Tree diagrams including conditional probability

What you'll learn

Tree diagrams represent sequential probability events and form a cornerstone of the AQA GCSE Mathematics Higher tier specification. This topic tests your ability to calculate probabilities across multiple stages, handle dependent events where earlier outcomes affect later probabilities, and apply conditional probability in real-world contexts. Questions worth 4-6 marks regularly appear on Paper 2 and Paper 3.

Key terms and definitions

Tree diagram — a visual representation showing all possible outcomes of two or more sequential events, with probabilities written on branches and outcomes at the end of each branch.

Conditional probability — the probability of an event occurring given that another event has already occurred, affecting the sample space or available outcomes.

Independent events — events where the outcome of one does not affect the probability of the other; probabilities remain constant across stages.

Dependent events — events where the outcome of one directly affects the probabilities of subsequent events; probabilities change between stages.

Branch — a line connecting stages in a tree diagram, labelled with the probability of that particular outcome.

Outcome — the final result at the end of a complete path through the tree, found by multiplying probabilities along that path.

Sample space — the set of all possible outcomes for an experiment or trial.

Without replacement — a sampling method where selected items are not returned to the population, creating dependent events and changing subsequent probabilities.

Core concepts

Structure of tree diagrams

A tree diagram consists of stages representing sequential events. Each stage branches into all possible outcomes for that event:

• The first stage begins from a single starting point
• Each branch represents one possible outcome
• Probabilities are written on branches (not at endpoints)
• All branches from the same point must sum to 1
• The diagram continues for as many stages as events in the problem
• Final outcomes appear at the right-hand end of the complete paths

When drawing tree diagrams, maintain clear spacing between branches and label every branch with both the outcome and its probability. Align stages vertically for clarity.

Calculating probabilities using tree diagrams

To find the probability of a specific outcome:

1. Identify the path through the tree that leads to that outcome
2. Multiply all probabilities along that path
3. This gives the probability of that complete sequence occurring

To find the probability when multiple paths lead to the desired outcome:

1. Calculate the probability for each individual path (multiply along each path)
2. Add the probabilities of all relevant paths together
3. This applies the addition rule: P(A or B) = P(A) + P(B) for mutually exclusive outcomes

Example calculation structure:

• P(Red then Blue) = P(Red) × P(Blue|Red)
• P(At least one Red) = P(Red, Red) + P(Red, Blue) + P(Blue, Red)

Independent events on tree diagrams

When events are independent, the probability of the second event remains unchanged regardless of the first event's outcome. Common contexts include:

• Tossing coins multiple times
• Rolling dice repeatedly
• Spinning spinners more than once
• Drawing cards with replacement (returning the card before the next draw)

For independent events, probabilities on the second stage are identical across all branches from the first stage. If a bag contains 3 red and 7 blue counters, and you replace the counter after each draw, every "second draw" branch shows P(Red) = 3/10 and P(Blue) = 7/10, regardless of what was drawn first.

Dependent events and conditional probability

Dependent events occur when the first outcome changes the conditions for the second event. The most frequent GCSE context is sampling without replacement:

• Selecting items from a bag, box, or group without returning them
• Choosing people from a team for different positions
• Drawing cards without replacing them in the deck

When items are removed, both the number of favourable outcomes and the total population change for subsequent events.

Calculating conditional probabilities:

If a bag contains 5 red and 3 blue counters (8 total), and you draw two without replacement:

First draw:

• P(Red) = 5/8
• P(Blue) = 3/8

Second draw depends on the first:

• If Red was drawn first: 4 red and 3 blue remain (7 total)
  • P(Red|first was Red) = 4/7
  • P(Blue|first was Red) = 3/7
• If Blue was drawn first: 5 red and 2 blue remain (7 total)
  • P(Red|first was Blue) = 5/7
  • P(Blue|first was Blue) = 2/7

Notice the denominator always decreases by 1 after the first draw. The numerator for the selected colour also decreases by 1, while other colours remain unchanged.

Reading conditional probability notation

AQA GCSE Mathematics uses the notation P(A|B), read as "the probability of A given B has occurred."

• The vertical bar | means "given that" or "conditional on"
• The event after the bar (B) is the condition that has already happened
• The event before the bar (A) is what you're finding the probability for

P(Green|Red drawn first) means "probability of drawing green on the second draw, given that red was drawn first."

Tree diagrams with more than two stages

Three-stage tree diagrams follow the same principles but with additional complexity:

• Each endpoint from stage 2 becomes a starting point for stage 3
• The number of final outcomes equals (outcomes per stage)^(number of stages)
• For two outcomes per stage across three stages: 2³ = 8 final outcomes
• Multiply along all three branches for each complete path
• Carefully track how each previous outcome affects subsequent probabilities

Three-stage trees appear less frequently but are worth significant marks (typically 5-7 marks) when they do appear.

Worked examples

Example 1: Without replacement (5 marks)

A box contains 7 chocolate biscuits and 5 plain biscuits. Sarah takes two biscuits at random without replacement.

(a) Complete the tree diagram. [3 marks]

                      Chocolate  ___
         Chocolate
        /          \
       /            Plain     ___
      /
Start
      \
       \            Chocolate  ___
        \          /
         Plain    
                  \ Plain     ___

(b) Calculate the probability that Sarah takes two biscuits of the same type. [2 marks]

Solution:

(a) First draw: 12 biscuits total

• P(Chocolate) = 7/12
• P(Plain) = 5/12

Second draw: 11 biscuits remain

After Chocolate drawn first: 6 chocolate, 5 plain remain

• P(Chocolate|Chocolate first) = 6/11
• P(Plain|Chocolate first) = 5/11

After Plain drawn first: 7 chocolate, 4 plain remain

• P(Chocolate|Plain first) = 7/11
• P(Plain|Plain first) = 4/11

(b) P(same type) = P(C, C) + P(P, P)

• P(C, C) = 7/12 × 6/11 = 42/132
• P(P, P) = 5/12 × 4/11 = 20/132
• P(same type) = 42/132 + 20/132 = 62/132 = 31/66

Example 2: Real-world context (6 marks)

The probability that it rains on a Saturday is 0.3. If it rains on Saturday, the probability that it rains on Sunday is 0.6. If it does not rain on Saturday, the probability that it rains on Sunday is 0.2.

(a) Draw a tree diagram to represent this information. [2 marks]

(b) Work out the probability that it rains on exactly one of the two days. [4 marks]

Solution:

(a)

Saturday              Sunday        Outcome
                      0.6  Rain     (R, R)
        0.3  Rain    /
                    \
                     0.4  No rain   (R, N)
Start
                     0.2  Rain      (N, R)
        0.7  No rain/
                    \
                     0.8  No rain   (N, N)

(b) "Exactly one day" means either (Rain, No rain) OR (No rain, Rain)

P(Rain Sat, No rain Sun) = 0.3 × 0.4 = 0.12

P(No rain Sat, Rain Sun) = 0.7 × 0.2 = 0.14

P(exactly one day rains) = 0.12 + 0.14 = 0.26

Example 3: Three outcomes per stage (4 marks)

A spinner has three equal sections: Red, Blue, and Green. James spins it twice.

Calculate the probability that he gets two different colours.

Solution:

Each outcome has probability 1/3 on each spin (independent events).

P(two different colours) = 1 − P(two same colours)

P(R, R) = 1/3 × 1/3 = 1/9

P(B, B) = 1/3 × 1/3 = 1/9

P(G, G) = 1/3 × 1/3 = 1/9

P(two same) = 1/9 + 1/9 + 1/9 = 3/9 = 1/3

P(two different) = 1 − 1/3 = 2/3

Alternative method: Count favourable outcomes directly

Total outcomes when spinning twice = 3 × 3 = 9

Same colour outcomes = 3 (RR, BB, GG)

Different colour outcomes = 9 − 3 = 6

P(two different) = 6/9 = 2/3

Common mistakes and how to avoid them

• Mistake: Adding probabilities along a single path instead of multiplying. Correction: Always multiply probabilities along branches in a path; only add probabilities when combining different paths that all satisfy the question requirement.

• Mistake: Forgetting to adjust probabilities after the first event in without-replacement problems. Correction: After each draw, reduce the total by 1 and reduce the numerator by 1 for the colour/item that was selected. Write the new total next to each stage.

• Mistake: Making all second-stage probabilities identical in a without-replacement scenario. Correction: Check whether the question says "with replacement" or "without replacement." Without replacement creates different probabilities on later branches depending on earlier outcomes.

• Mistake: Writing outcomes at branch endpoints instead of probabilities. Correction: Probabilities belong on the branches themselves; outcome labels (like "Red" or "Heads") go at the far right end or above/below the branches.

• Mistake: Using the original population total for second-stage probabilities. Correction: In without-replacement problems, the denominator changes. If you start with 10 items and remove 1, the second stage has denominator 9, not 10.

• Mistake: Calculating P(A or B) by multiplying when paths should be added. Correction: "Or" means addition (for mutually exclusive outcomes). "And" or "followed by" means multiplication along a path.

Exam technique for Tree diagrams including conditional probability

• Command word "Complete": You must fill in missing probabilities on a partially drawn tree diagram. Check that probabilities from the same point sum to 1. Show at least one calculation if space permits (1 mark for method, 1 mark for correct values).

• Command word "Calculate" or "Work out": Show your complete method. Write out the multiplication for each path, then show addition if combining paths. For 2-3 mark questions, correct answer without working may not earn full marks.

• Drawing your own tree diagram: Use a ruler for neatness. Label every branch clearly. If the question provides specific information (like "the probability it rains is 0.4"), make sure your tree includes every given value. Trees with missing or unclear labels lose marks even if calculations are correct.

• Fraction or decimal answers: Unless specified, either form is acceptable, but give fractions in simplest form. If the question uses decimals throughout, give your answer as a decimal. If it uses fractions, match that format. For probabilities, answers between 0 and 1 are expected; an answer greater than 1 indicates an error.

Quick revision summary

Tree diagrams visualise sequential probability events across multiple stages. Multiply probabilities along each path; add probabilities of different paths for "or" scenarios. Independent events maintain constant probabilities across stages. Dependent events (typically without-replacement contexts) require recalculating probabilities after each stage—reduce both numerator and denominator appropriately. Conditional probability notation P(A|B) means the probability of A given B occurred. Always check branches from the same point sum to 1, show full working, and express final answers in simplest form.