Tree diagrams GCSE Questions and Worked Answers

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Start with the meaning

What you need to know about tree diagrams

Toss a fair coin twice and record the results in order. The possibilities are HH, HT, TH and TT, where H means heads and T means tails. HT and TH are different records because the order differs. Each first toss can be followed by either result on the second toss. A tree draws these choices as branches, so every complete route records one possible sequence.

See the idea first

Follow one sequence from left to right

A probability describes how likely an outcome is, from 0 (impossible) to 1 (certain). A fair coin has probability 1/2 for each result. After either first result, the next independent toss still has two equally likely outcomes. Each complete route has probability 1/2 × 1/2 = 1/4.

Read a complete route from left to right. HT means heads first, then tails. Each final route has probability ½ × ½ = ¼.

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.

Find one ordered sequence

What the problem asks: For two fair coin tosses, find the probability of heads then tails.

How to solve it: Follow the H branch and then its T branch. Half of a half is a quarter, so multiply 1/2 by 1/2.

Find an outcome with more than one route

What the problem asks: Find the probability of exactly one head in two fair tosses.

How to solve it: The separate routes are HT and TH. They cannot both describe the same pair of tosses, so add their probabilities: 1/4 + 1/4 = 1/2.

Draw without replacement

What the problem asks: A bag contains 3 red and 2 blue counters. Draw two without putting the first back. What changes?

How to solve it: After red, 2 red and 2 blue remain: use 2/4 and 2/4. After blue, 3 red and 1 blue remain: use 3/4 and 1/4. Each second branch describes the bag after its own first result.

Find at least one success

What the problem asks: Find the chance of at least one head in three fair tosses.

How to solve it: The only excluded sequence is TTT, with probability 1/8. Subtract from 1 to get 7/8. This complement method saves listing seven successful routes.

A reliable routine

For a sequence with branch probabilities

A later branch gives the chance of its outcome after the earlier results on that route. Multiplying gives the fraction that survives every stage. Adding works for distinct complete routes because those sequences are mutually exclusive: one trial cannot end on two routes.

Draw a stage for each event, label the outcomes and attach probabilities to branches.
Check that branches leaving each individual node sum to 1. Update remaining counts when needed.
Identify complete paths matching the wording, including order and ‘exactly’ versus ‘at least’.
Multiply along each selected path, then add distinct matching paths, or subtract the excluded outcome from 1.

Check: Equal numbers of branches do not guarantee equal probabilities. Replacement restores the bag's counts; without replacement, later probabilities depend on the first draw. Independence means the earlier outcome does not change a later event's probability.

Fully worked

Tree diagrams GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

One independent path

2 marks

Question

A fair coin is tossed and a fair six-sided die is rolled independently. Find P(heads then a 6).

The branches have probabilities 1/2 and 1/6.

P(H\text{ and }6)=\frac12\times\frac16=\frac1{12}

Both conditions must occur in the same trial.

Example 2

Exactly one success

3 marks

Question

A player scores each shot with probability 0.7, independently. Find the probability of exactly one score in two shots.

A miss has probability 0.3. The routes are score–miss and miss–score.

P(SM)=0.7(0.3)=0.21

P(MS)=0.3(0.7)=0.21

P(\text{exactly one})=0.21+0.21=0.42

Do not count the two-score route.

Example 3

With replacement

3 marks

Question

A bag contains 4 red and 6 blue counters. Draw a counter, replace it and mix the bag, then draw again. Find P(two red).

Replacing the first counter restores 4 red out of 10 for the second draw.

P(RR)=\frac4{10}\times\frac4{10}

=\frac{16}{100}=\frac4{25}

The second probability stays the same under this random-draw model.

Example 4

Without replacement

3 marks

Question

The same bag contains 4 red and 6 blue counters. Draw twice without replacement. Find P(two red).

After a red is removed, 3 red remain among 9 counters.

P(RR)=\frac4{10}\times\frac3{9}

=\frac{12}{90}=\frac2{15}

Both the red count and total decrease on this route.

Example 5

Either order without replacement

4 marks

Question

A bag contains 3 red and 2 blue counters. Draw two without replacement. Find the probability of one of each colour.

There are two possible orders.

P(RB)=\frac35\times\frac24=\frac3{10}

P(BR)=\frac25\times\frac34=\frac3{10}

Add the distinct routes.

P(\text{one of each})=\frac3{10}+\frac3{10}=\frac35

The second branches refer to different remaining bags.

The bag before the second draw
First draw	Red left	Blue left	Total left
Red	2	2	4
Blue	3	1	4

Example 6

At least one

3 marks

Question

An alarm fails a test with probability 0.1. Three tests are independent. Find the probability of at least one failure.

The opposite event is no failures: all three tests pass. A pass has probability 0.9.

P(\text{all pass})=0.9^3=0.729

P(\text{at least one failure})=1-0.729=0.271

This includes one, two or three failures.

10 original questions · total 28 marks

Tree diagrams GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 33 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed

1

Complete a node

1 mark

The only branches leaving a node are success with probability 0.65 and failure. Find the failure probability.

Show worked answer

1-0.65=0.35

The outcomes at this node are exhaustive and mutually exclusive, so their probabilities sum to 1.

2

Two heads

2 marks

A fair coin is tossed twice independently. Find P(HH).

Show worked answer

P(HH)=\frac12\times\frac12=\frac14

Follow both H branches.

3

Different coin results

3 marks

Two fair independent coin tosses are made. Find the probability that their results differ.

Show worked answer

The routes are HT and TH.

P(HT)=\frac14,\quad P(TH)=\frac14

P(\text{different})=\frac14+\frac14=\frac12

4

Biased coin

2 marks

A coin has probability 0.6 of heads on each independent toss. Find P(TT).

Show worked answer

Each tail has probability 1 − 0.6 = 0.4.

P(TT)=0.4^2=0.16

5

Exactly one head

3 marks

For the same coin with P(H) = 0.6, find P(exactly one head in two independent tosses).

Show worked answer

P(HT)=0.6(0.4)=0.24

P(TH)=0.4(0.6)=0.24

P(\text{exactly one})=0.48

6

Replacement

3 marks

A bag contains 2 green and 3 yellow counters. Draw twice with replacement and mixing. Find P(two green).

Show worked answer

P(GG)=\frac25\times\frac25=\frac4{25}

The original counts are restored before the second draw.

7

No replacement

3 marks

A bag contains 2 green and 3 yellow counters. Draw twice without replacement. Find P(two green).

Show worked answer

After green, 1 green remains out of 4.

P(GG)=\frac25\times\frac14=\frac1{10}

8

Both orders

4 marks

A bag contains 2 green and 3 yellow counters. Draw twice without replacement. Find P(one of each colour).

Show worked answer

P(GY)=\frac25\times\frac34=\frac3{10}

P(YG)=\frac35\times\frac24=\frac3{10}

P(\text{one of each})=\frac35

Include both orders.

9

Complement

3 marks

A player succeeds with probability 0.8 on each independent attempt. Find P(at least one success in three attempts).

Show worked answer

The only excluded route is three failures, each with probability 0.2.

P(FFF)=0.2^3=0.008

P(\text{at least one success})=1-0.008=0.992

10

A condition narrows the outcomes

Higher only4 marks

A fair coin is tossed twice. Given that at least one result is heads, find the probability that both are heads.

Show worked answer

The condition excludes TT. The remaining equally likely routes are HH, HT and TH. Only one of these three is HH. The vertical bar in the probability notation below means ‘given that’: only outcomes meeting the stated condition are counted.

P(HH\mid\text{at least one H})=\frac{P(HH)}{P(\text{at least one H})}

=\frac{1/4}{3/4}=\frac13

The denominator refers to the restricted set, not all four original routes.

Examiner-style feedback

Common tree diagrams mistakes

Adding along a path

One full path needs every stage to happen, so multiply its conditional branch probabilities.

Forgetting the other order

‘One of each’ may be RB or BR. ‘Red then blue’ asks only for RB.

Keeping the old bag total

Without replacement, one counter has gone. Update both the total and the appropriate colour count.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

A route records outcomes in order.
Branches from each node total 1.
Multiply along routes; add distinct matching routes.
Use updated counts and read ‘at least’ carefully.

Quick answers

Tree diagrams FAQ

Does without replacement mean Higher only?

No. Dependent combined events and probability trees can appear across tiers. The explicit conditional-probability extension here is labelled Higher.

Why can I add two routes without subtracting an overlap?

Different complete sequences cannot occur in the same trial, so the routes are mutually exclusive. For overlapping events in general, simply adding would double-count the overlap.

Build connected skills

What to revise next

Prepare the calculations

Fractions

Multiply and add probability fractions accurately.

Revise fractions

Organise events another way

Venn diagrams

See unions, intersections and complements.

Revise venn diagrams

Content standards

Curriculum and rights review

P6–P8: probability trees, independent and dependent events across tiers; the explicit conditional-probability question uses Higher P9. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references

AQA GCSE Maths subject content