Mathematics • Year 7 • Unit 4 • Lesson 14

Probability — Real World

Apply P(event) = favourable ÷ total and the complement rule to real situations: raffle tickets, classroom name draws, library books, weather forecasts and dice games.

Apply · Real-World Maths

1. Word problems

Each scenario asks you to identify the sample space and apply the probability formula. Write probabilities as fractions AND decimals where possible.

1.1 — School raffle. The Year 7 raffle has 200 tickets. You buy 8 tickets.

(a) What is the sample space size?
(b) What is P(you win the prize) as a fraction and a decimal?
(c) What is P(you do NOT win)?    3 marks

Stuck on (c)? Use the complement rule: P(not win) = 1 − P(win).

1.2 — Classroom name draw. A class has 25 students: 14 girls and 11 boys. The teacher draws one name at random to answer a question.

(a) Find P(girl).
(b) Find P(boy).
(c) Verify that P(girl) + P(boy) = 1.    3 marks

Stuck on (c)? Add your two fractions and check whether they total 1.

1.3 — Library shelf. A library shelf has 30 books: 12 fiction, 15 non-fiction, 3 reference. A book is selected at random.

(a) Find P(fiction).
(b) Find P(non-fiction).
(c) Find P(not reference).    3 marks

Stuck on (c)? Use the complement rule, OR add fiction + non-fiction.

1.4 — Weather forecast. A forecast says the probability of rain tomorrow is 0.65.

(a) What is P(no rain)?
(b) Is rain more or less likely than no rain? Justify.
(c) The same forecast last week said 0.4. Was that wetter or drier than tomorrow's forecast?    3 marks

Stuck on (b)? Compare 0.65 to 0.5 (the "even chance" mark).

1.5 — Dice game. A board game asks a player to roll a die and "win" if they roll a 5 or a 6.

(a) Find P(win) by listing the favourable outcomes.
(b) Find P(lose) using the complement rule.
(c) If 60 players play this game once each, about how many would you expect to win? Use P(win) × 60.    4 marks

Stuck on (c)? Expected wins = probability × number of trials.

2. Explain your thinking

Communication matters. Use full sentences. 4 marks

2.1 A friend says: "I rolled three sixes in a row, so my next roll is definitely going to be a six too." Explain (i) what fraction of the time you would expect to roll a six on any single roll of a fair die, (ii) why the past three sixes do NOT change the probability of the next roll, and (iii) what P(rolling a six on the 4th roll) actually is.

Stuck? Each roll is independent — the die has no "memory" of past rolls.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — School raffle

(a) Total tickets = 200, so sample space size = 200.
(b) P(you win) = 8 / 200 = 1/25 = 0.04.
(c) P(not win) = 1 − 1/25 = 24/25 = 0.96.

1.2 — Classroom name draw

(a) P(girl) = 14/25 = 0.56.
(b) P(boy) = 11/25 = 0.44.
(c) Check: 14/25 + 11/25 = 25/25 = 1 ✓.

1.3 — Library shelf

(a) P(fiction) = 12/30 = 2/5 = 0.4.
(b) P(non-fiction) = 15/30 = 1/2 = 0.5.
(c) P(not reference) = 1 − 3/30 = 27/30 = 9/10 = 0.9.

1.4 — Weather forecast

(a) P(no rain) = 1 − 0.65 = 0.35.
(b) Rain is more likely than no rain because 0.65 > 0.5 (the even-chance mark).
(c) Last week's forecast (0.4) was DRIER than tomorrow's (0.65) — a lower probability of rain.

1.5 — Dice game

(a) Favourable = {5, 6} = 2 outcomes. P(win) = 2/6 = 1/3 ≈ 0.333.
(b) P(lose) = 1 − 1/3 = 2/3 ≈ 0.667.
(c) Expected wins ≈ 1/3 × 60 = 20 winners.

2.1 — Explain your thinking (sample response)

(i) On any single roll of a fair die, P(six) = 1/6 — there is one favourable face out of 6 equally likely outcomes. (ii) Past rolls do NOT change the probability of the next roll because each roll is independent. The die has no "memory" — it does not know what was rolled before. Three sixes in a row is unusual but does not influence the next roll. (iii) P(rolling a six on the 4th roll) is still exactly 1/6 ≈ 0.167, the same as any other single roll.

Marking: 1 for stating P(six) = 1/6; 1 for explaining independence ("die has no memory"); 1 for explicitly stating P(6 on 4th roll) = 1/6; 1 for clear sentences.