Tree Diagrams

SPEC 3.1, 3.2 • PROBABILITY

Visualising sequential events and calculating compound probabilities

Why Tree Diagrams?

Tree diagrams provide a systematic way to visualise sequential events — situations where one event follows another.

Tree Diagram

A branching diagram that shows all possible outcomes of successive events and their associated probabilities.

When to Use

Use tree diagrams when you have two or more stages of events happening in sequence (e.g., drawing twice from a bag, tossing multiple coins).

Key Advantage

Tree diagrams make it easy to see ALL possible outcomes and calculate their probabilities systematically.

Structure of a Tree Diagram

Key Rules

Branches from any node must sum to 1

Each branch is labelled with its probability

Outcomes are listed at the end of each path

The Multiplication Rule

To find the probability of a complete path through the tree, multiply along the branches.

AND Rule

P (A \cap B) = P (A) \times P (B ∣ A)

Multiply probabilities along a single path

Example: A bag has 3 red and 2 blue balls. Two are drawn without replacement. Find P(both red).

Important

The second probability P(B|A) may depend on what happened first. Here, after taking 1 red, only 2 red remain out of 4 total.

The Addition Rule

To find the probability of multiple outcomes, add across the final probabilities.

OR Rule

P (outcome_{1} OR outcome_{2}) = sum of path probabilities

Example: Same bag (3 red, 2 blue). Find P(exactly one red) when drawing twice without replacement.

Remember

Multiply along branches (AND) • Add across paths (OR)

Worked Example01

Two fair coins are tossed. Draw a tree diagram and find:

(a) P(two heads)

(b) P(exactly one head)

(c) P(at least one head)

a)

P(HH) = 0.5 × 0.5 = 0.25 or ¼

b)

P(exactly one H) = P(HT) + P(TH) = 0.25 + 0.25 = 0.5 or ½

c)

P(at least one H) = P(HH) + P(HT) + P(TH) = 0.75 or use 1 − P(TT) = 1 − 0.25 = 0.75 or ¾

Independent vs Dependent Events

Independent Events

Probabilities stay the same at each stage.

Example: Coin tosses, dice rolls, drawing with replacement

P(B|A) = P(B)

Dependent Events

Probabilities change based on previous outcomes.

Example: Drawing cards without replacement

P(B|A) ≠ P(B)

Key Phrase

"Without replacement" signals dependent events — the total decreases, changing the probabilities on later branches.

Worked Example02

A bag contains 4 red and 6 blue counters. A counter is drawn, its colour noted, and replaced. A second counter is then drawn. Find P(both same colour).

1

With replacement: probabilities stay the same. P(R) = 4/10 = 0.4, P(B) = 6/10 = 0.6

2

P(RR) = 0.4 × 0.4 = 0.16

3

P(BB) = 0.6 × 0.6 = 0.36

4

P(same colour) = P(RR) + P(BB) = 0.16 + 0.36 = 0.52

Worked Example03

A bag contains 4 red and 6 blue counters. Two counters are drawn without replacement. Find P(both same colour).

1

First draw: P(R) = 4/10, P(B) = 6/10. Without replacement: total becomes 9, and counts change.

2

P(RR) = (4/10) × (3/9) = 12/90

3

P(BB) = (6/10) × (5/9) = 30/90

4

P(same colour) = 12/90 + 30/90 = 42/90 = 7/15

"At Least One" Problems

When asked for "at least one", you have two approaches:

Method 1: Direct

Add all paths that satisfy "at least one"

P(≥1) = P(exactly 1) + P(exactly 2) + ...

Method 2: Complement

Subtract "none" from 1

P (\geq 1) = 1 - P (none)

When to Use Complement

The complement method is usually faster when there are many ways to get "at least one" but only one way to get none.

Worked Example04

A fair die is rolled 3 times. Find the probability of getting at least one six.

1

P(six) = 1/6, P(not six) = 5/6. Events are independent (dice rolls).

2

Using complement: P(at least one six) = 1 − P(no sixes)

3

P(no sixes in 3 rolls) = (5/6) × (5/6) × (5/6) = (5/6)³ = 125/216

4

P(at least one six) = 1 − 125/216 = 91/216 ≈ 0.421

Note: The direct method would require adding P(exactly 1) + P(exactly 2) + P(exactly 3) — much longer!

Three or More Branches

Sometimes the first event has more than two outcomes. The tree extends with multiple branches from the start.

Key Points

Branches from start must sum to 1: 0.1 + 0.6 + 0.3 = 1

Each set of second-stage branches must also sum to 1

Multiply along each path, add paths for combined outcomes

Worked Example05

Bill travels to work by car with probability 0.1, by bus with probability 0.6, or by train. The probabilities of being late are 0.55 (car), 0.3 (bus), and 0.05 (train).

(a) Find P(Bill is late).

(b) Given Bill is late, find P(he travelled by bus).

1

P(Train) = 1 − 0.1 − 0.6 = 0.3. Now build the tree and find late outcomes:

a)

P(Late) = P(Car ∩ L) + P(Bus ∩ L) + P(Train ∩ L) = 0.055 + 0.18 + 0.015 = 0.25

b)

P(Bus | Late) = P(Bus ∩ Late) / P(Late) = 0.18 / 0.25 = 0.72 or 18/25

Three-Stage Trees

For three successive events, extend the tree to a third level. The same rules apply:

Extended Trees

Multiply along all three branches for each outcome

Add the probabilities of all paths that satisfy your condition

Each set of branches from a node must still sum to 1

Be Systematic

With 3 stages, you get 2³ = 8 outcomes (for binary choices). List them methodically and check your probabilities sum to 1.

Worked Example06

In a game, a player has probability 0.6 of winning each round. Find the probability that in 3 rounds, the player wins exactly 2.

1

P(Win) = 0.6, P(Lose) = 0.4. Rounds are independent.

2

Ways to win exactly 2: WWL, WLW, LWW (highlighted paths)

3

Each path: 0.6 × 0.6 × 0.4 = 0.144

4

P(exactly 2 wins) = 0.144 × 3 = 0.432

Alternative: 3 × (0.6)² × (0.4)¹ = 3 × 0.36 × 0.4 = 0.432

Trees and Conditional Probability

Tree diagrams can help find conditional probabilities by identifying relevant outcomes.

Using Trees for P(A|B)

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

Use the tree to find both

P (A \cap B)

and

P (B)

Strategy

Draw the tree and calculate all outcome probabilities

Identify outcomes where B occurs → add these for P(B)

Identify outcomes where both A and B occur → this is P(A ∩ B)

Divide: P(A|B) = P(A ∩ B) / P(B)

Worked Example07

Box A contains 3 red and 2 blue balls. Box B contains 4 red and 6 blue balls. A fair coin is tossed: heads means pick from A, tails from B. Given that a red ball is picked, find the probability that it came from Box A.

1

P(A) = P(B) = 0.5 (fair coin). P(R|A) = 3/5, P(R|B) = 4/10 = 2/5

2

P(A and R) = 0.5 × (3/5) = 3/10 = 0.3

3

P(B and R) = 0.5 × (2/5) = 2/10 = 0.2

4

P(Red) = P(A and R) + P(B and R) = 0.3 + 0.2 = 0.5

5

P(A | Red) = P(A and R) / P(Red) = 0.3 / 0.5 = 0.6 or 3/5

Worked Example08

A factory has two machines, A and B. Machine A produces 60% of items and has a 2% defect rate. Machine B produces 40% of items and has a 5% defect rate.

(a) Find the probability that a randomly selected item is defective.

(b) Given that an item is defective, find the probability it was made by Machine B.

1

P(A) = 0.6, P(B) = 0.4. P(D|A) = 0.02, P(D|B) = 0.05

2

P(A and D) = 0.6 × 0.02 = 0.012

3

P(B and D) = 0.4 × 0.05 = 0.020

a)

P(Defective) = P(A and D) + P(B and D) = 0.012 + 0.020 = 0.032

b)

P(B | D) = P(B and D) / P(D) = 0.020 / 0.032 = 20/32 = 5/8 = 0.625

Worked Example09

A bag contains 6 green beads and 4 yellow beads. A bead is taken at random, the colour is recorded and it is not replaced. A second bead is then taken from the bag.

Given that both balls are the same colour, find the probability that they are both yellow.

1

Draw tree: P(G₁) = 6/10, P(Y₁) = 4/10. Without replacement: after G → 5G, 4Y in 9; after Y → 6G, 3Y in 9

"Given same colour" restricts the sample space to only GG and YY outcomes. We find P(YY) ÷ P(GG or YY).

Answer: 2/7

10

Worked Example

A genetic condition is present in 4% of a population. A test is developed to detect it.

If a person has the condition, the test is positive with probability 0.9

If a person does not have the condition, the test is positive with probability 0.02

(a) Find the probability that a randomly selected person tests negative.

(b) Given that a person tests negative, find the probability they have the condition.

1

Let C = has condition, T⁺ = tests positive, T⁻ = tests negative

P(C) = 0.04, P(C') = 0.96, P(T⁺|C) = 0.9, P(T⁺|C') = 0.02

Part (b) asks for P(has condition | negative test). This is the false negative rate - very low (0.4%) because most negatives are true negatives from the 96% without the condition.

Answers: (a) 0.9448 (b) 0.00423

11

Worked Example

A box contains 7 sweet flavoured and 3 sour flavoured jelly beans. Two jelly beans are taken one after the other and eaten.

Find the probability that both jelly beans are sweet given that at least one of them is sweet.

1

Draw tree diagram for without replacement:

"At least one sweet" eliminates only the SoSo path. We need P(SwSw) ÷ P(all paths except SoSo).

Answer: 1/2

Practice

Question 1: A bag contains 5 green and 3 yellow balls. Two balls are drawn without replacement. Find the probability that both balls are the same colour.

1

First draw: P(G) = 5/8, P(Y) = 3/8

2

Without replacement: If G first → P(G) = 4/7, P(Y) = 3/7. If Y first → P(G) = 5/7, P(Y) = 2/7

3

P(GG) = (5/8) × (4/7) = 20/56

4

P(YY) = (3/8) × (2/7) = 6/56

5

P(same colour) = 20/56 + 6/56 = 26/56 = 13/28

Practice

Question 2: Three independent tests are conducted. Each test has probability 0.7 of success. Find the probability of at least two successes.

1

P(S) = 0.7, P(F) = 0.3. Need P(exactly 2) + P(exactly 3)

2

P(exactly 3) = P(SSS) = 0.7³ = 0.343

3

P(exactly 2): SSF, SFS, FSS → 3 ways, each = (0.7)²(0.3) = 0.147

4

P(exactly 2) = 3 × 0.147 = 0.441

5

P(at least 2) = 0.343 + 0.441 = 0.784

Practice

Question 3: In a medical test, 1% of the population has a disease. The test correctly identifies 95% of those with the disease and 90% of those without. A person tests positive. Find the probability they actually have the disease.

1

P(D) = 0.01, P(D') = 0.99. P(+|D) = 0.95, P(+|D') = 0.10 (false positive)

2

P(D and +) = 0.01 × 0.95 = 0.0095

3

P(D' and +) = 0.99 × 0.10 = 0.099

4

P(+) = 0.0095 + 0.099 = 0.1085

5

P(D|+) = 0.0095 / 0.1085 = 0.0876 ≈ 8.8%

Despite the positive test, there's only about 9% chance of having the disease — because the disease is rare!

Summary

Tree Diagram Types

TypeWhen to Use

Two-branch, two-stageTwo binary events (e.g., win/lose twice)

Three-branch first stageThree initial options (e.g., Car/Bus/Train)

Three-stage treesThree sequential events (8 outcomes)

Conditional Probability from Trees

Given ConditionMethod

Given B happenedP(A|B) = P(A ∩ B) / P(B)

Given same outcomeRestrict to matching paths only

Given at least oneUse 1 − P(none) for denominator

Exam Skills

Branches from each node must sum to 1

Without replacement: denominators decrease by 1

Medical tests: distinguish false positive/negative rates

"At least one" → complement method: 1 − P(none)