P8: Tree Diagrams
Calculate the probability of independent and dependent combined events using tree diagrams
📋 Key Concepts
Tree Diagrams: A visual way to show all possible outcomes of combined events. Each branch represents an outcome with its probability.
Key Rules
| Rule | Description |
|---|---|
| Branches sum to 1 | P(A) + P(not A) = 1 at each branching point |
| Multiply along | Multiply probabilities along branches for combined probability |
| Add across | Add probabilities of different ways to get the same outcome |
| Check total = 1 | All final outcomes should sum to 1 |
📝 Drawing Tree Diagrams
Structure: Start on the left. Each branch shows an outcome with its probability written on it. Each branch splits into further branches for subsequent events.
Example 1
A fair coin is flipped twice. Draw a tree diagram.
Tree diagram:
┌── H (1/2) ──→ HH (1/4)
H (1/2)
│ └── T (1/2) ──→ HT (1/4)
Start
│ ┌── H (1/2) ──→ TH (1/4)
T (1/2)
└── T (1/2) ──→ TT (1/4)
Probabilities: P(HH) = 1/2 × 1/2 = 1/4
Check: 1/4 + 1/4 + 1/4 + 1/4 = 1 ✓
📝 Independent Events
Independent: The outcome of one event doesn't affect the probability of another. Probabilities stay the same at each branch.
Example 2
A bag contains 3 red and 7 blue counters. A counter is drawn, replaced, and another drawn. Find P(at least one red).
Solution:
P(red) = 3/10, P(blue) = 7/10
┌── R (3/10) ──→ RR (9/100)
R (3/10)
│ └── B (7/10) ──→ RB (21/100)
Start
│ ┌── R (3/10) ──→ BR (21/100)
B (7/10)
└── B (7/10) ──→ BB (49/100)
P(at least one red) = P(RR) + P(RB) + P(BR)
= 9/100 + 21/100 + 21/100 = 51/100 = 0.51
Alternative: 1 - P(BB) = 1 - 49/100 = 51/100
📝 Dependent Events
Dependent: The outcome of one event affects the probability of another. Probabilities change after the first event (without replacement).
Example 3
A bag contains 3 red and 7 blue counters. Two counters are drawn WITHOUT replacement. Find P(one of each colour).
Solution:
First draw: P(R) = 3/10, P(B) = 7/10
After red: 2R, 7B left → P(R) = 2/9, P(B) = 7/9
After blue: 3R, 6B left → P(R) = 3/9, P(B) = 6/9
┌── R (2/9) ──→ RR (6/90)
R (3/10)
│ └── B (7/9) ──→ RB (21/90)
Start
│ ┌── R (3/9) ──→ BR (21/90)
B (7/10)
└── B (6/9) ──→ BB (42/90)
P(one of each) = P(RB) + P(BR)
= 21/90 + 21/90 = 42/90 = 7/15
📝 Using Tree Diagrams for Calculation
Method: Multiply along branches, then add probabilities for the outcomes you want.
Example 4
Weather forecast: P(rain) = 0.3. If it rains, P(bus late) = 0.6. If not rain, P(bus late) = 0.2. Find P(bus late).
Solution:
┌── Late (0.6) ──→ RL (0.18)
Rain (0.3)
│ └── On time (0.4) ──→ RT (0.12)
Start
│ ┌── Late (0.2) ──→ NL (0.14)
No rain (0.7)
└── On time (0.4) ──→ NT (0.56)
P(bus late) = P(RL) + P(NL) = 0.18 + 0.14 = 0.32
📝 Three or More Events
Example 5
A coin is flipped 3 times. Find P(exactly 2 heads).
Solution:
Tree has 8 endpoints: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Each has probability (1/2)³ = 1/8
Exactly 2 heads: HHT, HTH, THH (3 outcomes)
P(exactly 2 heads) = 3 × 1/8 = 3/8
🧮 Worked Examples — Exam Style
Tree diagrams: branches multiply ALONG the path, results ADD across paths. Without replacement, the second-stage fractions change.
A coin is flipped and a die rolled. Tree or space: P(heads and an even number)?
P(H) × P(even) = ½ × ½ = 1/4
M1 for multiplying along the branch, A1 for 1/4. Independent events: the second probability ignores the first.
P(rain Sat) = 0.6, P(rain Sun) = 0.6, independent. Find P(rain on exactly one day).
RS + S̄Su: 0.6 × 0.4 + 0.4 × 0.6 = 0.24 + 0.24 = 0.48
M1 for the two paths, A1 for 0.48. 'Exactly one' = the two mixed paths — add them.
Exam wording to expect: "complete the tree diagram", "find the probability that …", with/without replacement pairs.
❓ Practice Questions
Q1: A biased coin has P(H) = 0.7. It's flipped twice. Find P(two heads).
Q2: A bag has 4 red and 6 blue balls. Two are drawn WITH replacement. Find P(both blue).
Q3: Same bag, but WITHOUT replacement. Find P(both blue).
Q4: A test has 80% accuracy for a disease. If someone has the disease, P(positive) = 0.8. If healthy, P(positive) = 0.2. 5% of people have the disease. Find P(positive test).
Q5: A coin is flipped 3 times. Find P(at least one tail).
Q6: A bag has 5 red and 3 blue balls. Two drawn WITHOUT replacement. P(one of each)?
Q7: A test is 95% reliable: P(positive|has bug) = 0.95, P(negative|no bug) = 0.95. If 10% have the bug, find P(positive result).
✅ Answers
- P(HH) = 0.7 × 0.7 = 0.49
- P(BB) = 6/10 × 6/10 = 36/100 = 0.36
- P(BB) = 6/10 × 5/9 = 30/90 = 1/3
- P(positive) = 0.05 × 0.8 + 0.95 × 0.2 = 0.04 + 0.19 = 0.23
- 1 - P(HHH) = 1 - 1/8 = 7/8
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. 0.7 × 0.7 = 0.49
Q2. 6/10 × 6/10 = 0.36
Q3. 6/10 × 5/9 = 30/90 = 1/3
Q4. 0.05 × 0.8 + 0.95 × 0.2 = 0.23 (using the page’s stated prior P(bug) = 0.05)
Q5. 1 − P(HHH) = 1 − 1/8 = 7/8
Q6. RB + BR = 5/8 × 3/7 + 3/8 × 5/7 = 15/56 + 15/56 = 15/28
Q7. 0.1 × 0.95 + 0.9 × 0.05 = 0.095 + 0.045 = 0.14
📌 Key Facts to Memorise
- Along the path: MULTIPLY · Across paths: ADD
- Branch sums at each stage = 1 — the completion check
- Without replacement: denominators drop (5/8 then 4/7)
- "At least one" → 1 − P(none)
- "Exactly one" → add the two mixed paths
🗺️ In the Exam
Tree diagrams: 4–6 marks on Papers 1/2 — the biggest probability marks. Draw the full tree (both stages, both outcomes) even when only one path is needed.
Links to: possibility spaces (P7), conditional probability (P9), mutual exclusivity (P4).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Always label branches with both the outcome AND probability
- Check branches from each point sum to 1
- For "without replacement" problems, update probabilities after each draw
- "At least one" = 1 - P(none) - use this shortcut!
- Check your final answer - all probabilities should sum to 1
🧠 Problem-Solving Strategies
Problem-Solving
For tree diagram problems: (1) Always label each branch with both outcome AND probability, (2) Check that branches from each node sum to 1, (3) Multiply along branches for combined probability, (4) Add across different paths to the same result, (5) For "at least" problems, use 1 − P(none), (6) For "without replacement", update probabilities after each draw.
Multi-Step Problem
A bag contains 4 red and 6 blue balls. Two balls are drawn WITHOUT replacement. Find P(both same colour) and P(at least one red).
Solution: P(RR) = 4/10 × 3/9 = 12/90. P(BB) = 6/10 × 5/9 = 30/90. P(same colour) = 12/90 + 30/90 = 42/90 = 7/15. P(at least one red) = 1 − P(BB) = 1 − 30/90 = 60/90 = 2/3.
⚠️ Common Errors
Watch Out!
1. Wrong: Adding probabilities along branches instead of multiplying Correct: Multiply along branches (AND rule), add across different paths (OR rule)
2. Wrong: Using the same probabilities for both draws in a "without replacement" problem Correct: After removing one item, the total decreases by 1 and the remaining count changes — update all probabilities
3. Wrong: Only counting one order (e.g. RB) when there are two ways to get one of each (RB and BR) Correct: P(one of each) = P(RB) + P(BR) — add both paths on the tree
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A bag contains 3 red, 5 blue and 2 green counters. Two counters are drawn at random without replacement. (a) Draw a tree diagram showing all outcomes. (b) Find P(both counters are the same colour). (c) Find P(at least one green counter).
Total = 10 counters.
(a) First draw: P(R)=3/10, P(B)=5/10, P(G)=2/10. After R: P(R)=2/9, P(B)=5/9, P(G)=2/9. After B: P(R)=3/9, P(B)=4/9, P(G)=2/9. After G: P(R)=3/8, P(B)=5/8, P(G)=1/8.
(b) P(same colour) = P(RR) + P(BB) + P(GG) = (3/10×2/9) + (5/10×4/9) + (2/10×1/8) = 6/90 + 20/90 + 2/80 = 6/90 + 20/90 + 2.25/90 = 28.25/90. Converting: 6/90 + 20/90 + 1/40 = 2/30 + 1/6 + 1/40 = 8/120 + 20/120 + 3/120 = 31/120.
(c) P(at least one green) = 1 − P(no green) = 1 − [P(RR) + P(RB) + P(BR) + P(BB)] = 1 − (6/90 + 15/90 + 15/90 + 20/90) = 1 − 56/90 = 34/90 = 17/45.
Mark scheme: M1 for correct first branch probabilities, M1 for correct second branch probabilities, A1 for tree diagram, M1 for same colour calculation, A1 for 31/120, M1 for complement method, A1 for 17/45
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A medical test for a disease has a 95% accuracy rate. 1% of the population has the disease. If someone has the disease, P(positive) = 0.95. If someone does not have the disease, P(positive) = 0.05.
(a) Draw a tree diagram and find P(positive test result).
(b) Find P(has disease | positive test). This is the "real" accuracy from the patient's perspective.
(c) Explain why most positive tests are actually false positives despite the 95% accuracy.
Answers: (a) P(positive) = 0.01×0.95 + 0.99×0.05 = 0.0095 + 0.0495 = 0.059. (b) P(disease|positive) = 0.0095/0.059 ≈ 0.161 (only 16.1%!). (c) Because the disease is rare (1%), the 5% false positive rate applied to 99% of people produces many more false positives than the true positives from the 1% who actually have the disease.
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