What is a tree diagram in the context of probability?

A tree diagram is a visual representation used in probability to map out all possible outcomes of a series of events. It helps in organizing and calculating the probabilities of different outcomes by branching out each possible event from a single starting point. This method is particularly useful in Cambridge IGCSE Mathematics for solving complex probability problems.

How do you construct a tree diagram for a probability problem?

To construct a tree diagram, start by identifying the initial event and draw a branch for each possible outcome. For each subsequent event, draw branches from the end of the previous branches, representing all possible outcomes of the new event. Label each branch with the probability of the outcome it represents. This structured approach helps in visualizing and calculating the total probability of different event sequences.

Why are tree diagrams useful in solving probability questions?

Tree diagrams are useful because they provide a clear and organized way to visualize all possible outcomes of a series of events. This makes it easier to calculate the probability of complex events by breaking them down into simpler, sequential steps. In the Cambridge IGCSE Mathematics curriculum, tree diagrams are a valuable tool for understanding and solving probability problems efficiently.

Can tree diagrams be used for dependent events in probability?

Yes, tree diagrams can be used for both independent and dependent events. For dependent events, the probability of each subsequent event may change based on the outcomes of previous events. In a tree diagram, this is represented by adjusting the probabilities on the branches accordingly. This allows students to accurately calculate the overall probability of sequences of dependent events, a key concept in the Cambridge IGCSE Mathematics syllabus.

What are some common mistakes to avoid when using tree diagrams in probability?

Common mistakes include not accounting for all possible outcomes, incorrectly labeling probabilities, and failing to adjust probabilities for dependent events. It's important to ensure that the sum of probabilities for each set of branches equals one. Additionally, careful attention should be paid to the sequence of events to avoid errors in calculations. Understanding these aspects is crucial for mastering tree diagrams in the Cambridge IGCSE Mathematics curriculum.

Why is "Not updating the second-draw probabilities for 'without replacement'" a common mistake in Tree Diagrams?

Why it happens: Treating both draws as identical. How to avoid it: Without replacement: total drops by 1; favourable count drops by 1 if it was drawn. Source: 0580/42 — every series

Why is "Adding probabilities along a single path" a common mistake in Tree Diagrams?

Why it happens: Confusing AND with OR. How to avoid it: ALONG a path: MULTIPLY (AND). ACROSS paths: ADD (OR).

Why is "Missing one of the valid paths" a common mistake in Tree Diagrams?

Why it happens: Listing only the obvious one (e.g. RB but not BR for 'one red'). How to avoid it: List ALL paths that satisfy the condition before adding.

Why is "Adding paths that aren't mutually exclusive" a common mistake in Tree Diagrams?

Why it happens: Confusing 'paths to the leaf' with 'general events'. How to avoid it: Different end-leaves are mutually exclusive by construction. Adding their probabilities is safe.

ProbabilityCambridge IGCSE Mathematics (0580)

Tree Diagrams

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Tree Diagrams — Cambridge IGCSE 0580 Maths Extended (2026)

Visualise multi-stage probability experiments. Multiply along a branch (AND), add between branches (OR). With-replacement keeps probabilities the same; without-replacement changes the denominator each stage.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E10.5 — Calculate the probability of combined events using tree diagrams (with and without replacement).

Drawing a tree diagram

Each stage of the experiment is a fan of branches; labels go on each branch.

Setup.

Identify the stages of the experiment (e.g. "first ball drawn", "second ball drawn").
Each stage spreads into branches — one per outcome.
Write the probability on each branch.
Branches from the same node sum to $1$ .

Worked (with replacement). A bag has $4$ red and $6$ blue balls. Draw two balls WITH REPLACEMENT.

Tree:

Stage 1: $R$ (prob $\dfrac{4}{10}$ ) or $B$ (prob $\dfrac{6}{10}$ ).
Stage 2 from $R$ : $R$ ( $\dfrac{4}{10}$ ) or $B$ ( $\dfrac{6}{10}$ ).
Stage 2 from $B$ : $R$ ( $\dfrac{4}{10}$ ) or $B$ ( $\dfrac{6}{10}$ ).

The probabilities at stage 2 are the same as stage 1 because the ball is replaced.

Worked (without replacement). Same bag, but DON'T replace.

Stage 1: $R$ ( $\dfrac{4}{10}$ ), $B$ ( $\dfrac{6}{10}$ ).
Stage 2 from $R$ : $R$ ( $\dfrac{3}{9}$ ), $B$ ( $\dfrac{6}{9}$ ). (One red gone, total down to $9$ .)
Stage 2 from $B$ : $R$ ( $\dfrac{4}{9}$ ), $B$ ( $\dfrac{5}{9}$ ).

Stages along the tree.
Branches at each node sum to $1$ .
Without replacement: update both numerator AND denominator.
Label each branch with its probability.

Multiply along a path (AND)

Probability of a complete sequence of outcomes = product of branch probabilities.

Each complete path from start to end represents a specific sequence of outcomes. Its probability is the PRODUCT of the branch probabilities along the way.

Worked (with replacement). Bag of $4R, 6B$ . Two draws WITH replacement. $P(\text{both red})$ :

Path: $R \to R$ .
$P = \dfrac{4}{10} \times \dfrac{4}{10} = \dfrac{16}{100} = \dfrac{4}{25}$ .

Worked (without replacement). Same bag. Two draws WITHOUT replacement. $P(\text{both red})$ :

Path: $R \to R$ .
$P = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$ .

Tip. "Both", "all three", "exactly this sequence" → multiply along ONE path.

Multiply branch probabilities along a path.
Path = sequence of outcomes.
Cambridge mark scheme awards the unsimplified product, then the final simplified value.

Add between paths (OR)

Probability of an event that can happen in multiple paths = sum of those path probabilities.

When an event can occur via several different sequences, ADD the probabilities of those sequences.

Worked (without replacement). Bag of $4R, 6B$ . Two draws without replacement. $P(\text{exactly one red})$ :

Two paths: $R \to B$ or $B \to R$ .
$R \to B$ : $\dfrac{4}{10} \times \dfrac{6}{9} = \dfrac{24}{90}$ .
$B \to R$ : $\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}$ .
Sum: $\dfrac{48}{90} = \dfrac{8}{15}$ .

Worked. Same bag. $P(\text{at least one red})$ :

Easier: complement. $P(\text{no red}) = P(BB) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}$ .
$P(\text{at least one red}) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$ .

Tip. "At least one X" → almost always use the complement: $1 - P(\text{none})$ .

Add path probabilities for OR events.
List ALL paths that satisfy the event.
'At least one' → complement is usually easier.

Three or more stages

Same rules: multiply along, add between. Just more branches.

For three or more stages, the tree has more layers but the rules are unchanged.

Worked. A bag has $3$ red and $5$ blue balls. Draw THREE balls without replacement. $P(\text{exactly$ 2 $red})$ :

Paths satisfying "exactly 2 red": $RRB$ , $RBR$ , $BRR$ .
$RRB$ : $\dfrac{3}{8} \times \dfrac{2}{7} \times \dfrac{5}{6} = \dfrac{30}{336}$ .
$RBR$ : $\dfrac{3}{8} \times \dfrac{5}{7} \times \dfrac{2}{6} = \dfrac{30}{336}$ .
$BRR$ : $\dfrac{5}{8} \times \dfrac{3}{7} \times \dfrac{2}{6} = \dfrac{30}{336}$ .
Sum: $\dfrac{90}{336} = \dfrac{15}{56}$ .

Tip. With many paths, organise systematically — list each sequence in order before computing. Don't lose track.

Same multiply-along, add-between rules.
List all paths that satisfy the event before computing.
Be systematic: don't miss a path.

How it’s examined

Tree diagrams appear every Paper 4 (5-8 marks) — typically a 2- or 3-stage problem with one part "with replacement" and one part "without". Examiner reports flag forgetting to update the denominator in without-replacement problems and missing a path when computing OR events.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E10.5); 0580/42 Oct/Nov 2024 — Q16 (tree, without replacement); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Tree Diagrams

Step-by-step solutions to past-paper-style questions on tree diagrams, written exactly the way a tutor would explain them at the board.

1Two draws with replacement
Extended• with replacement
Question
A bag has $3$ red and $5$ blue balls. Two are drawn with replacement. Find $P(\text{exactly one red})$ .
Step-by-step solution
1. Step 1
  $P(R) = \dfrac{3}{8}$ , $P(B) = \dfrac{5}{8}$ on each draw.
2. Step 2
  Two paths give exactly one red: $RB$ or $BR$ .
  $P(\text{exactly one R}) = \dfrac{3}{8} \times \dfrac{5}{8} + \dfrac{5}{8} \times \dfrac{3}{8}$
3. Step 3
  Compute.
  $= \dfrac{15}{64} + \dfrac{15}{64} = \dfrac{30}{64} = \dfrac{15}{32}$
Answer
$\dfrac{15}{32}$
2Two draws without replacement
Extended• Adapted from 0580/42 May/Jun 2024 Q11• without replacement
Question
A bag has $4$ red and $6$ blue balls. Two are drawn without replacement. Find $P(\text{at least one red})$ .
Step-by-step solution
1. Step 1
  Easier: complement = $P(\text{both blue})$ .
  $P(BB) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}$
2. Step 2
  Subtract from 1.
  $P(\text{at least one R}) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$
Answer
$\dfrac{2}{3}$
3Three-stage tree
Extended• three stage
Question
A coin is flipped three times. Find $P(\text{exactly 2 heads})$ .
Step-by-step solution
1. Step 1
  Three paths: $HHT$ , $HTH$ , $THH$ .
2. Step 2
  Each has probability $(1/2)^{3} = 1/8$ .
  $P = 3 \times \dfrac{1}{8} = \dfrac{3}{8}$
Answer
$\dfrac{3}{8}$

Key Formulae — Tree Diagrams

The formulae you need to memorise for tree diagrams on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Tree-diagram rules
$\text{Multiply ALONG branches; add ACROSS distinct paths}$
When to use
Every tree diagram problem.
Branch sum
$\text{Probabilities on branches from any node sum to } 1$
When to use
Sanity check — if your branches don't add to 1, you've made an error.

Key Definitions and Keywords — Tree Diagrams

Definitions to memorise and the exact keywords mark schemes credit for tree diagrams answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Tree diagram
Examiner keyword
A branching diagram showing all possible outcomes of a sequence of events with their probabilities.
Path
Examiner keyword
A sequence of branches from the root to a leaf — represents one possible outcome.
Conditional probability
Examiner keyword
Probability on the second branch depends on what happened on the first (without replacement).

Common Mistakes and Misconceptions — Tree Diagrams

The traps other students keep falling into on tree diagrams questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Not updating the second-draw probabilities for 'without replacement'
0580/42 — every series
Why it happens
Treating both draws as identical.
How to avoid it
Without replacement: total drops by 1; favourable count drops by 1 if it was drawn.
✕Adding probabilities along a single path
Why it happens
Confusing AND with OR.
How to avoid it
ALONG a path: MULTIPLY (AND). ACROSS paths: ADD (OR).
✕Missing one of the valid paths
Why it happens
Listing only the obvious one (e.g. $RB$ but not $BR$ for 'one red').
How to avoid it
List ALL paths that satisfy the condition before adding.
✕Adding paths that aren't mutually exclusive
Why it happens
Confusing 'paths to the leaf' with 'general events'.
How to avoid it
Different end-leaves are mutually exclusive by construction. Adding their probabilities is safe.

Practice questions

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Tree Diagrams — frequently asked questions

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Tree Diagrams

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Two draws with replacement

2Two draws without replacement

3Three-stage tree

Tree-diagram rules

Branch sum

Tree diagram

Path

Conditional probability

✕Not updating the second-draw probabilities for 'without replacement'

✕Adding probabilities along a single path

✕Missing one of the valid paths

✕Adding paths that aren't mutually exclusive

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Tree Diagrams — frequently asked questions