What is a tree diagram in the context of Edexcel IGCSE Mathematics?

A tree diagram is a graphical representation used in probability and statistics to map out all possible outcomes of a sequence of events. In Edexcel IGCSE Mathematics, tree diagrams help students visualize and calculate the probabilities of combined events by branching out each possible outcome.

How do you construct a tree diagram for probability problems?

To construct a tree diagram, start by identifying the sequence of events. Draw a branch for each possible outcome of the first event. From each of these branches, draw further branches for the outcomes of the second event, and so on. Label each branch with the probability of that outcome. This helps in calculating the probability of combined events by multiplying the probabilities along the branches.

Why are tree diagrams useful in solving probability questions?

Tree diagrams are useful because they provide a clear visual representation of all possible outcomes and their probabilities. This makes it easier to understand complex probability problems, especially when dealing with multiple events. They help in systematically calculating the probability of different scenarios by following the branches of the diagram.

Can tree diagrams be used for both independent and dependent events?

Yes, tree diagrams can be used for both independent and dependent events. For independent events, the probability of each event does not affect the others, and you simply multiply the probabilities along the branches. For dependent events, the probability of subsequent events changes based on previous outcomes, and this is reflected in the probabilities assigned to the branches.

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

Common mistakes include not accounting for all possible outcomes, incorrectly calculating 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 1 and to carefully follow the sequence of events to avoid errors in the diagram.

Why is "Not reducing probabilities (without replacement)" a common mistake in Tree Diagrams?

Why it happens: Treating like independent events. How to avoid it: Without replacement: BOTH numerator AND denominator change after first pick. Source: 4MA1 — examiner reports 2022-2024

Why is "Forgetting to add multiple paths" a common mistake in Tree Diagrams?

Why it happens: Computing only RB but missing BR. How to avoid it: 'Exactly one' usually has TWO paths: RB and BR. Add them.

Why is "Using wrong second-pick probability" a common mistake in Tree Diagrams?

Why it happens: Not tracking what's been removed. How to avoid it: Label each branch carefully. 1st R → 2 red of 7 left. 1st B → 3 red of 7 left.

Why is "Branch probabilities not summing to 1" a common mistake in Tree Diagrams?

Why it happens: Calculation error. How to avoid it: From any branching point, probabilities must sum to 1. Always check.

Statistics and ProbabilityEdexcel IGCSE Mathematics (4MA1)

Tree Diagrams

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Tree Diagrams Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

Tree diagrams visualise sequential probability. Multiply along branches, add the relevant paths. With/without replacement is a guaranteed Higher Tier topic.

What you’ll learn

Mapped to the Cambridge IGCSE 4MA1 syllabus (2026 onwards).

6.5 — Construct a tree diagram for a two-stage event.
6.5 — Calculate probabilities by multiplying along branches and adding outcomes.
6.5 — Solve problems with replacement and without replacement.

Building a tree diagram

Each branch = an outcome. Each branch labelled with its probability.

A tree diagram shows all outcomes of a sequence of events.

Structure:

Start from a single point.
First "level" of branches: outcomes of the first event.
From each first-level node, draw second-level branches: outcomes of the second event.
Each branch has a probability label.

Example. Two coins.

       0.5  H ─── 0.5  H  (HH)
            │     0.5  T  (HT)
            │
       0.5  T ─── 0.5  H  (TH)
                  0.5  T  (TT)

Rule. Probabilities of branches from any single node SUM to 1.

Worked example. Bag: 3 red, 5 blue. Pick two WITHOUT replacement.

First level: $P(R_1) = 3/8$ , $P(B_1) = 5/8$ .

After picking red: 2 red and 5 blue (7 total) remain. After picking blue: 3 red and 4 blue (7 total) remain.

Second level:

After $R_1$ : $P(R_2|R_1) = 2/7$ , $P(B_2|R_1) = 5/7$ .
After $B_1$ : $P(R_2|B_1) = 3/7$ , $P(B_2|B_1) = 4/7$ .

Note: the second-level probabilities DEPEND on what happened first. With replacement, they would all be the same.

Edexcel tip. Lay out the tree clearly, with probabilities written ON the branches. Mark schemes credit a fully-labelled tree even before computation.

Branches show outcomes.
Each branch = probability.
Branches from a node sum to 1.
Without replacement → second probabilities change.

Multiply along branches; add the paths

AND multiplies down. OR adds across.

Two rules:

Multiply along a path (AND): the probability of a single full outcome is the product of probabilities along its branches.
Add the path probabilities (OR): if multiple paths give the desired outcome, sum their probabilities.

Worked example. Bag: 3 red, 5 blue. Without replacement. $P(\text{exactly one red})$ ?

Two paths give exactly one red: $R_1B_2$ and $B_1R_2$ .

$P(R_1B_2) = (3/8)(5/7) = 15/56$ .
$P(B_1R_2) = (5/8)(3/7) = 15/56$ .
Total: $15/56 + 15/56 = 30/56 = 15/28$ .

Worked example. Same bag. $P(\text{both same colour})$ ?

Two paths: $R_1R_2$ and $B_1B_2$ .

$P(R_1R_2) = (3/8)(2/7) = 6/56 = 3/28$ .
$P(B_1B_2) = (5/8)(4/7) = 20/56 = 5/14$ .
Total: $3/28 + 10/28 = 13/28$ .

Worked example. $P(\text{at least one red})$ ?

Direct: many paths. EASIER via complement: $P(\text{at least one red}) = 1 - P(\text{both blue}) = 1 - 5/14 = 9/14$ .

Edexcel tip. "At least one" almost always uses complement. Save time.

Multiply along branches.
Add path probabilities.
'Exactly one' → typically two paths.
'At least one' → use complement.

See the full worked example for tree diagrams →

With vs without replacement

With: independent. Without: probabilities change.

With replacement. After the first pick, the item is put BACK. Total stays the same. Probabilities for the second pick are the SAME as the first.

Worked example. Bag: 4 red, 6 blue. With replacement. $P(\text{both red})$ ?

$P(R_1) = 4/10$ . After replacing, $P(R_2) = 4/10$ .
$P(\text{both red}) = (4/10)(4/10) = 16/100 = 4/25$ .

Without replacement. After the first pick, the item is NOT put back. Total decreases. Whatever was picked is unavailable.

Worked example. Same bag, without replacement.

$P(R_1) = 4/10$ . After taking a red: 3 red, 6 blue, 9 total. $P(R_2|R_1) = 3/9 = 1/3$ .
$P(\text{both red}) = (4/10)(3/9) = 12/90 = 2/15$ .

Spot the difference. With replacement gives $4/25 \approx 0.16$ . Without gives $2/15 \approx 0.133$ . Without is smaller because the second probability is reduced.

Edexcel tip. Read the question carefully. Keywords:

"With replacement" / "replaces" → independent, probabilities unchanged.
"Without replacement" / "does not replace" → conditional, probabilities reduced.

If the situation is something like "two students are chosen at random" — without replacement (you can't pick the same student twice).

With: probabilities unchanged.
Without: probabilities reduce.
Read the question carefully.
'Two students chosen' → usually without replacement.

How it’s examined

Tree diagrams + without replacement is THE classic Higher Tier 6-mark question. Examiner reports flag: (1) wrong second-pick denominator, (2) only computing one path when two are needed, (3) not using complement for 'at least one'.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

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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.

1Tree with replacement (independent)
Higher• tree, with replacement
Question
Bag: 4 red, 6 blue. Pick two WITH replacement. Find $P(\text{both red})$ .
Step-by-step solution
1. Step 1
  $P(R) = 4/10 = 0.4$ each pick (independent).
2. Step 2
  Multiply along R-R branch: $0.4 \times 0.4 = 0.16$ .
Answer
$P(\text{both red}) = 0.16$
2Tree without replacement
Higher• Adapted from 4MA1/2H May/Jun 2024 Q19• tree, without replacement
Question
Bag: 3 red, 5 blue. Pick two WITHOUT replacement. Find $P(\text{exactly one red})$ .
Step-by-step solution
1. Step 1
  1st red probabilities: $3/8$ . 1st blue: $5/8$ .
2. Step 2
  After red taken: 2 red, 5 blue. After blue: 3 red, 4 blue.
3. Step 3
  Two paths give exactly one red: RB and BR.
4. Step 4
  $P(\text{RB}) = (3/8)(5/7) = 15/56$ . $P(\text{BR}) = (5/8)(3/7) = 15/56$ .
5. Step 5
  Total: $15/56 + 15/56 = 30/56 = 15/28$ .
Answer
$P(\text{exactly one red}) = 15/28$
Examiner tip
Multiply along branches; add the relevant outcomes. Mark scheme: M1 for tree with correct probabilities, M1 for multiplying, M1 for adding paths, A1 for value.
3At least one (use complement)
Higher• complement, tree
Question
Bag: 3 red, 5 blue. Pick two without replacement. Find $P(\text{at least one red})$ .
Step-by-step solution
1. Step 1
  Complement: $P(\text{at least one red}) = 1 - P(\text{no red}) = 1 - P(\text{both blue})$ .
2. Step 2
  $P(\text{both blue}) = (5/8)(4/7) = 20/56 = 5/14$ .
3. Step 3
  $1 - 5/14 = 9/14$ .
Answer
$P(\text{at least one red}) = 9/14$
Examiner tip
'At least one' problems → use complement to save time.

Key Formulae — Tree Diagrams

The formulae you need to memorise for tree diagrams on the Pearson Edexcel IGCSE 4MA1 paper, with every variable defined in plain English and a note on when to use it.

Multiply along branches
$P(\text{path}) = p_1 \times p_2$
$p_1, p_2$
probabilities along the branches
When to use
AND down a single path.
Add the relevant outcomes
$P(\text{event}) = \sum P(\text{path}_i)$
When to use
OR — multiple paths lead to the same outcome.

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 Pearson Edexcel IGCSE 4MA1 sitting.

Tree diagram
Examiner keyword
Branching diagram showing all possible outcomes of a sequence of events with their probabilities.
With replacement
Item put back after first pick. Probabilities UNCHANGED for second pick. Independent.
Without replacement
Examiner keyword
Item NOT put back. Probabilities CHANGE for second pick. Conditional.

Common Mistakes and Misconceptions — Tree Diagrams

The traps other students keep falling into on tree diagrams questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Not reducing probabilities (without replacement)
4MA1 — examiner reports 2022-2024
Why it happens
Treating like independent events.
How to avoid it
Without replacement: BOTH numerator AND denominator change after first pick.
✕Forgetting to add multiple paths
Why it happens
Computing only RB but missing BR.
How to avoid it
'Exactly one' usually has TWO paths: RB and BR. Add them.
✕Using wrong second-pick probability
Why it happens
Not tracking what's been removed.
How to avoid it
Label each branch carefully. 1st R → 2 red of 7 left. 1st B → 3 red of 7 left.
✕Branch probabilities not summing to 1
Why it happens
Calculation error.
How to avoid it
From any branching point, probabilities must sum to 1. Always check.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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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

1Tree with replacement (independent)

2Tree without replacement

3At least one (use complement)

Multiply along branches

Add the relevant outcomes

Tree diagram

With replacement

Without replacement

✕Not reducing probabilities (without replacement)

✕Forgetting to add multiple paths

✕Using wrong second-pick probability

✕Branch probabilities not summing to 1

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Tree Diagrams — frequently asked questions