Tree Diagrams Independent Events — AQA GCSE Mathematics

SPEC P8 FOUNDATION + HIGHER

Tree Diagrams: Independent Events

Multiply along a path, add between paths — the core rules for calculating combined probabilities

Why Tree Diagrams?

Sample space grids and Venn diagrams work well for one-stage problems. When an experiment has two or more stages — like flipping a coin then rolling a die — a tree diagram shows every possible sequence clearly.

What a tree diagram does

Breaks a multi-stage experiment into branches at each stage.

Every possible path from root to tip gives one combined outcome.

Multiplying the probabilities along a path gives the probability of that combined outcome.

This lesson vs the next

Lesson 5.03 (this lesson): Independent events — the probabilities on each stage don't change. Lesson 5.04 (Higher tier only): Dependent events — "without replacement", where probabilities DO change between stages.

Independent Events

Two events are independent when the outcome of one does not affect the probability of the other.

Classic independent scenarios

Flipping two separate coins.

Rolling a die and flipping a coin.

Drawing a counter, replacing it, then drawing again.

Rain on two separate days (if stated independent).

The key signal: "with replacement"

If a question says "with replacement", the draws are independent — probabilities stay the same each time. If it says "without replacement", draws are dependent (that's Lesson 5.04).

P8 spec point

AQA requires you to "know the underlying assumptions" — which means you should be able to state why two events are independent (e.g. "because the counter is replaced before the second draw").

Anatomy of a Tree Diagram

Every tree diagram has the same basic structure. Learn the parts, and every tree question becomes the same problem.

Parts of a tree diagram

Nodes (dots): points where the tree branches.

Branches (lines): each carries one possible outcome of one stage.

Probability labels: written on each branch.

Outcome labels: written at the end of each branch.

Path: one complete route from start to final outcome.

Building a Tree — Coin Flipped Twice

A fair coin is flipped twice. P(heads) = 0.5, P(tails) = 0.5 — the flips are independent, so the second flip has the same probabilities regardless of the first.

Check the probabilities

All four path probabilities sum to 1: 0.25 + 0.25 + 0.25 + 0.25 = 1. ✓ Every tree diagram should give a total of 1 across all paths.

Branches at Each Node Sum to 1

At every branching point in the tree, the probabilities going out must sum to 1. This is the "exhaustive outcomes" rule from Lesson 5.01 applied to each stage.

Why this matters

Finding the complement probability is often the missing branch. If P(success) = 0.3 at some stage, then P(not success) = 1 − 0.3 = 0.7 — fill it in automatically.

Two-branch node

Weighted coin: P(H) = 0.7.

Then P(T) = 1 − 0.7 = 0.3. Branches sum: 0.7 + 0.3 = 1. ✓

Multi-branch node (3+ branches)

3-colour spinner: P(Red) = 0.2, P(Blue) = 0.3, P(Green) = ?

Green = 1 − (0.2 + 0.3) = 0.5. Branches sum: 0.2 + 0.3 + 0.5 = 1. ✓

Trees are not always binary

Nodes can have 2, 3, 4, or more branches — whatever matches the outcomes of that stage. AQA uses 3-colour counter trees regularly. The multiply-along-path and add-between-paths rules work identically regardless of how many branches there are.

Examiner alert

Before you calculate anything, check the branches at each node sum to 1. If they don't, one label is wrong — every downstream calculation will be wrong too. Full marks get lost to a single mis-labelled branch.

Multiply Along a Path (AND)

A path is one complete route from the start of the tree to a final outcome. The probability of that combined outcome is the product of the branch probabilities along the path.

P(first event AND second event) = P(first) × P(second)

Why "AND" means multiply

For two independent events to both happen, you need the first, and then the second. Probabilities of a sequence compound — which mathematically is multiplication.

Ex

Coin tossed twice — P(HH)

Path: start → H (0.5) → H (0.5).

P(H AND H) = 0.5 × 0.5 = 0.25

Same for any specific path: HT, TH, TT all give 0.25.

Add Between Paths (OR)

Many events can happen via more than one path. To find the probability of such an event, add the path probabilities.

P(path 1 OR path 2) = P(path 1) + P(path 2)

Why "OR" means add

Different paths through the tree are mutually exclusive (they can't happen at the same time). From Lesson 5.01, for mutually exclusive events, P(A or B) = P(A) + P(B).

Ex

Coin tossed twice — P(exactly one head)

"Exactly one head" happens via two paths: HT or TH.

P(HT) = 0.5 × 0.5 = 0.25

P(TH) = 0.5 × 0.5 = 0.25

P(exactly one H) = 0.25 + 0.25 = 0.5

The Core Rule: AND = ×, OR = +

Every tree-diagram question reduces to one decision per step: is it an AND (along one path) or an OR (across paths)? Get this right and the rest is arithmetic.

Signal word in questionOperationWhy

"Both", "and", "then", "followed by"× (multiply)Combining sequential outcomes (one path)

"Either", "or", "one of", "exactly one"+ (add)Combining separate paths

"At least one"1 − P(none)Complement shortcut (next slide)

Spec P8 note

AQA explicitly tests "knowing when to add and when to multiply two or more probabilities". The rules scale to any number of events: for three independent flips, P(HHH) = 0.5 × 0.5 × 0.5 = 0.125. Multiplying along the path works no matter how many stages.

The Formal Definition of Independence

The formula P(A B) = P(A) × P(B) isn't just a shortcut — it's the mathematical definition of what it means for two events to be independent. This is the "underlying assumption" the P8 spec asks you to know.

A, B independent ↔ P(A B) = P(A) × P(B)

When to use the formula directly

For P(A and B) with independent events, go straight to the formula — no tree needed. For "exactly one" or "at least one" questions, a tree (or the 1 − P(none) shortcut) is usually cleaner since multiple paths must be combined.

Ex

Coin and die — formula used directly

A fair coin is flipped and a fair die is rolled. Find P(heads AND a 6).

These are independent (one doesn't affect the other). P(heads) = 1/2. P(6) = 1/6.

P(heads 6) = 12 × 16 = 112

Other representations (P8 spec)

The same P(A B) calculation works via any representation you've met: the tree diagram, the sample-space grid from 5.02 (count the favourable cells for "H and 6"), or a two-way table. The formula is the fastest for simple AND questions; trees are best when a question has several path-based sub-parts.

MCE Worked Example — Part A

Ex

Two fair dice are rolled. Find P(two sixes).

M — Method: Independent events (one die doesn't affect the other). Build a tree with branches "6" and "not 6" for each die, then multiply along the path for two 6s.

C — Calculate: Follow the highlighted path (6 then 6):

P(two sixes) = 16 × 16 = 136

E — Evaluate: P(two sixes) = 1/36. Sense-check: all four path probabilities sum to 1/36 + 5/36 + 5/36 + 25/36 = 36/36 = 1. ✓

MCE Worked Example — Part B

Ex

Same two dice. Find P(exactly one six).

M — Method: "Exactly one six" happens via two different paths: (6, not 6) OR (not 6, 6). Multiply along each path, then ADD the paths.

Four outcomes of the two-dice tree — the two matching paths are highlighted

C — Calculate: Add the two highlighted path probabilities.

P(exactly one 6) = 536 + 536 = 1036 = 518

E — Evaluate: P(exactly one six) = 5/18.

The "exactly one" trap

Many students forget the second path and give only 5/36. "Exactly one" means either order — so you must include both (6, not 6) and (not 6, 6).

"At Least One" — The Complement Shortcut

"At least one" questions can have many winning paths. A faster method uses the complement rule from 5.01:

P(at least one) = 1 − P(none)

Why the shortcut works

"At least one" and "none" are complementary events — they cover every possibility between them. P(at least one) + P(none) = 1, so P(at least one) = 1 − P(none).

Ex

Same two dice. Find P(at least one six).

Method 1 (long): Add all winning paths: P(6,6) + P(6,not 6) + P(not 6,6) = 1/36 + 5/36 + 5/36 = 11/36.

Method 2 (shortcut): Find P(no sixes) and subtract from 1:

P(no sixes) = 56 × 56 = 2536

P(at least one 6) = 1 − 2536 = 1136

Both methods give the same answer — but the shortcut has just one multiplication and one subtraction, versus three path calculations.

MCE Practice Question

The probability it rains on any given Saturday is 0.3. The probability it rains on any given Sunday is 0.4. Rain on one day is independent of rain on the other.

(a) Complete a tree diagram for rain on Saturday and rain on Sunday. [2 marks]

(b) Find the probability it rains on both days. [2 marks]

(c) Find the probability it rains on exactly one of the two days. [3 marks]

(d) Find the probability it rains on at least one of the two days. [3 marks]

Total: [10 marks]

M

Events are stated independent. Build a tree: Sat (rain 0.3, no rain 0.7), Sun (rain 0.4, no rain 0.6). (b) Multiply along one path. (c) Add the two "exactly one" paths. (d) Use the complement shortcut 1 − P(neither).

C

(a) Sat branches: 0.3, 0.7. Sun branches from each: 0.4, 0.6.

(b) P(rain both) = 0.3 × 0.4 = 0.12.

(c) P(rain Sat, not Sun) + P(not Sat, rain Sun) = (0.3 × 0.6) + (0.7 × 0.4) = 0.18 + 0.28 = 0.46.

(d) P(neither) = 0.7 × 0.6 = 0.42. P(at least one) = 1 − 0.42 = 0.58.

E

(b) 0.12, (c) 0.46, (d) 0.58. Sense-check all four paths: 0.12 + 0.46 + 0.42 = 1.00. ✓

Examiner Alert — AND vs OR Confusion

The most common mistake

Students add when they should multiply (along a path) or multiply when they should add (between paths).

SituationOperationExample

Two stages of one path (e.g. H then H)×P(HH) = 0.5 × 0.5 = 0.25

Two different paths for the same event+P(one H) = P(HT) + P(TH) = 0.5

Quick test

Ask: "Does this outcome happen via one path (sequence of events, both happening)? → MULTIPLY. Or does it happen via multiple paths (any one of several sequences)? → ADD."

Examiner Alert — Missing Paths & Concept Traps

"Exactly one" needs BOTH orders

If a question asks for P(exactly one 6), students often calculate just P(6, not 6) = 5/36 and stop — missing P(not 6, 6) = 5/36. The real answer is 5/36 + 5/36 = 10/36 = 5/18.

"At least one" misread as "exactly one"

"At least one" includes "all of them" too, not just "one of them". P(at least one 6) = P(one 6) + P(two 6s), or use the shortcut 1 − P(no 6s).

Mutually exclusive ≠ independent

These are different concepts, and in general opposites. If A and B are mutually exclusive (both with non-zero probability), they are not independent — knowing A happened tells you B didn't.

Mutually exclusive: P(A B) = 0

Independent: P(A B) = P(A) × P(B)

Assuming independence without checking

The P(A) × P(B) formula works only if A and B are independent. For questions using "without replacement", the first outcome does affect the second — that's Lesson 5.04.

Common Errors Summary

Students often writeCorrect versionWhy it matters

Add along a pathMultiply along a pathAND = × — sequential events compound

Multiply across pathsAdd across pathsOR = + — mutually exclusive paths sum

P(exactly one X) = one path onlyAdd both orders"Exactly one" has 2 paths on a 2-stage tree

Branches don't sum to 1Check each node sums to 1Applies to 2, 3, 4+ branches alike

P(A B) = P(A) + P(B)= P(A) × P(B) for independent eventsAND is multiply, not add

"Mutually exclusive" = "independent"These are different conceptsM.E.: P(A B)=0. Indep: P(A B)=P(A)×P(B)

Use independent formula without replacementCheck: is the first outcome replaced?Without replacement needs 5.04 method

✓ Checkpoint: What You Should Now Be Able to Do

Core Skills

Identify independent events and "with replacement" scenarios.

Build a tree diagram with any number of branches per node (2, 3, or more).

Check that branches at each node sum to 1.

Multiply along a single path for P(A AND B) type events.

Add between paths for P(A OR B) type events.

Apply the independence definition P(A B) = P(A) × P(B).

Use the complement P(at least one) = 1 − P(none) shortcut.

Distinguish mutually exclusive (P(A B) = 0) from independent (P(A B) = P(A) × P(B)).

Key formulae

P(A B) = P(A) × P(B) (independent events)

P(A OR B) via tree = P(path 1) + P(path 2) + ...

P(at least one) = 1 − P(none)

Next lesson (Higher tier only)

Lesson 5.04 extends tree diagrams to dependent events — "without replacement" — where the probabilities on the second branch change based on the first outcome. This also covers conditional probability.