Independent Events and Tree Diagrams

This lesson covers independent events, the multiplication rule, and the use of tree diagrams — essential topics for the AQA GCSE Mathematics specification. Tree diagrams are one of the most versatile tools for solving probability problems and appear frequently in exam questions at both Foundation and Higher tier.

Independent Events

Two events are independent if the outcome of one event does not affect the outcome of the other.

Examples

Events	Independent?	Reason
Flipping a coin and rolling a die	Yes	The coin result does not affect the die
Picking a card from a pack, replacing it, then picking again	Yes	Replacement means the pack is the same each time
Picking a card, not replacing it, then picking again	No	The pack has changed — outcomes are affected
The weather today and the result of a die roll	Yes	These are completely unrelated events

The Multiplication Rule (AND Rule)

For independent events A and B:

P(A and B) = P(A) x P(B)

Worked Example 1

A fair coin is flipped and a fair die is rolled. What is the probability of getting heads AND a 6?

P(heads) = 1/2
P(6) = 1/6

P(heads and 6) = 1/2 x 1/6 = 1/12

Exam Tip: The word "and" in probability means multiply. The word "or" means add (for mutually exclusive events). Remember: AND = Multiply, OR = Add.

Tree Diagrams

A tree diagram shows all possible outcomes of a sequence of events, along with their probabilities. Each "branch" represents one possible outcome.

Rules for Tree Diagrams

The probabilities on branches from the same point must add up to 1.
To find the probability of a specific path, multiply along the branches.
To find the probability of multiple paths (an "or" situation), add the probabilities of the relevant paths.

Worked Example 2 — Two Coin Flips

A fair coin is flipped twice. Draw a tree diagram and find the probability of getting exactly one head.

graph LR
    S[ ] -->|1/2 H| A[H]
    S -->|1/2 T| B[T]
    A -->|1/2 H| C[HH = 1/4]
    A -->|1/2 T| D[HT = 1/4]
    B -->|1/2 H| E[TH = 1/4]
    B -->|1/2 T| F[TT = 1/4]

Paths with exactly one head: HT and TH

P(exactly one head) = 1/4 + 1/4 = 1/2

Tree Diagrams with Unequal Probabilities

Worked Example 3

The probability that Anya wins a game is 0.7. She plays two games. Find the probability that she wins exactly one game.

graph LR
    S[ ] -->|0.7 Win| A[Win]
    S -->|0.3 Lose| B[Lose]
    A -->|0.7 Win| C[WW = 0.49]
    A -->|0.3 Lose| D[WL = 0.21]
    B -->|0.7 Win| E[LW = 0.21]
    B -->|0.3 Lose| F[LL = 0.09]

Paths with exactly one win: WL and LW

P(exactly one win) = 0.21 + 0.21 = 0.42

Check: All path probabilities should add to 1: 0.49 + 0.21 + 0.21 + 0.09 = 1.00 (correct).

Exam Tip: Always check that the probabilities at the end of your tree diagram add up to 1. If they do not, you have made an error somewhere.

Without Replacement (Dependent Events)

When items are taken without replacement, the probabilities change for the second event. The events are not independent.

Worked Example 4

A bag contains 5 red and 3 blue balls. Two balls are taken at random without replacement. Find the probability that both balls are red.

First pick:

P(red) = 5/8
P(blue) = 3/8

Second pick (given first was red — 4 red and 3 blue left, 7 total):

P(red) = 4/7
P(blue) = 3/7

graph LR
    S[ ] -->|5/8 Red| A[Red]
    S -->|3/8 Blue| B[Blue]
    A -->|4/7 Red| C[RR = 20/56]
    A -->|3/7 Blue| D[RB = 15/56]
    B -->|5/7 Red| E[BR = 15/56]
    B -->|2/7 Blue| F[BB = 6/56]

P(both red) = 5/8 x 4/7 = 20/56 = 5/14

Worked Example 5

Using the same bag (5 red, 3 blue, without replacement), find the probability of getting one of each colour.

Paths with one of each colour: RB and BR

P(one of each) = 15/56 + 15/56 = 30/56 = 15/28

Exam Tip: "Without replacement" is a key phrase. When you see it, the probabilities on the second set of branches must change. The denominator decreases by 1 because there is one fewer item to choose from.

"At Least One" Problems

These problems are most efficiently solved using the complement rule combined with a tree diagram.

Worked Example 6

The probability that a bus is on time is 0.8. Jordan catches a bus on Monday and Tuesday (independent journeys). Find the probability that at least one bus is on time.

P(at least one on time) = 1 - P(neither on time)

P(not on time) = 1 - 0.8 = 0.2

P(neither on time) = 0.2 x 0.2 = 0.04

P(at least one on time) = 1 - 0.04 = 0.96

Exam Tip: For "at least one" questions, always consider using the complement: P(at least one) = 1 - P(none). This is almost always quicker than adding up all the individual paths.

Three-Event Tree Diagrams

Tree diagrams can extend to three or more events.

Worked Example 7

A biased coin has P(heads) = 0.6. It is flipped three times. Find P(exactly two heads).

The paths with exactly two heads are: HHT, HTH, THH

P(HHT) = 0.6 x 0.6 x 0.4 = 0.144
P(HTH) = 0.6 x 0.4 x 0.6 = 0.144
P(THH) = 0.4 x 0.6 x 0.6 = 0.144

P(exactly two heads) = 0.144 + 0.144 + 0.144 = 0.432

Independent Events and Tree Diagrams

Independent Events and Tree Diagrams