AQA GCSE Mathematics Paper 2 (Foundation), June 2025: Question 19

Probability Tree Diagrams: Independent Events

📋 Specification Focus

What this question tests

This question assesses your core foundation in probability tree diagrams for independent trials:

Recalling that the probabilities on any set of branches from a single node must add up to 1 .
Completing missing branches using the complementary rule: P(not A) = 1 - P(A) .
Recognising independent events where the outcome of Spinner 1 has no impact on the probabilities for Spinner 2.
Applying the multiplication rule along branches to calculate combined outcomes: P(A and B) = P(A) × P(B) .

Part (a) • 2 Marks

Question 19 (a): Complete the Tree Diagram

Completing missing branch probabilities for two independent spinners

Diagram layout to complete:

Spinner 1: Upper branch is Red (3/8) → Lower branch is Green (___)
Spinner 2 (after Red): Upper branch is Red (1/5) → Lower branch is Green (___)
Spinner 2 (after Green): Upper branch is Red (1/5) → Lower branch is Green (___)

✅ Correct Values

Spinner 1 Green: 5/8 (or equivalent decimal 0.625 / 62.5% )
Spinner 2 Green (top): 4/5 (or 0.8 / 80% )
Spinner 2 Green (bottom): 4/5 (or 0.8 / 80% )

Mark Scheme Breakdown:
• B1: 5/8 on spinner 1 (green)
• B1: 4/5 on spinner 2 (green) in both positions

📐 Step-by-Step Calculations

Spinner 1 Green branch:
The sum of outcomes from the starting point must be 1.
1 - 3/8 = 8/8 - 3/8 = 5/8
Spinner 2 Green branches:
The second spin is independent, so both pairs sum to 1.
1 - 1/5 = 5/5 - 1/5 = 4/5

💡 Key Knowledge

Sum to 1: The branches splitting from any single node represent all possible mutually exclusive outcomes, so their probabilities must sum to 1.
Independent Events: The outcome of Spinner 1 does not affect Spinner 2. Therefore, Spinner 2 has the exact same probabilities ( 1/5 and 4/5 ) regardless of whether Spinner 1 was Red or Green.

❌ Common Traps

Different denominators: Changing the denominator on Spinner 2 (treating it like non-replacement ball/card questions). Spinners always retain their original total sections.
Missing a branch: Only filling in one of the two 4/5 branches on Spinner 2 and leaving the other blank. Both must be filled to secure the mark.
Writing ratios: Writing odds like 3:5 instead of a probability fraction 5/8 .

Part (b) • 1 Mark

Question 19 (b): Combined Probability

Work out the probability that both spinners land on red

✅ Correct Answer

Answer: 3/40

Also accepted: 0.075 or 7.5% .

Mark Scheme:
• B1: 3/40 or equivalent fraction, decimal or percentage.
Examiner note: Ignore any incorrect attempt to simplify after the correct answer 3/40 is seen.

📐 Step-by-Step Calculation

Trace the required path:
Follow branch Red on Spinner 1, then branch Red on Spinner 2.
Multiply along the path ("AND" rule):
P(Red and Red) = P(Spinner 1 Red) × P(Spinner 2 Red)
Calculate:
3/8 × 1/5 = (3 × 1) / (8 × 5) = 3/40

🧠 Exam Technique

Across = Multiply: When moving along branches to find combined events (e.g. Red and Red), multiply the fractions.
Down = Add: If asked for alternative routes (e.g. "at least one Red"), add the final probabilities of the separate paths together.
Leave unsimplified: You do not need to reduce fractions unless explicitly instructed. 3/40 cannot be simplified further anyway, but leaving it as standard avoids silly arithmetic slips.

❌ Common Errors

Adding instead of multiplying: Students often incorrectly calculate 3/8 + 1/5 = 15/40 + 8/40 = 23/40 . Remember: consecutive events multiply!
Cross-multiplying: Confusing fraction multiplication with solving proportional equations. Simply multiply numerators straight across and denominators straight across.