Tree Diagrams

This lesson covers how to draw and use tree diagrams, applying the AND and OR rules, and calculating combined probabilities for independent events and dependent events (without replacement). Tree diagrams are one of the most important probability tools on the Edexcel GCSE Mathematics (1MA1) exam, appearing on both Foundation and Higher tier papers.

What Is a Tree Diagram?

A tree diagram shows all possible outcomes of two or more successive events. Each "branch" represents a possible outcome, and the probability is written on the branch.

Standard Tree-Diagram Template

A typical two-stage tree diagram (e.g. a bag containing red and blue counters, picking twice) looks like this:

graph LR
    Start["Start"] -->|P(R)| R1["Red (1st)"]
    Start -->|P(B)| B1["Blue (1st)"]
    R1 -->|P(R after R)| RR["Red, Red"]
    R1 -->|P(B after R)| RB["Red, Blue"]
    B1 -->|P(R after B)| BR["Blue, Red"]
    B1 -->|P(B after B)| BB["Blue, Blue"]

Rules to apply with this template:

Each pair of branches from the same point must sum to 1.
Multiply along a single path to get the probability of that joint outcome (AND).
Add the probabilities of separate paths to get the probability of "one path OR another" (OR).

Key Rules

Rule	When to use	How to calculate
AND rule (multiplication)	Both events happen in sequence	Multiply along the branches
OR rule (addition)	Either one outcome or another	Add the final probabilities

Important Properties

The probabilities on each set of branches from the same point must add up to 1.
To find the probability of a particular path, multiply along that path.
To find the probability of one outcome or another, add the relevant path probabilities.

Independent Events — With Replacement

Two events are independent if the outcome of one does not affect the outcome of the other. When items are replaced, the probabilities stay the same for each trial.

Worked Example 1

A bag contains 4 red balls and 6 blue balls. A ball is chosen at random, its colour is noted, and it is replaced. A second ball is then chosen. Find:

(a) P(both red) (b) P(one of each colour)

Tree Diagram:

First pick:

P(Red) = 4/10 = 2/5
P(Blue) = 6/10 = 3/5

Second pick (same probabilities because of replacement):

P(Red) = 2/5
P(Blue) = 3/5

Paths:

Path	Calculation	Probability
Red, Red	$\frac{2}{5} \times \frac{2}{5}$	4/25
Red, Blue	$\frac{2}{5} \times \frac{3}{5}$	6/25
Blue, Red	$\frac{3}{5} \times \frac{2}{5}$	6/25
Blue, Blue	$\frac{3}{5} \times \frac{3}{5}$	9/25

Check: 4/25 + 6/25 + 6/25 + 9/25 = 25/25 = 1 ✓

(a) P(both red) = 4/25 = 4/25

(b) P(one of each) = P(Red, Blue) + P(Blue, Red) = 6/25 + 6/25 = 12/25

Edexcel Exam Tip: Always label your tree diagram branches clearly with the event and its probability. Marks are awarded for the correct diagram as well as the final answer.

Dependent Events — Without Replacement

When items are not replaced, the probabilities change for the second event because the total number of items has decreased.

Worked Example 2

A bag contains 4 red balls and 6 blue balls. A ball is chosen at random and not replaced. A second ball is then chosen. Find:

(a) P(both red) (b) P(at least one blue)

Tree Diagram:

First pick:

P(Red) = 4/10 = 2/5
P(Blue) = 6/10 = 3/5

Second pick (probabilities change — one ball has been removed):

If first was Red (3 red and 6 blue remain, total 9):

P(Red) = 3/9 = 1/3
P(Blue) = 6/9 = 2/3

If first was Blue (4 red and 5 blue remain, total 9):

P(Red) = 4/9
P(Blue) = 5/9

Paths:

Path	Calculation	Probability
Red, Red	$\frac{2}{5} \times \frac{1}{3}$	2/15
Red, Blue	$\frac{2}{5} \times \frac{2}{3}$	4/15
Blue, Red	$\frac{3}{5} \times \frac{4}{9}$	12/45 = 4/15
Blue, Blue	$\frac{3}{5} \times \frac{5}{9}$	15/45 = 1/3

Check: 2/15 + 4/15 + 4/15 + 5/15 = 15/15 = 1 ✓

(a) P(both red) = 2/15

(b) "At least one blue" means everything except "both red". P(at least one blue) = 1 − P(both red) = 1 − 2/15 = 13/15

Three-Stage Tree Diagrams

Worked Example 3

A fair coin is tossed three times. Find P(exactly two heads).

Tree Diagram (abbreviated):

The 8 equally likely outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

Each path has probability (1/2)³ = 1/8.

Paths with exactly 2 heads: HHT, HTH, THH = 3 paths

P(exactly 2 heads) = 3 × 1/8 = 3/8

Worked Example 4

The probability that a machine produces a defective item is 0.05. Three items are produced. Find P(exactly one is defective).

Let D = defective, G = good. $P(D)$ = 0.05, $P(G)$ = 0.95.

Paths with exactly one defective:

DGG: 0.05 × 0.95 × 0.95 = 0.045125
GDG: 0.95 × 0.05 × 0.95 = 0.045125
GGD: 0.95 × 0.95 × 0.05 = 0.045125

P(exactly one defective) = 3 × 0.045125 = 0.135375

The AND/OR Rules Summarised

AND Rule (Multiplication Rule)

P(A and B) = $P(A) × P(B)$ (if A and B are independent)

P(A and B) = $P(A)$ × P(B given A) (if A and B are dependent)

OR Rule (Addition Rule)

P(A or B) = $P(A) + P(B)$ (if A and B are mutually exclusive)

P(A or B) = $P(A) + P(B)$ − P(A and B) (general case)

Edexcel Exam Tip: On tree diagrams, AND = multiply along branches, OR = add between paths. This is the most important principle to remember.

Worked Example 5 — Full Exam-Style Question

A box contains 3 strawberry (S) and 7 chocolate (C) sweets. Priya takes a sweet at random and eats it. She then takes another sweet at random.

(a) Draw a tree diagram to represent this information. (b) Find the probability that Priya eats two chocolate sweets. (c) Find the probability that she eats one of each flavour.

Solution:

(a) Tree Diagram:

First sweet:

$P(S) = \frac{3}{10}$
$P(C) = \frac{7}{10}$

Second sweet (after first eaten — without replacement):

After S: 2S and 7C remain (total 9)

$P(S) = \frac{2}{9}$
$P(C) = \frac{7}{9}$

After C: 3S and 6C remain (total 9)

$P(S) = \frac{3}{9}$ = 1/3
$P(C) = \frac{6}{9}$ = 2/3

(b) P(CC) = $\frac{7}{10} \times \frac{6}{9} = \frac{42}{90}$ = 7/15

(c) P(one of each) = P(SC) + P(CS) = ( $\frac{3}{10} \times \frac{7}{9}$ ) + ( $\frac{7}{10} \times \frac{3}{9}$ ) = 21/90 + 21/90 = 42/90 = 7/15

Check: P(SS) = $\frac{3}{10} \times \frac{2}{9} = \frac{6}{90}$ = 1/15 Total: 1/15 + 7/15 + 7/15 = 15/15 = 1 ✓

Worked Example 6 — With Replacement (independent events)

A fair spinner has 3 equal sections coloured red, yellow and green. The spinner is spun twice. Find:

(a) P(both green) (b) P(at least one red)

Solution:

Each spin is independent and the probabilities stay the same: $P(R) = \frac{1}{3}$ , $P(Y) = \frac{1}{3}$ , $P(G) = \frac{1}{3}$ .

(a) P(both green) = 1/3 × 1/3 = 1/9

(b) P(at least one red) = 1 − P(no red on either spin)

P(no red on one spin) = 2/3. P(no red on two spins) = $\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$ .

P(at least one red) = 1 − 4/9 = 5/9

Tree Diagrams

Tree Diagrams

What Is a Tree Diagram?

Standard Tree-Diagram Template

Key Rules

Important Properties

Independent Events — With Replacement

Worked Example 1

Dependent Events — Without Replacement

Worked Example 2

Three-Stage Tree Diagrams

Worked Example 3

Worked Example 4

The AND/OR Rules Summarised

AND Rule (Multiplication Rule)

OR Rule (Addition Rule)

Worked Example 5 — Full Exam-Style Question

Worked Example 6 — With Replacement (independent events)

Worked Example 7 — With Replacement (exam practice)

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