Probability

Relative frequency and expected outcomes

11 past-paper questions, easiest first. Try each one, then reveal the answer.

Foundation & Higher8 of 11 from Foundation papers

Revision guide by The Full Marks

Relative frequency and expected outcomes: what you need to know

Relative frequency is a probability estimated from an experiment rather than worked out in theory. You use it, and theoretical probability, to predict how many times something should happen, which is called the expected frequency.

Key facts

Relative frequency $= \frac{\text{number of times it happened}}{\text{total number of trials}}$ .
Expected frequency $=$ probability $\times$ number of trials.
The more trials you do, the more reliable a relative frequency is as an estimate. It gets closer to the true probability.
If results are far from what a fair dice or coin would give over many trials, it may be biased.
Frequency trees split a total into branches; the numbers on each pair of branches add up to the number before them.

Expected number of outcomes

Find the probability (from theory, a table or a relative frequency).
Multiply it by the number of trials.
It is an estimate, so it does not have to be a whole number; do not round unless the question asks.

Estimating a probability from results

Find the total number of trials.
Divide the frequency of the outcome by that total.
Use the largest experiment available: it is the most reliable.

Worked examples

Example 1 Foundation · grade 2

A fair dice is rolled 120 times. How many times would you expect it to land on 5?

Show the solution

$P(5) = \frac{1}{6}$
Expected number = $\frac{1}{6} \times 120 = 20$

Example 2 Foundation · grade 3

80 people are asked if they wear glasses. 35 are men. 12 of the men and 20 of the women wear glasses. (a) Complete a frequency tree. (b) One person is chosen at random. Find the probability they wear glasses.

Show the solution

(a) 80 splits into men 35 and women $80 - 35 = 45$
Men: glasses 12, no glasses $35 - 12 = 23$
Women: glasses 20, no glasses $45 - 20 = 25$
(b) Glasses: $12 + 20 = 32$ , so $P = \frac{32}{80} = \frac{2}{5}$

Example 3 Foundation · grade 5

Priya spins a biased spinner 100 times. It lands on 1 eighteen times, 2 twenty-five times, 3 forty-two times and 4 fifteen times.
(a) Estimate the probability it lands on 3.
(b) She spins it 250 more times. Estimate how many times it lands on 3.
(c) Estimate the probability it lands on 3 on two spins in a row.

Show the solution

(a) Relative frequency = $\frac{42}{100} = 0.42$
(b) $0.42 \times 250 = 105$ times
(c) The spins are independent, so multiply: $0.42 \times 0.42 = 0.1764$

Common mistakes

Using the theoretical probability (for example $\frac{1}{4}$ for four sections) when the spinner is biased. Use the results instead.
Giving a probability when the question asks 'how many times'. Expected frequency is a count, not a fraction.
Saying 'do more trials' without saying why: more trials make the estimate more reliable.
Averaging two relative frequencies to combine experiments. Add the successes and add the trials instead, e.g. $\frac{7 + 30}{10 + 50} = \frac{37}{60}$ .

Exam tips

When asked whose results are more reliable, pick the one with more trials and say that is the reason.
In frequency tree questions, check each pair of branches adds up to the total above it before you answer the probability part.

Questions people ask

What is the difference between relative frequency and probability?

Probability is what should happen in theory; relative frequency is what did happen in an experiment. With lots of trials they get close to each other.

How do you calculate expected frequency?

Multiply the probability of the outcome by the number of times the experiment is done. For example, $0.3 \times 50 = 15$ .

Exam questions: Relative frequency and expected outcomes (11, easiest first)

Relative frequency and expected outcomes4 marksGrade 2CalculatorFoundation

November 2018 · Paper 3 Foundation · Q14

Complete a probability table for a spinner, find the least likely outcome and estimate a frequency from 60 spins.

Show answer

(a)	$0.3$	1 mark
(b)	$4$	1 mark
(c)	$12$	2 marks

Worked solution Checked against the mark scheme

(a) Probabilities add to 1: $1 - (0.2 + 0.4 + 0.1) = 0.3$

(b) The smallest probability is 0.1, which is for 4.

Watch out: In (b), writing the probability 0.1 instead of the number 4.

Relative frequency and expected outcomes5 marksGrade 3CalculatorFoundation

November 2023 · Paper 3 Foundation · Q15

Complete a frequency tree for holiday sales and find a probability from it.

Show answer

(a)	Families 64 (UK 57, abroad 7); couples 36 (UK 25, abroad 11)	3 marks
(b)	$\frac{57}{64}$	2 marks

Worked solution Checked against the mark scheme

(a) Couples $= 100 - 64 = 36$ . Couples abroad $= 11$ , so couples UK $= 36 - 11 = 25$ .

Families abroad $= 18 - 11 = 7$ , so families UK $= 64 - 7 = 57$ .

(b) Not abroad means UK: $\frac{57}{64}$

Watch out: In (b), dividing by 100 instead of by the 64 families.

Relative frequency and expected outcomes2 marksGrade 3CalculatorOn both tiers

November 2023 · Paper 3 Foundation · Q21

Estimate how often a biased spinner lands on one letter over many spins.

Show answer

Answer	80	2 marks

Worked solution Checked against the mark scheme

Expected number $=$ probability × number of spins

$0.4 \times 200 = 80$

Watch out: Giving the answer as a probability, 80/200.

Relative frequency and expected outcomes2 marksGrade 3CalculatorFoundation

November 2020 · Paper 3 Foundation · Q21

Estimate the expected number of heads for a fair coin and say which of two experiments is more reliable.

Show answer

(a)	25	1 mark
(b)	Simon, because he made more throws	1 mark

Worked solution Checked against the mark scheme

(a) $P(\text{heads}) = \frac{1}{2}$ , so $\frac{1}{2} \times 50 = 25$

(b) Simon: more trials give a more reliable estimate of the probability.

Watch out: In (b), just stating the number 100 without explaining why more throws helps.

Relative frequency and expected outcomes4 marksGrade 3No calculatorFoundation

November 2019 · Paper 1 Foundation · Q15

Fill in a frequency tree from given facts about males, females, sitting and standing, then find a probability from it.

Show answer

(a)	Male 45 (sit 16, stand 29); female 75 (sit 14, stand 61)	3 marks
(b)	$\frac{29}{120}$	1 mark

Worked solution Checked against the mark scheme

(a) Males: $120 - 75 = 45$

Males sitting: $45 - 29 = 16$

Females sitting: $30 - 16 = 14$

Females standing: $75 - 14 = 61$

(b) 29 of the 120 people are males who stand, so the probability is $\frac{29}{120}$ .

Watch out: Putting all 30 sitters on the female branch.

Relative frequency and expected outcomes2 marksGrade 4No calculatorOn both tiers

November 2022 · Paper 1 Foundation · Q23

Use survey results to estimate how many of one car model a factory should build.

Show answer

Answer	$7500$	2 marks

Worked solution Checked against the mark scheme

15 out of 80 people chose B.

$\frac{15}{80} \times 40\,000 = 7500$

Watch out: Dividing 40 000 by 4 models.

Relative frequency and expected outcomes2 marksGrade 4No calculatorFoundation

June 2019 · Paper 1 Foundation · Q17

Decide which of two sets of coin throws gives a better probability estimate and combine them into one estimate.

Show answer

((i))	Maxine, because she threw the coin more times	1 mark
((ii))	$\frac{37}{60}$	1 mark

Worked solution Checked against the mark scheme

(i) Maxine: she made more throws (50 against 10), and more trials give a more reliable estimate.

(ii) Combine all the results: $7 + 30 = 37$ Tails out of $10 + 50 = 60$ throws.

Estimate $= \frac{37}{60}$

Watch out: In (ii), averaging the two fractions $\frac{7}{10}$ and $\frac{30}{50}$ instead of pooling the results.

Relative frequency and expected outcomes4 marksGrade 5CalculatorHigher

June 2023 · Paper 2 Higher · Q10

Use relative frequency from a table to estimate the chance of two sixes in a row, and say how many throws would make the estimate more reliable.

Show answer

(a)	$\frac{225}{3600} = \frac{1}{16}$	3 marks
(b)	$n$ must be greater than 60	1 mark

Worked solution Checked against the mark scheme

(a) P(6) $\approx \frac{15}{60} = \frac{1}{4}$

P(6 both times) $= \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} = 0.0625$

(b) More throws give a more reliable estimate, so $n$ must be greater than 60.

Watch out: In (b), just saying 'n must be bigger' without comparing it to 60.

Relative frequency and expected outcomes4 marksGrade 5No calculatorHigher

November 2022 · Paper 1 Higher · Q10

Use a spinner's results to estimate the chance of two fives in a row and the percentage of ones expected over many spins.

Show answer

(a)	$\frac{1}{16}$	2 marks
(b)	$15\%$	2 marks

Worked solution Checked against the mark scheme

(a) A 5 came up 10 times out of 40, so $P(5) \approx \frac{10}{40} = \frac{1}{4}$

Two fives: $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

(b) A 1 came up 6 times out of 40: $\frac{6}{40} = 0.15 = 15\%$

Watch out: Using $\frac{1}{5}$ because the spinner has five sides, ignoring the bias.

Relative frequency and expected outcomes5 marksGrade 5No calculatorFoundation

Specimen · Paper 1 Foundation · Q25

Judge which set of coin results gives the best probability estimate, test a claim about heads and tails, and estimate the chance of two heads.

Show answer

(a)	Sharif, because he threw the coin the most times	1 mark
(b)	No: Paul's own results give heads twice as often, but overall 300 heads to 100 tails is three times as likely	2 marks
(c)	$\frac{9}{16}$	2 marks

Worked solution Checked against the mark scheme

(a) Sharif threw 160 times, more than anyone else, so his results are the most reliable.

(b) Paul got $80 : 40 = 2 : 1$ , but all results together give $300$ heads and $100$ tails, $3 : 1$ . So heads is about three times as likely, not twice: Paul is not correct.

P(two heads) $= \frac34 \times \frac34 = \frac{9}{16}$

Watch out: In (c), adding 3/4 + 3/4 instead of multiplying.

Relative frequency and expected outcomes5 marksGrade 5No calculatorHigher

Specimen · Paper 1 Higher · Q6

Choose the most reliable set of coin results, test a claim about heads being twice as likely, and estimate the probability of two heads.

Show answer

(a)	Sharif, because he threw the coin the most times	1 mark
(b)	No: over all the results heads is 3 times as likely as tails (300 heads, 100 tails); only Paul's own results give 2 to 1	2 marks
(c)	$\frac{9}{16}$	2 marks

Worked solution Checked against the mark scheme

(a) Sharif threw the coin 160 times, more than anyone else, so his results are the most reliable.

(b) Altogether there were 300 heads and 100 tails, a ratio of 3 : 1. Heads is about three times as likely, not twice, so Paul is wrong.

P(two heads) $= \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$

Watch out: In (c), adding the probabilities instead of multiplying them.

Also tested in these questions

Their main topic is different, but part of each one tests this topic.

November 2024 · Paper 1 Foundation · Q11: Fill in a two-way table about favourite film types and find a probability from it.
November 2022 · Paper 3 Higher · Q19: Estimate how often a green counter is picked in repeated trials using a probability and a ratio.
June 2022 · Paper 1 Foundation · Q17: Complete a frequency tree from given totals and find a probability from it.
November 2019 · Paper 1 Higher · Q11: Use the proportion of yellow counters in a sample to estimate how many counters are in the bag.
June 2019 · Paper 2 Foundation · Q16: Compare probabilities given as a decimal, fraction and percentage, judge a claim about heads and tails, and estimate an expected number of heads.