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.

Cambridge IGCSE Mathematics (0580)

Tree Diagrams

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Short Notes - Tree Diagrams

Detailed Study Notes

Detailed notes on Probability for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

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.

At a glance

Each branch = one outcome of one stage.
Probabilities on branches from the SAME node sum to $1$ .
Multiply along a path (events in sequence: AND).
Add between paths (alternative paths: OR).
With replacement: branch probabilities stay constant.
Without replacement: numerators and denominators decrease 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 $B$ : $R$ ( $\dfrac{4}{9}$ ), $B$ ( $\dfrac{5}{9}$ ).

Without replacement, the stage-2 denominator falls to 9 and the numerator drops for whichever colour was taken.

Stage-2 branch	With replacement	Without replacement
after R → R	$\tfrac{4}{10}$	$\tfrac{3}{9}$
after R → B	$\tfrac{6}{10}$	$\tfrac{6}{9}$
after B → R	$\tfrac{4}{10}$	$\tfrac{4}{9}$
after B → B	$\tfrac{6}{10}$	$\tfrac{5}{9}$

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.

The probability of a whole sequence is the product of every branch probability along that 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: $\dfrac{48}{90} = \dfrac{8}{15}$ .

Two different paths give 'exactly one red' — multiply along each, then add.

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: $\dfrac{90}{336} = \dfrac{15}{56}$ .

Path	Calculation	Probability
$RRB$	$\tfrac{3}{8} \times \tfrac{2}{7} \times \tfrac{5}{6}$	$\tfrac{30}{336}$
$RBR$	$\tfrac{3}{8} \times \tfrac{5}{7} \times \tfrac{2}{6}$	$\tfrac{30}{336}$
$BRR$	$\tfrac{5}{8} \times \tfrac{3}{7} \times \tfrac{2}{6}$	$\tfrac{30}{336}$
Total		$\tfrac{90}{336} = \tfrac{15}{56}$

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.

Quick recap

Tree: stages → branches → paths.
Branches from one node sum to $1$ .
Multiply along a path (AND).
Add between paths (OR).
Without replacement: numerator AND denominator decrease.
'At least one' → complement.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Tree diagram — branching diagram representing successive stages of an experiment.
With replacement — the item drawn is returned; probabilities stay constant.
Without replacement — the item is kept out; both numerator and denominator decrease at each stage.
Conditional probability — probability of one event given another has occurred.

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.

Take this whole topic with you

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.

Question type:

Question patterns to master — Tree Diagrams

Almost every tree diagrams exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single named path is wanted — find $P(\text{first green, second red})$ or $P(\text{all three red})$ — one branch from root to leaf.

How to approach it

Multiply the branch probabilities along that one path, updating the favourable count and total at each stage for 'without replacement' (or after any stated rule change).

Common trap

Examiner reports flag not updating the second-draw probabilities without replacement, and missing a rule such as 'add an extra red' that changes the next stage.

Multi-step problem

Recognise it by

Several paths satisfy the condition — exactly one, at least one, the same colour — or a conditional probability must be built from the tree.

How to approach it

List every path that satisfies the condition, multiply along each, then add across the distinct end-leaves; for 'at least one' use $1 - P(\text{none})$ , and for a conditional probability sum the marginal across all relevant first-draw outcomes.

Common trap

Examiner reports flag missing a valid path (e.g. $RB$ but not $BR$ ), adding probabilities along a single path instead of multiplying, and using the first-draw probability as the marginal $P(R_{2})$ .

1Two draws with replacement
ExtendedMulti-step problem• 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
ExtendedMulti-step problem• 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
ExtendedMulti-step problem• 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
$\dfrac{3}{8}$
4Two-stage tree (coin and die)
CoreDirect calculation• independent, two stage
Question
A fair coin is flipped and a fair six-sided die is rolled. Find the probability of obtaining a head AND a number greater than $4$ .
Step-by-step solution
1. Step 1
  $P(H) = \dfrac{1}{2}$ on the first branch.
2. Step 2
  Numbers $> 4$ on the die: $\{5, 6\}$ so $P(>4) = \dfrac{2}{6} = \dfrac{1}{3}$ .
3. Step 3
  Independent — multiply along the path.
  $P = \dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$
Answer
$\dfrac{1}{6}$
5Two-day weather tree (context)
CoreMulti-step problem• context, two day
Question
On any day, the probability that it rains is $0.4$ . Find the probability that it rains on exactly one of two consecutive days.
Step-by-step solution
1. Step 1
  Days are independent. $P(R) = 0.4$ and $P(R') = 0.6$ .
2. Step 2
  Two paths give 'exactly one rain': $RR'$ or $R'R$ .
  $P = (0.4)(0.6) + (0.6)(0.4) = 0.24 + 0.24$
3. Step 3
  Sum.
  $P = 0.48$
Answer
$0.48$
6Probability of a specific path
ExtendedDirect calculation• specific path, without replacement
Question
A bag contains $5$ red and $3$ green sweets. Two are taken at random without replacement. Find the probability that the first sweet is green and the second is red.
Step-by-step solution
1. Step 1
  First draw: $P(G) = \dfrac{3}{8}$ .
2. Step 2
  After removing a green, the bag has $5$ red and $2$ green ( $7$ total). $P(R \mid G) = \dfrac{5}{7}$ .
3. Step 3
  Multiply along the $G \to R$ path.
  $P(GR) = \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}$
Answer
$\dfrac{15}{56}$
7At-least-one via complement (three trials)
ExtendedMulti-step problem• Adapted from 0580/42 Oct/Nov 2023 Q10• at least one, complement
Question
The probability that a basketball player scores on any free throw is $0.7$ . She takes three independent free throws. Find the probability that she scores at least once.
Step-by-step solution
1. Step 1
  Direct calculation has many paths. Use the complement: $P(\text{at least one}) = 1 - P(\text{no scores})$ .
2. Step 2
  $P(\text{miss}) = 0.3$ . Three independent misses.
  $P(\text{no scores}) = 0.3^{3} = 0.027$
3. Step 3
  Subtract from 1.
  $P(\text{at least one}) = 1 - 0.027 = 0.973$
Answer
$0.973$
Examiner tip
The examiner report consistently flags students computing the eight tree branches separately and making arithmetic slips. The complement is faster and safer for 'at least one' questions.
8Sum of paths — same first colour
ExtendedMulti-step problem• Adapted from 0580/42 May/Jun 2023 Q9• sum of paths, without replacement
Question
A bag contains $4$ blue and $6$ yellow balls. Two are taken at random without replacement. Find the probability that both balls are the same colour.
Step-by-step solution
1. Step 1
  Two valid paths: $BB$ or $YY$ .
2. Step 2
  Path $BB$ .
  $P(BB) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90}$
3. Step 3
  Path $YY$ .
  $P(YY) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90}$
4. Step 4
  Add (mutually exclusive end-leaves).
  $P(\text{same colour}) = \dfrac{12}{90} + \dfrac{30}{90} = \dfrac{42}{90} = \dfrac{7}{15}$
Answer
$\dfrac{7}{15}$
9Three-stage tree (drawing three balls without replacement)
ExtendedDirect calculation• three stage, without replacement
Question
A bag contains $4$ red and $2$ white balls. Three balls are drawn one at a time without replacement. Find the probability that all three are red.
Step-by-step solution
1. Step 1
  Multiply along the $RRR$ path, updating the bag at each stage.
  $P(RRR) = \dfrac{4}{6} \times \dfrac{3}{5} \times \dfrac{2}{4}$
2. Step 2
  Compute the numerator and denominator.
  $= \dfrac{24}{120} = \dfrac{1}{5}$
Answer
$\dfrac{1}{5}$
Examiner tip
The examiner report notes that candidates often fail to update both the favourable count AND the total at each stage. Track the bag carefully after each draw.
10Conditional probability from a tree
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q11• conditional, challenge
Question
A bag has $5$ red and $3$ green balls. Two balls are drawn without replacement. Given that the second ball is red, find the probability that the first ball was green.
Step-by-step solution
1. Step 1
  Use Bayes-style logic: $P(G_{1} \mid R_{2}) = \dfrac{P(G_{1} \cap R_{2})}{P(R_{2})}$ .
2. Step 2
  $P(G_{1} \cap R_{2})$ is the path $GR$ .
  $P(GR) = \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56}$
3. Step 3
  $P(R_{2})$ comes from both paths ending in red: $RR$ + $GR$ .
  $P(R_{2}) = \dfrac{5}{8} \times \dfrac{4}{7} + \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{20}{56} + \dfrac{15}{56} = \dfrac{35}{56} = \dfrac{5}{8}$
4. Step 4
  Divide.
  $P(G_{1} \mid R_{2}) = \dfrac{15/56}{35/56} = \dfrac{15}{35} = \dfrac{3}{7}$
Answer
$\dfrac{3}{7}$
Examiner tip
The examiner report flags that very few candidates compute $P(R_{2})$ correctly — most just use $\dfrac{5}{8}$ from the first draw and forget that the marginal $P(R_{2})$ must sum across both first-draw outcomes.
11Unequal first- and second-stage probabilities (Challenge)
ChallengeDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q11• unequal stages, challenge
Question
A box contains $3$ red and $7$ blue counters. A counter is drawn at random; if it is red, it is replaced AND an extra red is added before the next draw. If blue, it is replaced and the box left unchanged. A second counter is then drawn. Find the probability that both counters are red.
Step-by-step solution
1. Step 1
  First draw: $P(R_{1}) = \dfrac{3}{10}$ .
2. Step 2
  If $R_{1}$ , the box now has $4$ red, $7$ blue — total $11$ . So $P(R_{2} \mid R_{1}) = \dfrac{4}{11}$ .
3. Step 3
  Multiply along the $R_{1}R_{2}$ path.
  $P(R_{1}R_{2}) = \dfrac{3}{10} \times \dfrac{4}{11} = \dfrac{12}{110} = \dfrac{6}{55}$
Answer
$\dfrac{6}{55}$
Examiner tip
Read the rules carefully. The examiner report notes that candidates routinely miss the 'add an extra red' instruction and use $\dfrac{3}{10}$ twice, which would only apply if the counter were simply replaced.

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.

Tree Diagrams — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Stage-2 branch

With replacement

Without replacement

after R → R

\tfrac{4}{10}

\tfrac{3}{9}

after R → B

\tfrac{6}{10}

\tfrac{6}{9}

after B → R

\tfrac{4}{10}

\tfrac{4}{9}

after B → B

\tfrac{6}{10}

\tfrac{5}{9}

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: $\dfrac{48}{90} = \dfrac{8}{15}$ .

Two different paths give 'exactly one red' — multiply along each, then add.

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: $1 - P(\text{none})$ .

Path

Calculation

Probability

RRB

\tfrac{3}{8} \times \tfrac{2}{7} \times \tfrac{5}{6}

\tfrac{30}{336}

RBR

\tfrac{3}{8} \times \tfrac{5}{7} \times \tfrac{2}{6}

\tfrac{30}{336}

BRR

\tfrac{5}{8} \times \tfrac{3}{7} \times \tfrac{2}{6}

\tfrac{30}{336}

Total

\tfrac{90}{336} = \tfrac{15}{56}

Tree Diagrams

Short Notes - Tree Diagrams

Detailed Study Notes

At a glance

What you’ll learn

Drawing a tree diagram

Multiply along a path (AND)

Add between paths (OR)

Three or more stages

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Tree Diagrams

1Two draws with replacement

2Two draws without replacement

3Three-stage tree

4Two-stage tree (coin and die)

5Two-day weather tree (context)

6Probability of a specific path

7At-least-one via complement (three trials)

8Sum of paths — same first colour

9Three-stage tree (drawing three balls without replacement)

10Conditional probability from a tree

11Unequal first- and second-stage probabilities (Challenge)

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

Tree Diagrams — frequently asked questions

Tree Diagrams

Short Notes - Tree Diagrams

Detailed Study Notes

At a glance

What you’ll learn

Drawing a tree diagram

Multiply along a path (AND)

Add between paths (OR)

Three or more stages

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Tree Diagrams

1Two draws with replacement

2Two draws without replacement

3Three-stage tree

4Two-stage tree (coin and die)

5Two-day weather tree (context)

6Probability of a specific path

7At-least-one via complement (three trials)

8Sum of paths — same first colour

9Three-stage tree (drawing three balls without replacement)

10Conditional probability from a tree

11Unequal first- and second-stage probabilities (Challenge)

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

Tree Diagrams — frequently asked questions