Mathematics • Year 8 • Unit 4 • Lesson 18
Tree Diagrams in the Real World
Apply tree diagrams and the multiplication rule to real scenarios: school commute, biased weather forecast, basketball free throws, raffle draws, and quiz multiple-choice guessing.
1. Word problems
Each problem uses tree-diagram skills from Lesson 18 — draw the tree, multiply along paths, add path probabilities for combined events. Show working.
1.1 — Two-day school commute. On any school day, Mia has P(catches bus on time) = 0.8 (and so P(misses bus) = 0.2). She travels two days in a row. Bus arrivals are independent.
(a) Draw the tree diagram. Show probabilities on each branch.
(b) Find P(on time both days).
(c) Find P(misses bus at least once across the two days). 4 marks
1.2 — Weather forecast. A weather model says P(rain tomorrow) = 0.3 and P(rain the day after) = 0.4, treating the two days as independent.
(a) Draw the tree (rain R / no rain N for each day).
(b) Find P(rain both days) and P(no rain either day).
(c) Verify the four endpoints sum to 1. 4 marks
1.3 — Basketball free throws. A Year 8 player has a free-throw success rate of 3/4. She shoots two free throws (independent).
(a) Draw the tree.
(b) Find P(both shots in), P(one in and one missed), P(both missed).
(c) Check that the three probabilities sum to 1. 4 marks
1.4 — Raffle. A raffle has 10 tickets: 3 winning, 7 losing. Two tickets are drawn without replacement.
(a) Find the stage-2 branch probabilities for each Stage-1 outcome.
(b) Find P(both winning).
(c) Find P(at least one winning ticket). 4 marks
1.5 — Quiz guessing. A two-question multiple-choice quiz: question 1 has 4 options, question 2 has 5 options. Eli guesses both at random.
(a) State P(correct on Q1) and P(correct on Q2).
(b) Find P(both correct) and P(both wrong).
(c) Find P(exactly one correct). 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate writes: "For drawing 2 marbles from a bag, I'll just use the same branch probabilities at each stage." In your own words explain (i) when their reasoning is correct (with replacement), (ii) when it is wrong (without replacement) and what changes, and (iii) give a worked example with 4 Red and 1 Blue marble, drawing 2 without replacement, showing how P(both Red) is different from the with-replacement case. Use the term without replacement somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Two-day commute
(a) Stage 1 and Stage 2 branches: P(on time) = 0.8, P(miss) = 0.2 at each stage (independent).
(b) P(on time both) = 0.8 × 0.8 = 0.64.
(c) P(at least one miss) = 1 − 0.64 = 0.36.
1.2 — Weather forecast
(a) Day-1 branches: R = 0.3, N = 0.7. Day-2 branches (same regardless of Day 1): R = 0.4, N = 0.6.
(b) P(rain both) = 0.3 × 0.4 = 0.12. P(no rain either) = 0.7 × 0.6 = 0.42.
(c) RR = 0.12; RN = 0.18; NR = 0.28; NN = 0.42. Sum = 0.12 + 0.18 + 0.28 + 0.42 = 1.00 ✓.
1.3 — Basketball free throws
(a) Each branch: P(hit) = 3/4, P(miss) = 1/4.
(b) P(both in) = 3/4 × 3/4 = 9/16. P(one in, one missed) = (3/4 × 1/4) + (1/4 × 3/4) = 3/16 + 3/16 = 6/16 = 3/8. P(both missed) = 1/4 × 1/4 = 1/16.
(c) Check: 9/16 + 6/16 + 1/16 = 16/16 = 1 ✓.
1.4 — Raffle without replacement
(a) Stage 1: P(W) = 3/10, P(L) = 7/10. Stage 2 if Stage 1 was W: P(W | W) = 2/9, P(L | W) = 7/9. Stage 2 if Stage 1 was L: P(W | L) = 3/9, P(L | L) = 6/9.
(b) P(both W) = 3/10 × 2/9 = 6/90 = 1/15.
(c) P(both L) = 7/10 × 6/9 = 42/90 = 7/15. P(at least one W) = 1 − 7/15 = 8/15.
1.5 — Quiz guessing
(a) P(Q1 correct) = 1/4; P(Q2 correct) = 1/5.
(b) P(both correct) = 1/4 × 1/5 = 1/20. P(both wrong) = 3/4 × 4/5 = 12/20 = 3/5.
(c) P(exactly one correct) = (1/4 × 4/5) + (3/4 × 1/5) = 4/20 + 3/20 = 7/20.
2.1 — Explain your thinking (sample response)
Your classmate's reasoning is correct only when the experiment is with replacement: the drawn item is returned before the next draw, so the bag is restored and the branch probabilities stay the same at every stage. When the experiment is without replacement, the item is NOT returned — both the denominator AND the numerator of the second-stage probability change depending on the first outcome. For a bag of 4 R and 1 B (5 marbles): with replacement, P(both R) = 4/5 × 4/5 = 16/25 = 0.64. Without replacement, after drawing 1 R, only 3 R and 1 B remain (4 marbles total), so the second-stage branch gives P(R | R) = 3/4. Therefore P(both R) = 4/5 × 3/4 = 12/20 = 3/5 = 0.60, which is LOWER than the with-replacement answer because we have used up one of the red marbles in the first draw.
Marking: 1 mark for "with replacement" condition; 1 mark for explaining what changes without replacement; 1 mark for correctly worked with-replacement value; 1 mark for correctly worked without-replacement value with comparison.