + 28 = 61
1 mark
A1 for \(\frac{33}{61}\)
oe; accept 0.541 (awrt 0.541)
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