Frequency Trees and Mutually Exclusive Events
A frequency tree splits a group into categories, then splits each category again. It is a GCSE-specific diagram that IGCSE textbooks skip, and a regular Foundation question. This page also covers the rule that probabilities must add up to 1.
1Frequency trees
A frequency tree shows actual numbers, not probabilities. Each pair of branches must add up to the number they came from.
2Completing a frequency tree
- Girls = 80 − 35 = 45.
- Boys who failed = 35 − 21 = 14.
- Girls who passed = 45 − 18 = 27.
3Probability from a frequency tree
- (a) Passed = 21 + 27 = 48, so P = 4880 = 35.
- (b) 2780.
- (c) Higher: there are 48 students who passed, and 21 of them are boys: 2148 = 716.
4Probabilities add up to 1
If a list of outcomes covers everything that can happen (the outcomes are exhaustive) and no two can happen together (they are mutually exclusive), their probabilities add up to 1.
| Colour | Red | Blue | Green | Yellow |
|---|---|---|---|---|
| Probability | 0.15 | 0.2 | 0.35 | x |
- 0.15 + 0.2 + 0.35 = 0.7.
- x = 1 − 0.7.
5The OR rule for mutually exclusive events
- Red and green can’t happen together, so add: 0.15 + 0.35.
- Using the addition rule for events that can happen together, such as “red” and “a prime number” on a dice. Use a Venn diagram or a list for those.
6Expected frequency
Expected number of times = probability × number of trials.
- 0.35 × 200.
- Writing probabilities on a frequency tree. It shows numbers.
- Pairs of branches that don’t add up to the number they came from.
- Using the grand total as the denominator for a conditional (“given that”) probability.
- Adding probabilities of events that are not mutually exclusive.
- Forgetting that the probabilities of all possible outcomes must total 1.
★Key Vocabulary Recap
- Frequency tree
- A diagram that splits a total into categories using actual counts.
- Mutually exclusive
- Events that can’t happen at the same time.
- Exhaustive
- A set of outcomes that covers every possibility.
- Expected frequency
- Probability × number of trials.
- Conditional probability
- A probability given that something else has happened.
- P1: record, describe and analyse the frequency of outcomes of probability experiments using tables and frequency trees.
- P3: relate relative expected frequencies to theoretical probability.
- P4: apply the property that the probabilities of an exhaustive set of mutually exclusive events sum to one.
- P9 (Higher): calculate and interpret conditional probabilities using expected frequencies with frequency trees.
Practice questions
Frequency Trees and Mutually Exclusive Events — Practice Questions
10 exam-style questions (27 marks), building from warm-up to grade 8–9. Show full working: Edexcel gives method marks even when the final answer is wrong.
- (2 marks)Grade 2A spinner can land on A, B or C. P(A) = 0.3 and P(B) = 0.4. Find P(C).
- (3 marks)Grade 360 people chose between cinema and bowling. 24 were adults. 9 adults chose cinema. 22 children chose bowling. Complete the frequency tree.
- (2 marks)Grade 3Using your tree from Question 2, a person is chosen at random. Find the probability that they chose cinema.
- (2 marks)Grade 4A bag contains red, blue and green counters. P(red) = 14 and P(blue) = 13. Find P(green).
- (3 marks)Grade 4A spinner is biased.(a) Find x. (b) Find P(red or blue). (c) The spinner is spun 200 times. How many times would you expect green?
Colour Red Blue Green Yellow Probability 0.15 0.2 0.35 x - (3 marks)Grade 5A dice is biased. The probabilities of 4, 5 and 6 are equal. P(1) = 0.1, P(2) = 0.2, P(3) = 0.4. Find P(6).
- (4 marks)Grade 5120 patients had a test. 48 of them have a condition. Of those with the condition, 30 tested positive. Of those without it, 60 tested negative.
(a) Draw a frequency tree. (b) How many patients tested positive?
- (2 marks)Grade 6Using Question 7, a patient with the condition is chosen at random. Find the probability that they tested positive.
- (3 marks)Grade 7Using your tree from Question 2, a person who chose cinema is selected at random. Find the probability that they are a child.
- (3 marks)Grade 7In a school, 36 students study Art. 14 of the Art students also study Music. One Art student is chosen at random. Find the probability that they also study Music, and explain why the events “studies Art” and “studies Music” are not mutually exclusive.
Worked solutions
Show worked solutions
Frequency Trees and Mutually Exclusive Events — Worked Solutions
Full worked solutions to all 10 questions (27 marks), set out the way an Edexcel examiner wants to see them.
- (2 marks)Grade 2A spinner can land on A, B or C. P(A) = 0.3 and P(B) = 0.4. Find P(C).Solution
- Probabilities add to 1: 1 − 0.3 − 0.4.
0.3 - (3 marks)Grade 360 people chose between cinema and bowling. 24 were adults. 9 adults chose cinema. 22 children chose bowling. Complete the frequency tree.Solution
- Adults bowling = 24 − 9 = 15.
- Children = 60 − 24 = 36.
- Children cinema = 36 − 22 = 14.
15, 36, 14 - (2 marks)Grade 3Using your tree from Question 2, a person is chosen at random. Find the probability that they chose cinema.Solution
- Cinema = 9 + 14 = 23.
- P = 2360.
2360
- (2 marks)Grade 4A bag contains red, blue and green counters. P(red) = 14 and P(blue) = 13. Find P(green).Solution
- 1 − 14 − 13 = 1212 − 312 − 412.
512 - (3 marks)Grade 4A spinner is biased.(a) Find x. (b) Find P(red or blue). (c) The spinner is spun 200 times. How many times would you expect green?
Colour Red Blue Green Yellow Probability 0.15 0.2 0.35 x Solution- (a) x = 1 − 0.7 = 0.3.
- (b) 0.15 + 0.2 = 0.35.
- (c) 0.35 × 200 = 70.
(a) 0.3 (b) 0.35 (c) 70 - (3 marks)Grade 5A dice is biased. The probabilities of 4, 5 and 6 are equal. P(1) = 0.1, P(2) = 0.2, P(3) = 0.4. Find P(6).Solution
- P(4) + P(5) + P(6) = 1 − 0.7 = 0.3.
- They are equal, so each is 0.3 ÷ 3.
0.1 - (4 marks)Grade 5120 patients had a test. 48 of them have a condition. Of those with the condition, 30 tested positive. Of those without it, 60 tested negative.
(a) Draw a frequency tree. (b) How many patients tested positive?Solution- Without the condition: 120 − 48 = 72. With it: 30 positive, 48 − 30 = 18 negative.
- Without it: 60 negative, so 72 − 60 = 12 positive.
- (b) Positive = 30 + 12 = 42.
(b) 42
- (2 marks)Grade 6Using Question 7, a patient with the condition is chosen at random. Find the probability that they tested positive.Solution
- Choose only from the 48 with the condition: 30 tested positive.
- 3048 = 58.
58 - (3 marks)Grade 7Using your tree from Question 2, a person who chose cinema is selected at random. Find the probability that they are a child.Solution
- Cinema total = 9 + 14 = 23.
- Children among them = 14.
- P = 1423.
1423 - (3 marks)Grade 7In a school, 36 students study Art. 14 of the Art students also study Music. One Art student is chosen at random. Find the probability that they also study Music, and explain why the events “studies Art” and “studies Music” are not mutually exclusive.Solution
- P = 1436 = 718.
- 14 students do both, so the events can happen together: they are not mutually exclusive.
718
Book a free trial class
Tell us a little about your requirement.