How to Solve: Probability Trees (Conditional & Independent)

How to Solve: Probability Trees (Conditional & Independent)

Exam Context: GCSE (9-1) & A-Level Maths (Edexcel/AQA/OCR) Score Impact: 4-6 marks per question (10-15% of paper). Mastering trees unlocks conditional probability, independent events, and real-world risk problems (e.g., medical testing, sports predictions).

Introduction

"Imagine you’re a doctor deciding whether to test a patient for a rare disease. The test isn’t perfect—it can give false positives. How do you calculate the chance the patient actually has the disease if the test comes back positive? Probability trees are the secret weapon for cracking this—and they’re worth easy marks if you follow the steps."

What You Need To Know First

1. Basic probability: P(A) = (Number of favourable outcomes) / (Total outcomes).
2. Independent vs. dependent events:
3. Independent: One event doesn’t affect the other (e.g., flipping a coin twice).
4. Dependent/Conditional: One event does affect the other (e.g., drawing cards without replacement).
5. Multiplication rule for "AND" probabilities: P(A and B) = P(A) × P(B|A).

Key Vocabulary

Formulas To Know

1. Multiplication Rule (AND)
2. Formula: P(A and B) = P(A) × P(B|A)
3. Variables:
   • P(A) = Probability of event A.
   • P(B|A) = Probability of B given A happened.

4. MEMORISE THIS (not always given on exam sheet).

5. Addition Rule (OR)

6. Formula: P(A or B) = P(A) + P(B) – P(A and B)

7. MEMORISE THIS (for non-mutually exclusive events).

8. Conditional Probability

9. Formula: P(B|A) = P(A and B) / P(A)

10. Given on exam sheet (but understand it!).

11. Independent Events Check

12. Formula: P(A and B) = P(A) × P(B) if and only if A and B are independent.
13. MEMORISE THIS (key for spotting independence).

Step-by-Step Method

Step 1: Draw the Tree Structure

• Start with a single node (the "root").
• From the root, draw branches for the first event (e.g., "Rain" or "No Rain").
• Label each branch with its probability (e.g., P(Rain) = 0.3).
• At the end of each branch, add a new node for the next event.
• Repeat for all events.

Step 2: Label All Probabilities

• For independent events, probabilities stay the same on each branch (e.g., P(Heads) = 0.5 every time).
• For conditional events, probabilities change based on the previous outcome (e.g., P(Second Red | First Red) = 4/12 if a card is drawn without replacement).
• Write probabilities on the branches, not at the nodes.

Step 3: Calculate Path Probabilities

• Multiply the probabilities along each path to find P(A and B and C...).
• Example: P(Rain and Umbrella) = P(Rain) × P(Umbrella | Rain).

Step 4: Answer the Question

• "Find P(A and B)": Multiply along the path.
• "Find P(A or B)": Add the relevant path probabilities (and subtract overlaps if needed).
• "Find P(B|A)": Use P(A and B) / P(A) (or read from the tree).

Step 5: Check for Independence (If Asked)

• Calculate P(A) × P(B) and compare to P(A and B).
• If equal → independent.
• If not equal → dependent.

Worked Examples

Example 1 – Basic (Independent Events)

Question: A fair coin is flipped twice. Draw a probability tree and find: a) P(Two Heads) b) P(One Head and One Tail)

Step 1: Draw the Tree - First flip: Heads (0.5), Tails (0.5). - Second flip: From each outcome, Heads (0.5), Tails (0.5).

Step 2: Label Probabilities - All branches = 0.5 (independent).

Step 3: Calculate Paths - P(HH) = 0.5 × 0.5 = 0.25 - P(HT) = 0.5 × 0.5 = 0.25 - P(TH) = 0.5 × 0.5 = 0.25 - P(TT) = 0.5 × 0.5 = 0.25

Step 4: Answer the Questions a) P(Two Heads) = P(HH) = 0.25 b) P(One Head and One Tail) = P(HT) + P(TH) = 0.25 + 0.25 = 0.5

What we did and why: - Drew a tree to visualise all outcomes. - Multiplied along paths for "AND" probabilities. - Added paths for "OR" probabilities (mutually exclusive here).

Example 2 – Medium (Conditional Events)

Question: A bag contains 3 red balls and 2 blue balls. Two balls are drawn without replacement. a) Draw a probability tree. b) Find P(Both Red). c) Find P(One Red and One Blue).

Step 1: Draw the Tree - First draw: Red (3/5), Blue (2/5). - Second draw: From Red, Red (2/4), Blue (2/4). From Blue, Red (3/4), Blue (1/4).

Step 2: Label Probabilities - First branch: P(Red) = 3/5, P(Blue) = 2/5. - Second branch: Probabilities change because balls aren’t replaced.

Step 3: Calculate Paths - P(RR) = (3/5) × (2/4) = 6/20 = 0.3 - P(RB) = (3/5) × (2/4) = 6/20 = 0.3 - P(BR) = (2/5) × (3/4) = 6/20 = 0.3 - P(BB) = (2/5) × (1/4) = 2/20 = 0.1

Step 4: Answer the Questions b) P(Both Red) = P(RR) = 0.3 c) P(One Red and One Blue) = P(RB) + P(BR) = 0.3 + 0.3 = 0.6

What we did and why: - Recognised conditional probability (balls not replaced). - Updated probabilities after each draw. - Added paths for "OR" (mutually exclusive).

Example 3 – Exam-Style (Disguised Problem)

Question: A factory makes light bulbs. 5% are faulty. A test catches 90% of faulty bulbs but also wrongly flags 2% of good bulbs as faulty. a) Draw a probability tree. b) Find P(Bulb is faulty and test says faulty). c) Find P(Test says faulty). d) Find P(Bulb is faulty given test says faulty).

Step 1: Define Events - F = Faulty, G = Good. - T+ = Test says faulty, T- = Test says good.

Step 2: Draw the Tree - First branch: P(F) = 0.05, P(G) = 0.95. - Second branch: - From F: P(T+|F) = 0.9, P(T-|F) = 0.1. - From G: P(T+|G) = 0.02, P(T-|G) = 0.98.

Step 3: Calculate Paths - P(F and T+) = 0.05 × 0.9 = 0.045 - P(F and T-) = 0.05 × 0.1 = 0.005 - P(G and T+) = 0.95 × 0.02 = 0.019 - P(G and T-) = 0.95 × 0.98 = 0.931

Step 4: Answer the Questions b) P(F and T+) = 0.045 c) P(T+) = P(F and T+) + P(G and T+) = 0.045 + 0.019 = 0.064 d) P(F|T+) = P(F and T+) / P(T+) = 0.045 / 0.064 ≈ 0.703 (70.3%)

What we did and why: - Translated words into events (F, G, T+, T-). - Used conditional probability (test accuracy depends on bulb status). - Applied Bayes’ Theorem (part d) without needing the formula—just the tree!

Common Mistakes

Exam Traps

1-Minute Recap (Night Before the Exam)

"Listen up—probability trees are easy marks if you follow the steps. Here’s the drill: 1. Draw the tree: Start with the first event, split into branches, label probabilities. 2. Update probabilities: If events are dependent (e.g., no replacement), change the numbers on the second branches. 3. Multiply along paths: For ‘AND’ probabilities, multiply the numbers on the branches. 4. Add paths for ‘OR’: If the question asks for ‘at least one’ or ‘either’, add the relevant path probabilities. 5. Conditional probability? Use P(A and B) / P(B) or read from the tree.

Watch out for: - ‘Without replacement’ = dependent events (update probabilities!). - ‘Given that’ = conditional probability (use the tree or formula). - Independence? Check if P(A) × P(B) = P(A and B).

You’ve got this. Draw the tree, label carefully, and show every step—examiners love method marks!