P8ProbabilityFoundation & HigherGrades 4–7

Tree Diagrams

A tree diagram shows every outcome of a sequence of events, with a probability on each branch. Multiplying along a path gives the probability of that exact sequence; adding the paths you want gives the probability of the event. Trees handle unequal probabilities and "without replacement" problems that sample space grids cannot.

45 min Video by Maths Genie AQA GCSE Maths

The big picture

Two rules run everything. $\textbf{AND}$ — this happens and then that happens — means multiply along the branches. $\textbf{OR}$ — one path or another — means add the path probabilities.

The branches from any single point always add up to 1. When items are not replaced, the second set of branches changes: there is one fewer item in total, and one fewer of whatever was taken first. That makes the events $\textbf{dependent}$ .

Probability Tree DiagramsCurated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

What you'll learn

Draw and complete tree diagrams for two or more events
Use the AND rule (multiply) and the OR rule (add)
Calculate probabilities for independent events
Solve problems with dependent events, such as sampling without replacement

Drawing a tree and the AND rule

Draw one set of branches for each event. Write the outcome at the end of each branch and its probability on the branch. Check each set adds to 1.

For independent events — where one does not affect the other — $P(A \text{ and } B) = P(A) \times P(B)$ .

Picking a counter twice with replacement,

P(\text{red}) = 0.3

. Multiply along each path; the four results add up to 1.

Example 1Using the tree above, find the probability of picking two blue counters.

1Follow B then B:

0.7 \times 0.7

Answer

0.49

Your turnSame method, new numbers — try it before moving on.

The probability that Kai is late on any day is 0.2, independently of other days. Find the probability he is late on both Monday and Tuesday.

Your answer

Show worked solution

Solution

1Late AND late: multiply.
2 $0.2 \times 0.2 = 0.04$ .

Answer: $0.04$

The OR rule

When more than one path gives the event you want, work out each path and add them.

"At least one" questions are often quickest with $1 - P(\text{none})$ .

Example 1Using the tree above, find the probability of picking one counter of each colour.

1Paths RB and BR:

0.3 \times 0.7 = 0.21

each.

2Add:

0.21 + 0.21

Answer

0.42

Your turnSame method, new numbers — try it before moving on.

A biased coin lands heads with probability 0.6. It is flipped twice. Find the probability of exactly one head.

Your answer

Show hint

Two paths: HT and TH.

Show worked solution

Solution

1 $P(HT) = 0.6 \times 0.4 = 0.24$ ; $P(TH) = 0.4 \times 0.6 = 0.24$ .
2Add: $0.48$ .

Answer: $0.48$

Example 2The probability a bulb is faulty is 0.1. Three bulbs are tested. Find the probability that at least one is faulty.

P(\text{none faulty}) = 0.9^3 = 0.729

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

Answer

0.271

Your turnSame method, new numbers — try it before moving on.

A fair coin is flipped 3 times. What is the probability of at least one head?

Your answer

Show worked solution

Solution

1 $P(\text{no heads}) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$ .
2 $1 - \frac{1}{8} = \frac{7}{8}$ .

Answer: $\frac{7}{8}$

Tip — Say the path out loud: "red AND blue, OR blue AND red". AND means multiply, OR means add.

Without replacement

If an item is not replaced, the second-stage probabilities change. Reduce the total by one, and reduce the count of the colour already taken by one.

Fractions are usually clearer than decimals here — and do not simplify the branch fractions until the end.

Example 1A bag has 5 red and 3 blue sweets. Two are taken without replacement. Find the probability both are the same colour.

1Both red:

\frac{5}{8} \times \frac{4}{7} = \frac{20}{56}

2Both blue:

\frac{3}{8} \times \frac{2}{7} = \frac{6}{56}

3Add:

\frac{26}{56} = \frac{13}{28}

Answer

\frac{13}{28}

Your turnSame method, new numbers — try it before moving on.

A drawer has 4 black and 6 white socks. Two are taken at random without replacement. What is the probability that both are black?

Your answer

Show worked solution

Solution

1 $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90}$ .
2 $= \frac{2}{15}$ .

Answer: $\frac{2}{15}$

Dependent versus independent

Without replacement, taking a red sweet first makes the bag slightly "less red", so the chance of a second red falls from $\frac{5}{8}$ to $\frac{4}{7}$ . The second event depends on the first. With replacement, the bag is restored and the events are independent.

Think like an examiner

Complete the tree fully before answering — tree completion is often worth marks on its own.
Show the multiplications and additions you use, e.g. $\frac{5}{8} \times \frac{4}{7} + \frac{3}{8} \times \frac{2}{7}$ .
In without-replacement questions, check that both the numerator and denominator change on the second branches.

Tree diagram rules

P(A \text{ and } B) = P(A) \times P(B)

AND: multiply along branches (independent events).

P(\text{path}_1 \text{ or } \text{path}_2) = P_1 + P_2

OR: add separate paths.

P(\text{at least one}) = 1 - P(\text{none})

The complement shortcut.

Watch out for these

Adding along a path:

0.3 + 0.7

Along a path, multiply.

Using

\frac{5}{8}

again for the second pick without replacement.

Second pick:

\frac{4}{7}

for red after a red.

Counting only one path for "one of each".

RB and BR are both "one of each" — add them.

Branch probabilities from one point adding to more than 1.

Each set of branches must sum to exactly 1.

Stretch yourself

A bag contains $n$ counters, of which 3 are red. Two counters are taken without replacement. The probability that both are red is $\frac{1}{15}$ . Find $n$ .

Hint — Write the probability as $\frac{3}{n} \times \frac{2}{n - 1}$ and set it equal to $\frac{1}{15}$ .

Reveal solution

Solution

\frac{6}{n(n - 1)} = \frac{1}{15}

, so

n(n - 1) = 90

, i.e.

n^2 - n - 90 = 0

. Factorising:

(n - 10)(n + 9) = 0

, so

n = 10

. Check:

\frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}

✓.

\blacksquare

Questions students ask

How do I know if events are independent?

If the outcome of the first cannot change the probabilities of the second — like replacing a counter, or separate coin flips — they are independent. Not replacing makes them dependent.

Can a tree have more than two branches at each stage?

Yes. A bag with three colours has three branches from each point; the probabilities still sum to 1.

Should I use fractions or decimals?

Use whatever the question uses. For without-replacement problems, fractions avoid rounding errors.

Key takeaways

Branches from one point sum to 1.
AND: multiply along the path.
OR: add the probabilities of separate paths.
At least one: $1 - P(\text{none})$ .
Without replacement: reduce the total and the chosen colour by one on the second branches.

Exam-style questions

Practise it the way it is examined

3 original questions · 10 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 44 marks

The probability that it rains on a Saturday is 0.4. The probability it rains on a Sunday is 0.3. The events are independent.

(a)Work out the probability that it rains on both days.[1]
Show mark scheme (a)
- B1 $0.12$
Answer: $0.12$
(b)Work out the probability that it rains on exactly one of the two days.[3]
Show mark scheme (b)
- M1 $0.4 \times 0.7$ or $0.6 \times 0.3$
- M1 $0.28 + 0.18$
- A1 $0.46$
Answer: $0.46$

Question 2Grade 63 marks

A box contains 6 milk and 4 dark chocolates. Sam takes two at random without replacement.

Work out the probability that Sam takes at least one dark chocolate.[3]
Show mark scheme
- M1 $P(\text{both milk}) = \frac{6}{10} \times \frac{5}{9}$
- M1 $1 - \frac{30}{90}$
- A1 $\frac{60}{90} = \frac{2}{3}$
Answer: $\frac{2}{3}$

Question 3Grade 73 marks

Priya either walks or cycles to school. The probability she cycles is 0.35. If she cycles, the probability she is late is 0.1. If she walks, the probability she is late is 0.2.

Work out the probability that Priya is not late on a given day.[3]
Show mark scheme
- M1Correct paths: $0.35 \times 0.9$ and $0.65 \times 0.8$
- M1 $0.315 + 0.52$
- A1 $0.835$
Answer: $0.835$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Along a path of a tree diagram, you…
- multiply
- add
- subtract
- divide
Show answer and explanation
Answer: multiply
AND means multiply.
Two fair coins: P(two heads)?
- $\frac{1}{4}$
- $\frac{1}{2}$
- $1$
- $\frac{3}{4}$
Show answer and explanation
Answer: $\frac{1}{4}$
$\frac{1}{2} \times \frac{1}{2}$ .
Branches from one point are 0.35 and $x$ . $x =$
- $0.65$
- $0.35$
- $0.75$
- $1.35$
Show answer and explanation
Answer: $0.65$
They sum to 1.
P(pass) $= 0.8$ for each of two independent tests. P(pass both)?
- $0.64$
- $1.6$
- $0.8$
- $0.16$
Show answer and explanation
Answer: $0.64$
$0.8 \times 0.8$ .
Same tests: P(fail both)?
- $0.04$
- $0.2$
- $0.4$
- $0.36$
Show answer and explanation
Answer: $0.04$
$0.2 \times 0.2$ .
3 red, 2 blue; take 2 without replacement. P(both red)?
- $\frac{3}{10}$
- $\frac{9}{25}$
- $\frac{6}{25}$
- $\frac{1}{2}$
Show answer and explanation
Answer: $\frac{3}{10}$
$\frac{3}{5} \times \frac{2}{4}$ .
P(win) $= 0.25$ each game, independent. P(at least one win in 2 games)?
- $0.4375$
- $0.5$
- $0.0625$
- $0.5625$
Show answer and explanation
Answer: $0.4375$
$1 - 0.75^2$ .
4 red, 1 green; take 2 without replacement. P(one of each)?
- $\frac{2}{5}$
- $\frac{1}{5}$
- $\frac{8}{25}$
- $\frac{4}{25}$
Show answer and explanation
Answer: $\frac{2}{5}$
$\frac{4}{5} \times \frac{1}{4} + \frac{1}{5} \times \frac{4}{4}$ .

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