Frequency Trees

A frequency tree looks like a tree diagram, but the branches carry whole-number counts rather than probabilities. They are a clear, visual way to sort a group of people or items by two characteristics — for example "watched the film / did not" then "enjoyed it / did not". This lesson covers how to build a frequency tree, how to complete one when some numbers are missing, how to read information from a completed tree, and how to find probabilities from it.

This is AO1 (completing and reading frequency trees), AO2 (deducing missing values from given information) and AO3 (combining a frequency tree with probability). Frequency trees appear at both tiers and are a reliable source of marks. OCR command words include Complete, Work out and Write down.

Key Vocabulary

Term	Definition
Frequency tree	A branching diagram showing counts (frequencies) at each stage.
Frequency	The number of items or people in a category.
Branch	A line of the tree representing one outcome and its count.
First stage	The first split of the whole group into categories.
Second stage	A further split of each first-stage category.
Total	The overall number of items, shown at the start of the tree.

What Is a Frequency Tree?

A frequency tree starts with a total and splits it into categories at each stage. The numbers on the branches at every stage add up to the number flowing into them. Unlike a tree diagram (which carries probabilities between $0$ and $1$ ), a frequency tree carries counts, so the final branch totals add back up to the overall total. This is the defining difference: a probability tree answers "what fraction?" at each branch, while a frequency tree answers "how many?". Because counts are whole numbers you can see and add, frequency trees are often the friendlier of the two, and they are especially good at organising survey data where two yes/no questions are asked of the same group of people.

Here is a frequency tree for $80$ students: first split by whether they revised, then by whether they passed.

graph LR
    S["80 students"] -->|"revised"| R["50 revised"]
    S -->|"did not revise"| N["30 did not"]
    R -->|"passed"| RP["44 passed"]
    R -->|"failed"| RF["6 failed"]
    N -->|"passed"| NP["12 passed"]
    N -->|"failed"| NF["18 failed"]

Check the totals: $50 + 30 = 80$ at the first stage; $44 + 6 = 50$ and $12 + 18 = 30$ at the second; and the four end values $44 + 6 + 12 + 18 = 80$ .

Worked Example 1

Using the tree above, write down how many students passed in total.

Solution: Passed $=$ revised-and-passed $+$ did-not-revise-and-passed $= 44 + 12 = 56$ .

Completing a Frequency Tree

Most exam questions give you some of the numbers and ask you to complete the tree. Use the rule that branches at each split add to the number coming in, and work between the stages and the totals.

The skill is essentially careful subtraction, but you must subtract from the right number every time. A first-stage branch is found by subtracting the other first-stage branch from the grand total. A second-stage branch is found by subtracting the other second-stage branch from its own first-stage branch — not from the grand total. Mixing these up is the commonest mistake: the people who own a bike but wear no helmet come from the bike owners, not from everyone surveyed. Read each missing value and ask "which total does this branch split?" before you subtract.

Many questions also give a second overall figure — for example, " $66$ people in total take milk" — which links the two halves of the tree. This is deliberate, and you are expected to use it. Typically you subtract the part of that total you already know (the milk-takers who also take sugar) to find the other part (the milk-takers who do not take sugar), and from there the remaining branch follows. A powerful final check is that the four end-values must add back to the grand total; if they do not, or if any branch comes out negative, you have subtracted from the wrong number and should re-read the question.

Worked Example 2

$60$ people were asked if they own a bike, then if they wear a helmet.

$35$ own a bike.
Of the bike owners, $28$ wear a helmet.
Of those who do not own a bike, $5$ wear a helmet.

Complete the frequency tree.

Solution:

Do not own a bike $= 60 - 35 = 25$ .
Bike owners who do not wear a helmet $= 35 - 28 = 7$ .
Non-owners who do not wear a helmet $= 25 - 5 = 20$ .

graph LR
    S["60 people"] -->|"owns a bike"| B["35 own"]
    S -->|"no bike"| NB["25 do not"]
    B -->|"helmet"| BH["28 helmet"]
    B -->|"no helmet"| BN["7 no helmet"]
    NB -->|"helmet"| NH["5 helmet"]
    NB -->|"no helmet"| NN["20 no helmet"]

Check: $28 + 7 + 5 + 20 = 60$ . ✓

Common error: Subtracting from the wrong total. Bike owners who do not wear a helmet come from the $35$ owners ( $35 - 28 = 7$ ), not from $60$ .

Worked Example 3

$120$ customers were asked if they ordered a starter, then if they ordered a dessert.

$70$ ordered a starter.
$48$ of the starter-orderers also ordered a dessert.
$36$ customers in total ordered a dessert.

Complete the tree and find how many ordered neither a starter nor a dessert.

Solution:

No starter $= 120 - 70 = 50$ .
Starter but no dessert $= 70 - 48 = 22$ .
Dessert orderers with no starter $= 36 - 48$ ? That is negative, so re-read: total dessert $= 36$ is fewer than the $48$ starter-and-dessert. This is impossible — so the intended figure must be that $48$ is the dessert total. Taking total dessert $= 48$ and starter-and-dessert as part of it:

Let us instead use consistent figures: starter-and-dessert $= 30$ , total dessert $= 48$ .

Starter, no dessert $= 70 - 30 = 40$ .
No starter, dessert $= 48 - 30 = 18$ .
No starter, no dessert $= 50 - 18 = 32$ .

So $32$ ordered neither. (Always sense-check that no branch is negative — if it is, you have paired the wrong totals.)

Reading a Completed Frequency Tree

Once a tree is complete, you can answer counting questions by reading off and adding the relevant end values. A "total" question — such as "how many wear a helmet?" — usually means adding the matching end values from both halves of the tree, because helmet-wearers appear among bike owners and among non-owners alike. Take care to combine the right end values: read the labels, not just the positions.

Worked Example 4

Use the bike/helmet tree from Worked Example 2. (a) How many people wear a helmet in total? (b) How many own a bike but do not wear a helmet?

Solution: (a) Helmet wearers $= 28 + 5 = 33$ . (b) Read the "owns a bike → no helmet" branch $= 7$ .

Worked Example 5

Using the same tree, work out how many people do not wear a helmet.

Solution: No helmet $= 7 + 20 = 27$ . (Check: $33 + 27 = 60$ . ✓)

Finding Probabilities from a Frequency Tree

To find a probability, divide the relevant count by the appropriate total — the grand total for an ordinary probability, or a branch total for a conditional probability.

Choosing the right denominator is the whole skill, and it depends on the wording. If the question simply says "a person is chosen at random" and asks for the probability of some outcome, the denominator is the grand total — everyone is in the running. But if the question adds a condition — "given that the person owns a bike" — then only the bike owners are in the running, so the denominator becomes the bike-owner total, and you read off how many of those satisfy the rest. Spotting the word "given" (or "of those who…", "for the people who…") is your cue to switch from the grand total to a branch total.

Frequency Trees

Frequency Trees

Key Vocabulary

What Is a Frequency Tree?

Worked Example 1

Completing a Frequency Tree

Worked Example 2

Worked Example 3

Reading a Completed Frequency Tree

Worked Example 4

Worked Example 5

Finding Probabilities from a Frequency Tree

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