Mathematics • Year 8 • Unit 4 • Lesson 16

Sample Space in the Real World

Apply the counting principle and systematic listing to real situations: cafeteria meal combos, board games, traffic light cycles, dice games, and a coin-flip phone-game tournament.

Apply · Real-World Maths

1. Word problems

Each problem uses sample space ideas from Lesson 16 — counting principle, systematic listing, and probability from outcomes. Show your working — a bare answer earns only half marks.

1.1 — Canteen meal deal. A school canteen meal-deal lets you pick one main (pie, wrap, sushi) and one drink (water, juice, soft drink, milk).

(a) Use the counting principle to find how many different meal combos are possible.
(b) List all combos as ordered pairs (main, drink).
(c) If a customer picks one combo at random, find P(wrap AND water). 4 marks

Stuck? mains × drinks = 3 × 4. Now fix each main and list every drink underneath.

1.2 — Snakes & Ladders. In Snakes & Ladders, two dice are rolled and the player moves the total number of squares. Pria needs a total of exactly 11 to land on the ladder square.

(a) State the size of the sample space for two dice.
(b) List all ordered pairs (d₁, d₂) where the sum is 11.
(c) Find P(sum = 11). Simplify your fraction. 3 marks

Stuck? Only two pairs of dice values sum to exactly 11 — find them and remember order matters.

1.3 — Traffic light combo. An intersection has two traffic lights. Each shows either Red (R), Amber (A), or Green (G). A traffic engineer wants to model every possible combination of the two lights at a moment in time.

(a) How many combined states are possible? Use the counting principle.
(b) List the full sample space as ordered pairs (light₁, light₂).
(c) Find P(at least one light shows Red). 3 marks

Stuck on (c)? "At least one Red" means (R, anything) OR (anything, R) — but don't double-count (R, R).

1.4 — Dice betting game. A fairground game asks you to roll two dice. You win a prize if the sum is 7 or 11.

(a) Use the sample space to find P(sum = 7) and P(sum = 11).
(b) Find P(winning), where winning means sum = 7 OR sum = 11.
(c) Out of 360 rolls, how many wins should the stallholder expect on average? 4 marks

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

1.5 — Phone game tournament. A phone-game tournament uses best-of-three coin flips to break a tie. The player who calls "heads" wins a flip when the coin shows H. Player A calls heads for all three flips.

(a) List the full sample space for three coin flips.
(b) Player A wins the tie-break if they win at least 2 flips. Find P(Player A wins).
(c) Is the game fair to both players? Justify in one sentence using your answer. 3 marks

Stuck on (b)? Win at least 2 flips = exactly 2 H OR all 3 H. Count outcomes in {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

2. Explain your thinking

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

2.1 Your friend claims: "Rolling two dice is the same as rolling one die twice and combining the values — there are only 21 different sums (from 2 to 12), so |S| = 21, not 36." In your own words explain (i) why the sample space has 36 outcomes not 21, (ii) why P(sum = 7) and P(sum = 2) are different even though both are "one possible sum", and (iii) give one specific example of two ordered pairs that produce the same sum but count as different outcomes. Use the term ordered pair somewhere in your answer.

Stuck? Revisit lesson § "Spot the Trap" — (1, 2) and (2, 1) are different ordered pairs even though they have the same sum.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Canteen meal deal

(a) |S| = 3 × 4 = 12.
(b) (pie, water), (pie, juice), (pie, soft drink), (pie, milk), (wrap, water), (wrap, juice), (wrap, soft drink), (wrap, milk), (sushi, water), (sushi, juice), (sushi, soft drink), (sushi, milk).
(c) Favourable = (wrap, water) → 1 outcome. P(wrap and water) = 1/12.

1.2 — Snakes & Ladders

(a) |S| = 6 × 6 = 36.
(b) Pairs summing to 11: (5, 6) and (6, 5) → 2 outcomes.
(c) P(sum = 11) = 2 / 36 = 1/18.

1.3 — Traffic light combo

(a) |S| = 3 × 3 = 9.
(b) (R, R), (R, A), (R, G), (A, R), (A, A), (A, G), (G, R), (G, A), (G, G).
(c) "At least one Red" = (R, R), (R, A), (R, G), (A, R), (G, R) → 5 outcomes. P(at least one Red) = 5/9.

1.4 — Dice betting game

(a) P(sum = 7) = 6 / 36 = 1/6. P(sum = 11) = 2 / 36 = 1/18.
(b) P(winning) = 6/36 + 2/36 = 8/36 = 2/9.
(c) Expected wins = (2/9) × 360 = 80 wins.

1.5 — Phone game tournament

(a) S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} — 8 outcomes.
(b) Player A wins with 2 or 3 heads: HHH, HHT, HTH, THH → 4 outcomes. P(A wins) = 4 / 8 = 1/2.
(c) Yes — fair. Each player has a 1/2 chance of winning, so neither has an advantage.

2.1 — Explain your thinking (sample response)

The sample space for two dice has 36 outcomes, not 21, because each die is a separate experiment and we must record every ordered pair (d₁, d₂). Although 21 different sums are possible, those sums are not equally likely — some sums can happen in many ways while others can happen in only one way. For example, the sum 2 can only be made by (1, 1), but the sum 7 can be made by (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), or (6, 1) — six different ordered pairs. So P(sum = 7) = 6/36 = 1/6, but P(sum = 2) = 1/36. The ordered pairs (1, 2) and (2, 1) both sum to 3 but they are different outcomes, because one means die 1 = 1 and die 2 = 2, while the other means die 1 = 2 and die 2 = 1. The denominator must always be 36 — the total number of ordered pairs in the sample space.

Marking: 1 mark for stating |S| = 36 with reasoning; 1 mark for explaining why different sums have different probabilities; 1 mark for a specific ordered-pair example; 1 mark for clear full-sentence answer that uses "ordered pair".