Frequency Trees
Frequency Trees
A frequency tree is used to show how a group of people/things can be broken up into certain categories.
Make sure you are happy with the following topics before continuing.
Constructing Frequency Trees
Data from experiments that have or more steps can be recorded using a frequency tree. For example:
people were given minutes to solve a puzzle.
- people who tried to solve the puzzle were under years old.
- people solved the puzzle.
- people aged and over did not solve the puzzle.
Complete the frequency tree using this information.
Step 1: of the people in question were under .
The total number of people is . We can now work out the following.
We can now add these to the frequency tree.
Step 2: ‘ people aged & over did not solve the puzzle’.
We know people aged and over did not solve the puzzle, so this can be added to the frequency diagram.
Next, we calculate how many people aged and over solved the puzzle.
Step 3: For the final two blank spaces, we can’t simply use the same technique as before, we have to look at the second bit of information:
We know ‘ people solved the puzzle’
We now know how many of the ‘ and over’ group solved the puzzle, so
Lastly, this number and the ‘under – didn’t solve’ group must add up to , so we get
Using Frequency Trees
We can now use the frequency trees we have created to answer probability questions.
Example: Using the frequency tree from the previous example, if you choose one of the under s at random, what is the probability that they did not solve the puzzle?
Leave your answer in its simplest form.
From our diagram we can see that there are under s, of which didn’t solve the puzzle. Therefore, the probability is
This fraction is now in its simplest form, so we’re done.
Frequency Trees Example Questions
Question 1: people travelled to an event by bus or by train.
people travelled by train.
Of the people who travelled by bus, were late.
people were late to the event.
Complete the frequency tree.
[3 marks]

Since there were people in total, of which travelled by train, we can calculate how many travelled by bus:
We are also told that of the total number of people who travelled by bus – which we now know is people – were late. If we know how many were late, we can easily work out how many who travelled by bus were not late:
So, that leaves two remaining bubbles in the tree diagram, the ones branching off from the people who took the train.
The question tells us that people in total were late to the event. Given that of those people who were late took the bus, we can calculate how many of these people who were late took the train:
Finally, if of the people who took the train were late, we can calculate how many took the train and were on time
The completed frequency tree should therefore look like this:

Question 2: people are either right-handed or left-handed.
of them are right-handed.
One third of left-handed people are right-footed.
people are left-footed.
a) Complete the frequency tree.
b) If a right-handed person is picked at random, what is the chance that they are right-footed?
c) If a person is picked at random, what is the chance that they are right-footed? Leave your answer in its simplest form.
[6 marks]

a) First of all, if there are people in total and are right-handed, we can perform a simple calculation to work out how many people are left-handed:
left-handed people
Secondly, we know that of the left-handed people, one third are right-footed. So the number of people who are left-handed and right-footed can be calculated as follows:
left-handed and right-footed people
Furthermore, since there are left-handed people in total, we can now work out how many of them are left-footed:
left-handed and left-footed people
So, we just have two remaining bubbles, the ones branching off from the right-handed people.
The final piece of information given to us in the question tells us that people in total are left-footed. Given that we now know that of the left-handed people are left-footed, we can calculate the number of people who are right-handed and left-footed:
right-handed and left-footed people
Finally, if of the right-handed people are left-footed, we can calculate how many are right-handed and right-footed
right-handed and right-footed people
The completed frequency tree should therefore look like this:

b) There are people who are right-handed, and of them are right-footed. So, the probability of picking a right-handed person who is also right-footed is
c) There are people in total. The number of right-footed people is:
Therefore, the probability of selecting a right-footed person is:
This fraction can be simplified to:
Question 3: At Grange Hill School for Boys, the year s took either French or Spanish for GCSE.
of the year group took French.
of the students who took French failed, which was students in total.
The total number of students who failed Spanish was twice the number of students who failed French.
a) Complete the frequency tree.
b) To the nearest whole number, what percentage of the whole year group passed French?
[4 marks]

a) We have very little information to go on here, but since we know that students failed French and that this was of the entire French group, we can work out how many students took French:
If students took French and failed, then the number that passed was:
We are also told that of the year group took French. This represents students. From this information, we can calculate the total number of students in the year group and then the number of students who took Spanish.
If
then
so
Therefore, there was a total of student in year that took French or Spanish.
Since of the year group took Spanish, the exact number that took Spanish can be calculated as follows:
We are told that the number of students who failed Spanish was twice the number that failed French. Since failed French, we can work out the number that failed Spanish:
Therefore the number that passed Spanish can be calculated as follows:
The completed frequency tree should therefore look like this:

b) We know that the total in the year group was students. Of these , passed French. As a percentage, this can be calculated as follows:
Question 4: Pupils at sports day were given a choice of a ham or cheese sandwich, which they could have on either white or brown bread.
of the pupils who chose a cheese sandwich opted for brown bread. children opted for a cheese sandwich on brown bread.
The number of pupils who chose a cheese sandwich and the number of pupils who chose a ham sandwich can be given as a ratio of . Of the people that chose a ham sandwich, opted for brown bread.
a) Draw a frequency tree to represent this information.
b) What is the probability of selecting a student at random who chose a ham sandwich on brown bread?
[4 marks]

a) The first fact that we are presented with is that of the pupils chose a cheese sandwich on brown bread and that this figure represents a total of pupils. From this information, we can work out the total number of pupils who had a cheese sandwich.
If
then
so
If pupils chose a cheese sandwich, and of them had brown bread, then we can easily calculate the number that had a cheese sandwich on white bread:
The next key piece of information we are given is that the ratio of pupils who chose a cheese sandwich to pupils who chose a ham sandwich is . From this information we can deduce that of these pupils had a cheese sandwich and of this pupils had a ham sandwich. (We are dealing in fifths here because the sum of the ratio is .)
If
then
so
Therefore we can conclude that a total of students chose a ham sandwich.
At this point we now also know that there was a total of pupils:
Finally, we know that of the people that chose a ham sandwich, opted for brown bread. We can now calculate exactly how many pupils this is since we have now worked out the number of students who opted for the ham sandwich:
The number of students who had a ham sandwich on white bread is simply the total that chose a ham sandwich minus the who chose a ham sandwich on brown bread:
The completed frequency tree should therefore look like this:

b) We know from the completed frequency tree that there was a total of pupils and that a total of pupils chose a ham sandwich on brown bread. The probability of selecting a pupil at random from the group who chose a ham sandwich on brown bread can be expressed as follows:
This can be simplified to:
This can again be simplified to:
Question 5: In a recent election, the votes for three parties, The Conservative Party, the Labour Party and the Green party, were shared in a ratio of .
The votes cast by men and by women for the Labour party were in a ratio of .
The total number of people who voted for the Green Party was .
The number of female votes received by the Green party was of the total number of female votes received by the Labour party.
The number of male votes received by the Conservative Party was more than the total of male votes received by the Labour party.
Draw a frequency tree to represent this information.
[4 marks]

In this question, we have a couple of ratios and some percentages, but the only precise figure we have to start with is the total number of people who voted for the Green Party.
We know that the votes were shared between the parties in the ratio of . This means that the Conservative Party received of the votes, the Labour party and the Green Party . (We are dealing in sixteenths here because the sum of the ratio is .)
If voted for the Green party and this represented of the total number of votes received, then we can work out the total number of votes:
If
then
so
So, if there was a total of votes, we can work out the number of votes received by the Conservative party and the Labour party.
The total number of votes received by the Conservative party was:
The total number of votes received by the Labour party was:
We know that votes cast by men and by women for the Labour party were in a ratio of . This means that of the votes were cast by men and by women. (We are dealing in fifths here because the sum of the ratio is .)
Since we know the total number of votes received by the Labour party and the fraction cast by men and women, we can work out the exact number of votes made by each gender.
The total number of votes cast by men was:
The total number of votes cast by women was:
We are also told that the number of female votes received by the Green party was of the total number of female votes received by the Labour party. Since the total number of female votes received by the Labour party was , we simply need to work out of this amount:
Since the Green party received votes in total, we can work out the number of male votes:
The final thing to work out is the number of male and female votes received by the Conservative Party. We know that the number of male votes was more than the total of male votes received by the Labour party. The total number of male votes received by the Labour party was , so we will need to increase this figure by , as follows:
If the Conservative party received votes in total, of which were male, then the number of female votes can be calculated as follows:
The completed frequency tree should, therefore, look like this:

Specification Points Covered
Probability – 1. record describe and analyse the frequency of outcomes of probability experiments using tables and frequency trees