## What you need to know

A frequency tree is used to show how a group of people/things can be broken up into certain categories. It will be best understood with an example.

Example: 120 people were given 3 minutes to solve a puzzle.

45 of the people who tried to solve the puzzle were under 18 years old.

78 of the people solved the puzzle.

32 of the people aged 18 and over did not solve the puzzle.

a) Complete the frequency tree below.

b) If you choose one of the under 18s at random, what is the probability that they did not solve the puzzle? Leave your answer in its simplest form.

Before filling any of the blanks, let’s understand what the frequency tree is saying.

The number at the far left is 120 – it represents the total. Then, that total of people is split into two categories, people aged under 18 and people aged 18 and over. Each of those two age groups is then itself separated into two groups: people who solved the puzzle and people who didn’t solve it.

The key fact to being able to fill in the missing pieces of a frequency tree is understand that

The numbers at the ends of the branches must add up to the value they branched off.

For example, the ’18 & over’ and ‘under 18’ bubbles must add up to 120. When answering frequency tree questions, you should fill in the “bubbles” as soon as you learn some new information.

a) Our first piece of info tells us that 45 of the people in question were under 18. Given that the total is 120, we can work out that

\text{‘18 and over’ people }=120-45=75So, we can now fill in the first two bubbles in the diagram – the ones branching off from 120.

Our second bit of info tells us that 78 people solved the puzzle. We’ll come back to this in a moment.

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Our third bit of info tells us that ’32 people aged 18 & over did not solve the puzzle’. Immediately, we can fill in the “18 & over – didn’t solve” bubble with this. Furthermore, we can now work out how many of the ’18 and over’ category did solve the puzzle (since they must add to 75):

\text{‘18 and over’ and ‘solved puzzle’ }=75-32=43For 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 again:

“78 people solved the puzzle”

Now that we know how many of the ’18 and over’ group solved the puzzle we can use this info to work out how many people are in ‘under 18 – solved’ group:

\text{‘under 18’ and ‘solved puzzle’ }=78-43=35Lastly, this number and the ‘under 18 – didn’t solve’ group must add up to 45, so we get

\text{‘under 18’ and ‘didn’t solve puzzle’}=45-35=10Thus, the completed frequency tree looks like the picture shown below.

b) If we pick someone from the under 18s at random, what is the probability that they didn’t solve the puzzle? From our diagram we can see that there are 45 under 18s, 10 of which didn’t solve the puzzle. Therefore, the probability is

\dfrac{10}{45}=\dfrac{2}{9}

This fraction is now in its simplest form, so we’re done.

### Example Questions

Firstly, there are 200 people in total and 58 travelled by train, so we get

\text{number travelled by bus }=200-58=142

Secondly, we’re told that of these ‘bus’ people – which we now know there are 142 of – 40 were late. That means

\text{travelled by bus and on time }=142-40=102

So, we just have two remaining bubbles: the ones branching off from the 58 people who took the train. The third bit of info in the question tells us that 71 people were late to the event. Given that 40 of those people took the bus, we get

\text{travelled by train and late }=71-40=31

Finally, if 31 of the 58 people who took the train were late, we must have that

\text{travelled by train and on time }=58-31=27

The completed frequency tree looks like what is shown below.

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2) 75 people are either right-handed or left-handed.

63 of them are right-handed.

One third of left-handed people are right-footed.

27 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.

a) Firstly, there are 75 people in total and 63 are right-handed, so we get

\text{left-handed people }=75-63=12

Secondly, we’re told that of the left-handed people, one third are right-footed. So, we get

\text{left-handed and right-footed people }=12\div 3=4

Furthermore, since 12 is the total number of left-handed people, we must have that

\text{left-handed and left-footed people }=12-4=8

So, we just have two remaining bubbles: the ones branching off from the 63 right-handed people. The last bit of info in the question tells us that 27 people are left-footed. Given that we now know 8 of the left-handed people are left-footed, we get

\text{right-handed and left-footed people }=27-8=19

Finally, if 19 of the 63 right-handed people are left-footed, we must have that

\text{right-handed and right-footed people }=63-19=44

The completed frequency tree looks like:

b) There are 63 people who are right-handed, and 44 of them are right-footed. So, the probability is

\dfrac{19}{63}

c) There are 75 people in total. The number of right-footed people is

44+4=48

Therefore, the probability in question is

\dfrac{48}{75}=\dfrac{16}{25}

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