Mathematics • Year 9 • Unit 4 • Lesson 13

Probability in the Real World

Use sample spaces, the probability formula and the complement rule in real situations — raffle tickets, weather forecasts, lottery odds, Uber arrivals and Year 9 class data. Then explain a common misconception in your own words.

Apply · Real-World Maths

1. Word problems

Each problem uses the formula P(E) = favourable ÷ total, the complement rule, or a sample space from Lesson 13. Show your working — a single final answer with no working only earns half marks.

1.1 — Year 9 fundraiser raffle. A school raffle has 250 tickets sold. You buy 6 tickets. A single winning ticket is drawn at random.

(a) Find P(you win).
(b) Find P(you do not win), once by direct count and once using the complement rule. 4 marks

Stuck? Favourable = your 6 tickets. Total = 250. Both methods for (b) should give the same fraction.

1.2 — Weather forecast. The Bureau of Meteorology forecasts P(rain on Saturday) = 0.65 and P(rain on Sunday) = 0.4. (Assume the days are described separately.)

(a) Find P(no rain on Saturday) and P(no rain on Sunday).
(b) Which day is more likely to be dry, and by how much (in percentage points)? 4 marks

Stuck? Use the complement rule on each day. "Dry" = "no rain".

1.3 — Year 9 class data. In a Year 9 cohort of 120 students: 70 study Music, 45 study Art and 30 study both.

(a) If a student is chosen at random, find P(student studies Music).
(b) Find P(student studies neither Music nor Art). 4 marks

Stuck? Use a Venn diagram in your head: Music only = 70 − 30 = 40; Art only = 45 − 30 = 15; both = 30. Neither = 120 − (40 + 15 + 30).

1.4 — Uber arrival times. Past data suggests that for a particular pickup spot, P(driver arrives within 5 minutes) = 0.72.

(a) Find P(driver takes longer than 5 minutes).
(b) Out of 200 bookings, about how many would you expect to take longer than 5 minutes? 4 marks

Stuck? Multiply the complement probability by the number of bookings to estimate the expected count.

1.5 — Lottery odds. The Oz Lotto Division 1 odds are about 1 in 45 million for a single game played.

(a) Write P(win Division 1 with one game) as a decimal (to 3 significant figures, in scientific notation).
(b) Find P(not winning Division 1 with one game) as a decimal to 8 decimal places. 4 marks

Stuck? 1 ÷ 45 000 000 ≈ 2.22 × 10⁻⁸. The complement is almost (but not quite) 1.

2. Explain your thinking

This question is about communication, not just answers. Use full sentences. 4 marks

2.1 A classmate insists "If you toss a fair coin and get 5 heads in a row, the next toss is much more likely to be a tail to even things out." In your own words, explain (i) why this is wrong, (ii) what the actual probability of a tail on the next toss is, and (iii) what kind of mistake this is (often called the "gambler's fallacy"). Mention "independent" somewhere in your answer.

Stuck? The coin has no memory. Each toss has a fresh sample space of {H, T}.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Raffle

(a) P(you win) = 6 ÷ 250 = 3/125 (= 0.024 = 2.4%).
(b) Direct: P(not win) = (250 − 6) ÷ 250 = 244 ÷ 250 = 122/125.
Complement: P(not win) = 1 − 3/125 = 125/125 − 3/125 = 122/125 ✓.

1.2 — Weather

(a) P(no rain Saturday) = 1 − 0.65 = 0.35. P(no rain Sunday) = 1 − 0.4 = 0.6.
(b) Sunday is more likely to be dry. Gap = 0.6 − 0.35 = 0.25 = 25 percentage points more chance of a dry day on Sunday.

1.3 — Year 9 cohort

(a) P(Music) = 70 ÷ 120 = 7/12.
(b) Music only = 40, Art only = 15, both = 30 → total taking at least one = 40 + 15 + 30 = 85. Neither = 120 − 85 = 35. P(neither) = 35 ÷ 120 = 7/24.

1.4 — Uber arrivals

(a) P(takes longer than 5 min) = 1 − 0.72 = 0.28.
(b) Expected count = 0.28 × 200 = 56 bookings.

1.5 — Oz Lotto

(a) P(win) = 1 ÷ 45 000 000 ≈ 2.22 × 10⁻⁸.
(b) P(not winning) = 1 − 2.22 × 10⁻⁸ ≈ 0.99999998.
So even buying one ticket gives you almost a 1.000 probability of not winning Division 1 — exactly why the lesson's "Real-World Anchor — Gambling and Insurance" warns that 1 in 45 million is essentially zero in everyday terms.

2.1 — Explain your thinking (sample response)

Each coin toss is independent — the coin has no memory of what happened before. The 5 heads in a row do not change the physics of the next toss, so the sample space is still {H, T} with each outcome equally likely. The actual probability of a tail on the next toss is still 1/2. This mistake is called the gambler's fallacy: believing that random results "balance out" over short runs. They balance out in the long run only because more trials swamp the early streak — not because the coin "owes" anyone a tail.

Marking: 1 mark for naming "independent"; 1 mark for stating P(T) = 1/2 on the next toss; 1 mark for explicit mention of "gambler's fallacy" or equivalent; 1 mark for a clear full-sentence explanation.