Maths · 9 Probability

9.2 Relative and expected frequencies

Before you start

The probability scale from 0 (impossible) to 1 (certain), and writing a probability as a fraction, a decimal or a percentage.
The probability of a single event when the outcomes are equally likely, e.g. P(a 4 on a fair dice) $= \frac{1}{6}$ .
P(not A) = 1 − P(A), and the fact that the probabilities of all the possible outcomes add up to 1.
Multiplying a whole number by a fraction or a decimal without a calculator, e.g. $200 \times \frac{3}{8}$ and $0.35 \times 180$ .
Reading a frequency table.

Syllabus objectives covered (2)

Understand relative frequency as an estimate of probability.
Calculate expected frequencies.

Part 1 of 6

Trials, outcomes, and what fair means

An experiment in probability is something you can repeat, such as spinning a spinner. Each go is one trial. What happens in a trial is its outcome. So one spin is a trial, and 'lands on red' is an outcome.
An event is an outcome, or a set of outcomes, that you are interested in. On a dice, 'an even number' is an event made of the three outcomes 2, 4 and 6.
A coin, dice or spinner is fair when every one of its possible results is equally likely: each face of the dice, each side of the coin, each section of the spinner. A fair dice gives each number with probability $\frac{1}{6}$ .
Biased means not fair. On a biased dice some numbers are more likely than others. For example, a 6 might come up more often than $\frac{1}{6}$ of the time.
At random means by chance alone, with nothing to make one result more likely than it should be. 'A counter is taken at random from a bag' means every counter in the bag has the same chance of being the one taken. You cannot predict the result of one random trial.
When the outcomes are equally likely, you find a probability by counting, as in 9.1. This is a theoretical probability.
$P(A) = \frac{number of outcomes in A}{total number of equally likely outcomes}$
Counting does not work when the outcomes are not equally likely. A biased dice, a drawing pin that lands point up or point down, a bus that is early or late: there is no list of equally likely outcomes to count. For these you need results from trials, and that is what relative frequency uses.

Part 2 of 6

Relative frequency

The frequency of an outcome is the number of times it happens. The relative frequency is that frequency as a fraction of the total number of trials.
$relative frequency = \frac{frequency of the outcome}{total number of trials}$
Example: a spinner is spun 80 times and lands on red 26 times. The relative frequency of red is $\frac{26}{80} = 0.325$ .
Divide by the total number of trials, not by the number of times the outcome did not happen. Here that is 80, not $80 - 26 = 54$ .
Relative frequency is an estimate of probability. It is the fraction of the trials in which the outcome happened, so it is your best guess at how likely the outcome is. Use it when you cannot count equally likely outcomes, or to test whether an object is fair.
Give a relative frequency as a fraction, a decimal or a percentage: $\frac{13}{40}$ , $0.325$ and $32.5 %$ are all correct. Do not give it as a ratio such as $26 : 80$ , or in words such as '26 out of 80'.
For an event made of more than one outcome, add the frequencies first. If the same spinner landed on blue 18 times, the relative frequency of 'red or blue' is $\frac{26 + 18}{80} = \frac{44}{80} = 0.55$ .
The relative frequencies of all the outcomes add up to 1, because the frequencies add up to the total number of trials. Use this to check a table, or to find a missing value.

Part 3 of 6

More trials give a better estimate

With a small number of trials, the relative frequency can be far from the probability. A fair coin thrown 10 times can give 7 heads, a relative frequency of $0.7$ , although the probability of a head is $0.5$ .
As the number of trials increases, the relative frequency tends to settle down and get closer to the probability. A short run of unusual results has less effect on a big total than on a small one.
graph diagram — not drawn yet
Axes: number of throws from 0 to 200 on the horizontal axis, relative frequency of heads from 0 to 1 on the vertical axis. A dashed horizontal line at 0.5 labelled 'probability = 0.5'. A line of joined points that starts at 1 after the first throw, swings widely between about 0.3 and 0.8 over the first 20 throws, then makes smaller and smaller swings and stays between 0.45 and 0.55 from about 120 throws to 200 throws.
The relative frequency jumps about when there are few throws, and settles near the probability as the number of throws grows.
So the most reliable estimate is the one from the most trials. If two people test the same coin, one with 20 throws and one with 500 throws, trust the 500.
To combine results from several people, add all the frequencies and add all the trials, then divide. This gives the estimate from the largest number of trials.
$combined estimate = \frac{total frequency}{total number of trials}$
Do not find the mean of the separate relative frequencies. That treats a result from 20 trials as being worth as much as a result from 500 trials. The mean gives the same answer only when everyone did the same number of trials.
Even with many trials, a relative frequency is an estimate, not the exact probability. A fair coin thrown 1000 times is unlikely to give exactly 500 heads, but the relative frequency will usually be close to $0.5$ .
A random trial has no memory. After five heads in a row, a fair coin still has probability $0.5$ of a head on the next throw. The relative frequency settles because the early results become a small part of a large total, not because the coin makes up for them.

Part 4 of 6

Is it fair? Using results to decide

To test whether a coin, dice or spinner is fair, compare your results with what a fair one would give. Work out the probability, or the expected number, for a fair object first.
A small difference is normal. A fair coin thrown 100 times might give 47 heads. The relative frequency $0.47$ is close to $0.5$ , so these results give no reason to think the coin is biased.
A large difference after many trials is evidence of bias. A dice is rolled 600 times and gives 160 sixes. A fair dice would give about $600 \times \frac{1}{6} = 100$ . The relative frequency $\frac{160}{600} \approx 0.267$ is much more than $\frac{1}{6} \approx 0.167$ , so the dice seems to be biased towards 6.
With only a few trials, you cannot decide. Two sixes in 6 rolls is a relative frequency of $\frac{1}{3}$ , but a fair dice can give this by chance. The answer is to do more trials.
There is no fixed cut-off between 'close' and 'far' at this level. Write the numbers down and say what they suggest: 'If the dice were fair, I would expect about 100 sixes. 160 is much more than 100, so the dice seems to be biased.'
Results give evidence, not proof. A very unusual set of results can still come from a fair object, so 'the results suggest the dice is biased' is a more accurate statement than 'the dice is definitely biased'.

Part 5 of 6

Expected frequency

If you know the probability of an outcome, you can predict how many times it will happen in a number of trials. This prediction is the expected frequency.
$expected frequency = probability \times number of trials$
Why multiply: a probability of $\frac{1}{6}$ means the outcome happens in about $\frac{1}{6}$ of the trials over a long run. So in 120 rolls of a fair dice you expect $\frac{1}{6}$ of 120 sixes, which is $120 \times \frac{1}{6} = 20$ .
Find the probability of the event first, then multiply. An even number on a fair dice has probability $\frac{3}{6} = \frac{1}{2}$ , so in 120 rolls you expect $120 \times \frac{1}{2} = 60$ even numbers.
An expected frequency is what happens on average, not a promise. Roll a fair dice 120 times and you may get a number of sixes a little more or a little less than 20.
An expected frequency does not have to be a whole number. For 25 throws of a fair coin, the expected number of heads is $25 \times 0.5 = 12.5$ . You cannot get 12.5 heads in one set of 25 throws, but 12.5 is the average over many sets. Give $12.5$ , and do not round it unless the question asks you to.
If a probability is missing from a table, find it first, because all the probabilities add up to 1. A spinner lands on A, B or C, with P(A) $= 0.5$ and P(B) $= 0.3$ . Then P(C) $= 1 - 0.5 - 0.3 = 0.2$ , and in 40 spins you expect $40 \times 0.2 = 8$ Cs.

Part 6 of 6

Estimating with relative frequency

When you do not know the probability, use a relative frequency in its place. Then multiply by the number of trials, as before.
Example: the spinner landed on red 26 times in 80 spins, so the estimate of P(red) is $0.325$ . In 500 more spins, the expected number of reds is $500 \times 0.325 = 162.5$ .
The same method estimates a number in a population. A population is the whole group, and a sample is the part of it that you check. In a random sample of 50 trees in a forest, 8 have a disease. The relative frequency is $\frac{8}{50} = 0.16$ . The forest has 2400 trees, so an estimate of the number with the disease is $0.16 \times 2400 = 384$ .
The sample must be random, so that every member of the population has the same chance of being in it. A sample of trees all from one corner of the forest might not represent the whole forest, and the estimate could be far out.
It works for a bag of counters too. A bag holds 12 counters. A counter is taken at random, its colour is noted, and it is put back. This is done 50 times, and blue comes up 19 times. An estimate of the number of blue counters is $\frac{19}{50} \times 12 = 4.56$ , so about 5.
A number of real objects must be whole, so round that kind of estimate to a whole number. An expected frequency, such as 162.5 reds, stays as it is.
You can also use the formula backwards. To find the number of trials, divide the expected frequency by the probability. A fair dice that is expected to give 45 sixes must be rolled $45 \div \frac{1}{6} = 270$ times.
$number of trials = \frac{expected frequency}{probability}$
To find a probability, divide the expected frequency by the number of trials. A biased coin that is expected to land heads 156 times in 240 throws has P(head) $= \frac{156}{240} = 0.65$ .

Worked examples

Example 1. A spinner has four equal sections coloured red, blue, green and yellow. Lina spins it 400 times. Her results are in the table. $\begin{matrix} Colour & red & blue & green & yellow \\ Frequency & 94 & 103 & 61 & \dots \end{matrix}$ (a) Find the frequency for yellow. (b) Find the relative frequency of green. (c) Find the relative frequency of red or blue. (d) Lina says the spinner is biased. Do her results support this? Explain your answer.

Show the working

(a) The frequencies add up to the total number of spins, 400.
$400 - (94 + 103 + 61) = 400 - 258 = 142$
(b) Green happened 61 times. Divide by the total number of trials, 400.
$\frac{61}{400} = 0.1525$
(c) 'Red or blue' is an event made of two outcomes, so add their frequencies first.
$\frac{94 + 103}{400} = \frac{197}{400} = 0.4925$
(d) Start with what a fair spinner would give. With four equal sections, each colour has probability $\frac{1}{4}$ , so in 400 spins you expect each colour about $400 \times \frac{1}{4} = 100$ times.
Red (94) and blue (103) are close to 100. Green (61) is far below 100 and yellow (142) is far above it, and 400 is a large number of spins.
The relative frequencies say the same: yellow is $\frac{142}{400} = 0.355$ , against $0.25$ for a fair spinner. So the results support Lina. The spinner seems to be biased, towards yellow and away from green.

Answer: (a) 142 (b) $0.1525$ (c) $0.4925$ (d) Yes. A fair spinner would give about 100 of each colour in 400 spins, but green came up only 61 times and yellow 142 times, so the spinner seems to be biased.

Example 2. Amir, Bea and Chen each roll the same dice and count the sixes. $\begin{matrix} Amir & Bea & Chen \\ Number of rolls & 20 & 50 & 130 \\ Number of sixes & 6 & 11 & 23 \end{matrix}$ (a) Whose results give the most reliable estimate of the probability of a six? Give a reason. (b) Use all the results to find the best estimate of the probability of a six. (c) Do the results show that the dice is biased?

Show the working

(a) The three relative frequencies are $\frac{6}{20} = 0.3$ , $\frac{11}{50} = 0.22$ and $\frac{23}{130} \approx 0.177$ . They are different, so you need to decide which one to trust.
Chen's estimate is the most reliable, because he did the most rolls. More trials give a better estimate.
(b) Combine the results: add the sixes, add the rolls, then divide.
$\frac{6 + 11 + 23}{20 + 50 + 130} = \frac{40}{200} = 0.2$
The trap is to take the mean of the three relative frequencies: $\frac{0.3 + 0.22 + 0.177}{3} \approx 0.232$ . That counts Amir's 20 rolls as much as Chen's 130 rolls, so it is not the best estimate.
(c) For a fair dice, P(six) $= \frac{1}{6} \approx 0.167$ . In 200 rolls you expect about $200 \times \frac{1}{6} \approx 33.3$ sixes. They got 40.
40 is more than 33.3, but for 200 rolls this is a small difference, and a fair dice can give it by chance. So the results do not show that the dice is biased. More rolls would give a clearer answer.

Answer: (a) Chen's, because he did the most rolls. (b) $0.2$ (c) No. $0.2$ is close to $\frac{1}{6} \approx 0.167$ , and a difference this small can happen by chance with a fair dice in 200 rolls.

Example 3. A biased spinner can land on red, blue, green or white. The table shows the probabilities. $\begin{matrix} Colour & red & blue & green & white \\ Probability & 0.35 & 0.2 & x & 2 x \end{matrix}$ (a) Find the value of $x$ . (b) The spinner is spun 300 times. Calculate the expected number of times it lands on white. (c) Calculate the expected number of times in the 300 spins that it does not land on red.

Show the working

(a) The spinner must land on one of the four colours, so the four probabilities add up to 1.
$0.35 + 0.2 + x + 2 x = 1$
Collect the terms: $0.55 + 3 x = 1$ , so $3 x = 0.45$ and $x = 0.15$ .
(b) Find the probability of white first: P(white) $= 2 x = 2 \times 0.15 = 0.3$ . Then multiply by the number of trials.
$300 \times 0.3 = 90$
(c) P(not red) $= 1 - 0.35 = 0.65$ . Multiply by the number of trials.
$300 \times 0.65 = 195$
Check: the expected number of reds is $300 \times 0.35 = 105$ , and $105 + 195 = 300$ , the total number of spins.

Answer: (a) $x = 0.15$ (b) 90 (c) 195

Example 4. A school has 840 students. In a random sample of 60 of them, 21 cycle to school. (a) Calculate an estimate of the probability that a student chosen at random from the school cycles to school. (b) Calculate an estimate of the number of students in the school who cycle to school. (c) At a second school, the probability that a student cycles to school is $0.15$ , and 102 students are expected to cycle. Find the number of students at the second school.

Show the working

(a) The sample is random, so its relative frequency is a sensible estimate for the whole school.
$\frac{21}{60} = 0.35$
(b) Use the relative frequency as the probability, and multiply by the size of the population, 840.
$0.35 \times 840 = 294$
This is an estimate. A different random sample of 60 students could give a slightly different relative frequency, and so a different answer.
(c) This time you know the probability and the expected frequency. Use the formula backwards: divide the expected frequency by the probability.
$102 \div 0.15 = 680$
Check forwards: $680 \times 0.15 = 102$ .

Answer: (a) $0.35$ (b) 294 students (c) 680 students

Where people go wrong

Mistake: Deciding that a coin or dice is biased from a few trials, e.g. '7 heads in 10 throws, so the coin is biased'.

Instead: Judge fairness only from a large number of trials. With 10 throws, say that more trials are needed.

A fair coin can give 7 heads in 10 throws by chance. A relative frequency is only a reliable estimate when the number of trials is large.

Mistake: Finding the mean of relative frequencies that come from different numbers of trials.

Instead: Add all the frequencies and all the trials, then divide: $\frac{6 + 11 + 23}{20 + 50 + 130} = 0.2$ .

The mean gives a result from 20 trials the same weight as a result from 130 trials.

Mistake: Treating an expected frequency as a result that must happen, or rounding it, e.g. changing $12.5$ expected heads to 13.

Instead: Give probability $\times$ number of trials exactly: $25 \times 0.5 = 12.5$ . The actual result of one experiment can be different.

An expected frequency is an average over many repeats of the experiment, not a prediction of one result.

Mistake: Believing that after a run of one result the other result is 'due', e.g. that a tail is more likely after five heads.

Instead: Each throw of a fair coin has P(head) = 0.5, whatever happened before.

A random trial has no memory. The relative frequency settles over many trials because the early results become a small part of the total, not because later results balance them out.

Mistake: Dividing by the wrong number for a relative frequency, e.g. 26 reds in 80 spins written as $\frac{26}{54}$ .

Instead: Divide by the total number of trials: $\frac{26}{80} = 0.325$ .

Relative frequency is the fraction of all the trials in which the outcome happened.

Check yourself

A drawing pin is dropped 150 times. It lands point up 87 times. (a) Find the relative frequency of the pin landing point up. (b) Why can you not find this probability by counting equally likely outcomes?
Answer
(a) $\frac{87}{150} = 0.58$ (b) There is no reason to think that 'point up' and 'point down' are equally likely, so there is nothing to count. The probability can only be estimated from trials.
Sami throws a coin 20 times and gets 12 heads. Tara throws the same coin 500 times and gets 260 heads. Whose results give the better estimate of the probability of a head? Write down that estimate.
Answer
Tara's, because she did more trials. Her estimate is $\frac{260}{500} = 0.52$ . (Sami's is $\frac{12}{20} = 0.6$ .)
Jo throws a coin 8 times and gets 6 heads. She says the coin must be biased. Comment on her statement.
Answer
Her results do not show this. 8 throws are too few to decide, and a fair coin can give 6 heads in 8 throws by chance. She needs many more throws.
A fair spinner has 8 equal sections. 3 of the sections are blue. The spinner is spun 200 times. Work out the expected number of times it lands on blue.
Answer
75, because P(blue) $= \frac{3}{8}$ and $200 \times \frac{3}{8} = 75$ .
A fair coin is thrown 25 times. Write down the expected number of heads. Can exactly this number of heads happen in one set of 25 throws?
Answer
$12.5$ . No: the number of heads in 25 throws must be a whole number. 12.5 is the average number of heads over many sets of 25 throws.
A bag holds 20 counters, some of them red. A counter is taken at random, its colour is noted, and it is put back. This is done 60 times, and red is taken 21 times. Work out an estimate of the number of red counters in the bag.
Answer
7. The relative frequency of red is $\frac{21}{60} = 0.35$ , and $0.35 \times 20 = 7$ .
How many times must you roll a fair dice to expect 30 sixes?
Answer
180, because $30 \div \frac{1}{6} = 30 \times 6 = 180$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Priya has a spinner with five sections numbered 1 to 5. She spins it 200 times. The table shows her results. $\begin{matrix} Number & 1 & 2 & 3 & 4 & 5 \\ Frequency & 38 & 41 & 36 & 23 & 62 \end{matrix}$

(a) Calculate the relative frequency of the spinner landing on an even number. 2 marks

Mark scheme

M1 (41 + 23) ÷ 200 oe
A1 0.32 oe, e.g. 8/25 or 32%

(b) Priya thinks that the spinner is fair. Do her results support this? Explain your answer. 2 marks

Mark scheme

B1 40 (or 0.2) seen as the expected frequency (or probability) of each number if the spinner were fair, or a statement that a fair spinner would give about the same frequency for each number
B1 No (the spinner seems biased), with a reason that compares the data with the fair value, e.g. 62 is much more than 40, or 23 is much less than 40, or 0.31 or 0.115 compared with 0.2, or 62 compared with 23 when these should be about equal

(c) Priya spins the spinner another 600 times. Use her results in the table to calculate an estimate of the number of times it lands on 5. 2 marks

Mark scheme

M1 62/200 × 600 oe, or 0.31 × 600
A1 186 cao

(d) State one way in which Priya could make her estimates of the probabilities more reliable. 1 mark

Mark scheme

B1 Spin the spinner more times oe, e.g. combine the 200 results with the results of the 600 extra spins

Question 2. Nadia's bus can be early, on time or late. The table shows the probability of each. $\begin{matrix} early & on time & late \\ Probability & 0.12 & 0.53 & \dots \end{matrix}$

(a) Find the probability that the bus is late. 1 mark No calculator

Mark scheme

B1 0.35 oe

(b) Nadia takes the bus on 180 days. Calculate the expected number of days on which the bus is late. 2 marks No calculator

Mark scheme

M1 180 × their (a)
A1 63, FT their (a)

(c) Over a different period, the bus is expected to be early on 30 days. Calculate the number of days in this period. 2 marks No calculator

Mark scheme

M1 30 ÷ 0.12 oe, e.g. 0.12n = 30
A1 250 cao

Question 3. A factory makes light bulbs. A quality checker tests a random sample of 250 bulbs and finds that 9 of them are faulty.

(a) Calculate the relative frequency of a faulty bulb in the sample. 1 mark

Mark scheme

B1 0.036 oe, e.g. 9/250 or 3.6%

(b) The factory makes 18 000 bulbs in one week. Calculate an estimate of the number of faulty bulbs made in that week. 2 marks

Mark scheme

M1 18 000 × their (a), or 9/250 × 18 000 oe
A1 648, FT their (a)

(c) Explain why your answer to part (b) is only an estimate. 1 mark

Mark scheme

B1 The relative frequency comes from a sample, so it is only an estimate of the probability; a different sample could give a different number oe

How this is examined

A missing probability from a table, then an expected frequency from it: A probability table with one gap (or an unknown such as $x$ ), followed by a Calculate part for the expected number in a given number of trials, worth about 3 marks in total 1 minus the sum of the other probabilities written down, e.g. $1 - 0.12 - 0.53 = 0.35$ , then that probability times the number of trials, e.g. $0.35 \times 180 = 63$ . If your probability from the first step is wrong, the multiplication still earns the method mark, and the answer can still earn the accuracy mark when the scheme marks it follow through (FT).
Using results from a sample or an experiment to estimate a number in a bigger group: A 2-mark Calculate: a relative frequency from a sample or from earlier trials, used for a population, a batch, or a further set of trials The relative frequency multiplied by the size of the bigger group, with the numbers shown, e.g. $\frac{62}{200} \times 600$ , then the value, $186$ . The multiplication earns the method mark even if the final value is wrong.
Relative frequency from a table of experiment results: A 1- or 2-mark Calculate after a results table, sometimes for an event made of two or more outcomes The frequency of the outcome over the total number of trials, e.g. $\frac{41 + 23}{200}$ , then $0.32$ , $\frac{8}{25}$ or $32 %$ . For an event made of several outcomes, add their frequencies first. A ratio such as $64 : 200$ is not a probability.
Deciding from results whether a coin, dice or spinner is fair: A 1- or 2-mark Explain or Comment after a results table, asking whether the results suggest the object is fair or biased What a fair object would give, e.g. 'about 40 of each number in 200 spins' or 'probability 0.2 each', then a number from the table compared with it and a conclusion: '62 is much more than 40, so the spinner seems biased'. A yes or no with no numbers does not explain anything.
Which estimate is most reliable, and why: A 1-mark State or Give after results from several people, or a question that asks how to make an estimate better The result from the most trials, with the reason 'more trials', or the statement 'do more trials'. To combine results, total frequency over total number of trials, not the mean of the relative frequencies.

Words to know

Trial: One go of an experiment, such as one spin of a spinner or one roll of a dice.
Outcome: A possible result of a trial, such as 'lands on red'.
Event: An outcome, or a set of outcomes, such as 'an even number' on a dice.
Frequency: The number of times an outcome happens.
Relative frequency: The frequency of an outcome divided by the total number of trials. It is an estimate of the probability.
Theoretical probability: A probability found by counting equally likely outcomes, without doing an experiment.
Fair: Every possible result is equally likely, e.g. each face of a fair dice has probability $\frac{1}{6}$ .
Biased: Not fair: some results are more likely than others.
At random: By chance alone. Choosing at random means every item has the same chance of being chosen.
Expected frequency: The number of times an outcome is expected to happen: probability × number of trials. It is an average, so it need not be a whole number.
Population: The whole group you want to know about.
Sample: The part of a population that you check. In a random sample, every member of the population has the same chance of being chosen.

All Maths lessons Next: 9.3 Probability of combined events