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GCSE Maths · Probability
Frequency trees GCSE Questions and Worked Answers
A frequency tree splits a total count into smaller groups. At every complete split, the child counts add back to their parent. Fill missing counts by subtraction, then use favourable count divided by the relevant total for probability.
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Start with the meaning
What you need to know about frequency trees
Sort 40 pupils into two groups: 24 take the bus and 16 walk. Then sort each group again into pupils who arrive on time and pupils who are late. A frequency tree draws this sorting as branches. Each number is a count of pupils; it is not yet a probability.
See the idea first
A split accounts for its whole parent group
In the bus group, 18 arrive on time and 6 are late, adding to 24. In the walking group, 14 are on time and 2 late, adding to 16. The final groups never overlap. Together they account for every pupil: 18 + 6 + 14 + 2 = 40. This is why missing counts can be found by subtraction.
From problem to method
Typical problems you need to be able to solve
These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.
Complete a missing branch
What the problem asks: Of 24 bus users, 18 are on time. How many are late?
How to solve it: The two groups fill the bus total, so late bus users = 24 − 18 = 6.
Combine final groups
What the problem asks: Using the bus and walk counts above, how many pupils arrive on time?
How to solve it: Add the two disjoint on-time groups: 18 bus + 14 walk = 32.
Calculate a probability
What the problem asks: One of all 40 pupils is chosen at random. Find the probability they walk and arrive late.
How to solve it: The final walk-and-late group contains 2 pupils. The choice is from all 40, so probability = 2/40 = 1/20.
A reliable routine
Complete and use a frequency tree
Use this when counts are split into mutually exclusive groups that cover their parent. Addition and subtraction work because each item belongs to exactly one child at a split.
- Write the overall total and identify what each branch means.
- Use parent total minus known child counts to complete each split.
- Add disjoint final groups if the question combines them.
- For probability, divide by the number of items that could actually be selected; check all final counts sum to the original total.
Check: Frequency-tree branches contain counts, not fractions that must sum to 1. Do not multiply branch counts along a route.
Fully worked
Frequency trees GCSE worked examples
Each solution explains the clue, why the method fits and how to check the result.
Example 1
First split
Question
A school surveys 70 pupils. 42 take art and the rest do not. Find the missing 'not art' count.
There are 28 pupils in the not-art branch. Check 42 + 28 = 70.
Example 2
Second split
Question
Of 42 art pupils, 17 also take music. The rest do not. Find the art-but-not-music count.
Subtract from the immediate parent 42, not from the whole school total.
Example 3
Complete both branches
Question
A survey has 60 pupils: 35 cycle and 25 walk. Of the cyclists 8 are late; of the walkers 4 are late. Find both on-time counts and check all final groups.
Cyclists on time: 35 − 8 = 27. Walkers on time: 25 − 4 = 21.
Every pupil is counted once.
Example 4
Combine a category
Question
A frequency tree has 27 on-time cyclists, 8 late cyclists, 21 on-time walkers and 4 late walkers. How many pupils are late?
The two late groups do not overlap:
There are 12 late pupils.
Example 5
Whole-sample probability
Question
A group has 27 on-time cyclists, 8 late cyclists, 21 on-time walkers and 4 late walkers. A pupil is chosen randomly from all of them. Find the chance they cycle and are on time.
Total = 27 + 8 + 21 + 4 = 60. Favourable group = 27.
Read one final group, then divide by the total.
Example 6
Probability within a branch
Question
Of 35 cyclists, 8 are late. A cyclist is chosen at random. Find the probability they are on time.
On-time cyclists = 35 − 8 = 27. The choice is restricted to cyclists, so
This is a conditional probability.
10 original questions · total 19 marks
Frequency trees GCSE exam-style questions
Try each question before opening its fully worked answer. The difficulty rises through the set.
Before you startAllow about 24 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed
Missing first branch
There are 90 people: 54 adults and the rest children. How many children?
Show worked answer
Missing second branch
Of 54 adults, 20 have a ticket. How many do not?
Show worked answer
Two complete splits
There are 36 children and 54 adults. 15 children and 20 adults have tickets. Find both no-ticket counts.
Show worked answer
Children without tickets: 36 − 15 = 21. Adults without tickets: 54 − 20 = 34.
Combine leaves
15 children and 20 adults have tickets. How many ticket holders are there?
Show worked answer
The groups are disjoint, so add the counts.
Probability from all
Among 90 people, 15 are children with tickets. One of all 90 is chosen randomly. Find the probability of choosing a child with a ticket.
Show worked answer
Complement
Among 90 people, 35 have tickets. Find the probability a randomly chosen person has no ticket.
Show worked answer
No ticket count = 90 − 35 = 55.
Check an error
A branch labelled 40 splits into counts 18 and 25, covering everyone without overlap. Explain the error.
Show worked answer
The children of a complete split must add to the parent, so at least one count is wrong.
Recover a parent
A group splits into 13 left-handed and 32 right-handed people, with everyone in exactly one category. Find the group total.
Show worked answer
Build counts from a fraction
A group has 80 pupils. Three fifths travel by bus. Of the bus users, 10 are late. Find bus users on time and non-bus users.
Show worked answer
Bus total:
Bus on time = 48 − 10 = 38. Non-bus = 80 − 48 = 32.
Restricted selection
There are 36 children, 15 with tickets. A child is selected randomly. Find the chance they do not have a ticket.
Show worked answer
No-ticket children = 36 − 15 = 21.
The denominator is the children group because the selection is restricted.
Examiner-style feedback
Common frequency trees mistakes
A second-level split belongs to its own branch, not automatically to the overall total.
A final leaf already counts the combined category. Multiplying counts along a path does not give its probability.
Add disjoint final groups; do not add a parent and its own children.
30-second recap
Choose, carry out, check
Translate the wording before you reach for a rule.
- Numbers are counts.
- Child groups add to their parent.
- Combine only disjoint groups.
- Probability uses the actual selection total.
Quick answers
Frequency trees FAQ
How is this different from a probability tree?
A frequency tree sorts a fixed set into counts. A probability tree puts probabilities on branches, often for successive random events.
Can a split have more than two branches?
Yes, provided the groups do not overlap and together cover the parent group.
Content standards
Curriculum and rights review
P1/P6 frequency trees across tiers. Conditional subgroup probability is explicitly labelled Higher as P9 extension. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.
Official specification references