Mathematics • Year 10 • Unit 4 • Lesson 16
Compound Events in the Real World
Apply Lesson 16's tree diagrams and rules (×, +) to scenarios from sport, traffic, retail and games. Decide each time: are the events independent or dependent? Mutually exclusive or not?
1. Word problems
Each scenario asks you to (i) classify the events and (ii) compute a probability. Show working. A bare answer earns at most half marks.
1.1 — Penalty shootout. An A-League striker scores penalties with probability 0.8. She takes two penalties in a shootout. Assume the kicks are independent.
(a) Draw a tree diagram for the two kicks.
(b) Find P(scores both kicks).
(c) Find P(scores exactly one kick). 3 marks
1.2 — Traffic lights on the way to school. Marcus passes two traffic lights on his ride to school. The lights are timed independently. The first light is green when he arrives 60% of the time; the second is green 70% of the time.
(a) Find P(both lights green).
(b) Find P(at least one light green).
(c) Lesson 16 warns "n(A or B) = n(A) + n(B)" is wrong. Use this warning to explain why P(at least one green) ≠ 0.60 + 0.70. 3 marks
1.3 — Canteen line. The school canteen sells 10 sandwiches: 4 chicken, 3 ham, 3 salad. Two students walk up and each buy one sandwich (so this is without replacement). The first student picks at random; the second then picks from the remaining nine.
(a) Are the two events independent or dependent? Justify in one sentence using Lesson 16's Key Term "with replacement".
(b) Find P(both buy chicken). 3 marks
1.4 — Two-die board game. In a board game, you roll two fair six-sided dice. You "win the round" if you roll a double (two of the same number) OR a sum of 7.
(a) List the outcomes for "double" and the outcomes for "sum of 7". How many are in each? Are the two events mutually exclusive?
(b) Use the correct addition rule to find P(double or sum of 7). 3 marks
1.5 — Two-question quiz. A Year 10 multiple-choice quiz has two questions. For Q1, four options (only one correct). For Q2, five options (only one correct). Maya guesses both at random.
(a) Find P(she gets both right).
(b) Find P(she gets at least one right).
(c) Draw a tree diagram showing the four outcomes and their probabilities, and confirm they sum to 1. 3 marks
2. Explain your thinking
Communication question — use full sentences. 4 marks
2.1 A classmate says: "P(rain on Saturday) = 0.4 and P(rain on Sunday) = 0.4, so P(rain on the weekend) = 0.4 + 0.4 = 0.8." Write a four-sentence reply that (i) names the rule the classmate has misused, (ii) explains why the rule does not apply here, (iii) shows the correct calculation if you assume the two days are independent, and (iv) finishes with a rule of thumb for when to add versus multiply probabilities.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Penalty shootout
(a) Tree: kick 1 → S (0.8) or M (0.2); kick 2 same.
(b) P(SS) = 0.8 × 0.8 = 0.64.
(c) P(exactly one) = P(SM) + P(MS) = 0.8×0.2 + 0.2×0.8 = 0.16 + 0.16 = 0.32.
1.2 — Traffic lights
(a) Independent → P(both green) = 0.60 × 0.70 = 0.42.
(b) P(none green) = 0.40 × 0.30 = 0.12. So P(at least one green) = 1 − 0.12 = 0.88.
(c) Adding 0.60 + 0.70 = 1.30 — a probability > 1, which is impossible. The error is the lesson's "n(A or B) = n(A) + n(B)" trap: it double-counts the case where both lights are green.
1.3 — Canteen line
(a) Dependent. The first student does not replace the sandwich, so the second student picks from a smaller bag — Lesson 16 Key Term: "with replacement" is the condition that keeps draws independent. Without replacement → dependent.
(b) P(both chicken) = 4/10 × 3/9 = 12/90 = 2/15 ≈ 0.133.
1.4 — Doubles and sum of 7
(a) Doubles: (1,1)(2,2)(3,3)(4,4)(5,5)(6,6) — 6 outcomes. Sum 7: (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) — 6 outcomes. A double has equal numbers, so its sum is even; 7 is odd. So no outcome is in both sets → events are mutually exclusive.
(b) P(double or sum 7) = 6/36 + 6/36 = 12/36 = 1/3.
1.5 — Random-guess quiz
P(right on Q1) = 1/4. P(right on Q2) = 1/5. Independent.
(a) P(both right) = 1/4 × 1/5 = 1/20.
(b) P(none right) = 3/4 × 4/5 = 12/20 = 3/5. So P(at least one right) = 1 − 3/5 = 2/5.
(c) Tree outcomes RR (1/20), RW (4/20), WR (3/20), WW (12/20) — they sum to 20/20 = 1. ✓
2.1 — Explain your thinking (sample response)
The classmate has used the addition rule for mutually exclusive events, but Saturday rain and Sunday rain are not mutually exclusive — it can rain on both days. So adding 0.4 + 0.4 double-counts the outcome "rain on both", and even worse it can give an answer over 1 if the values are bigger. Assuming the two days are independent, P(no rain either day) = 0.6 × 0.6 = 0.36, so P(rain on the weekend) = 1 − 0.36 = 0.64. Rule of thumb: add probabilities only when the events cannot both occur; multiply probabilities when the events are independent and you want them both to occur.
Marking: 1 mark per part (i)-(iv).