Mathematics • Year 8 • Unit 4 • Lesson 16
Sample Space — Mixed Challenge
Pull together every idea from Lesson 16: the counting principle, systematic listing, probability from outcomes, and arrays for two-stage experiments. Six mixed problems, one "find the mistake", and one open-ended design challenge.
1. Mixed problems — choose the right move
Each question uses a different idea from Lesson 16. Show your working. 3 marks each
1.1 A coin is flipped, a die is rolled, and a 4-sector spinner (A, B, C, D) is spun. How many outcomes are in the combined sample space? Use the counting principle.
1.2 Two dice are rolled. Find P(both dice show the same number) by listing the favourable outcomes.
1.3 A bag contains 1 Red, 1 Blue, 1 Green, and 1 Yellow marble. Two marbles are drawn one after the other with replacement (the first is returned before the second draw). How many outcomes are in the sample space? Then find P(both Red).
1.4 A coin is flipped three times. Find P(at least one tail). (Hint: use the complement and the sample space of 8 outcomes.)
1.5 Two dice are rolled. Draw a 6 × 6 array (or describe it) listing all 36 ordered pairs (d₁, d₂). Use it to find P(sum is a multiple of 4). Show the favourable outcomes.
1.6 A 4-sided die (faces 1, 2, 3, 4) and a 5-sided die (faces 1, 2, 3, 4, 5) are rolled together. (a) State |S|. (b) Find P(both dice show the same number).
2. Find the mistake
A student attempted this problem. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Problem: Two standard dice are rolled. Find P(sum = 6).
Line 1: Sample space: |S| = 6 + 6 = 12 outcomes.
Line 2: Pairs summing to 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes.
Line 3: P(sum = 6) = 5 / 12.
Line 4: This is approximately 0.42 or 42%.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full and give the correct probability.
Stuck? Two dice means MULTIPLY the stage outcomes (6 × 6), not add them. Check what the counting principle says.3. Open-ended challenge — design a fair game
This question has many valid answers. 4 marks
3.1 Your job: design a fairground game that uses two dice. Each player pays $1 to play. They win $3 if their event occurs, otherwise they lose their $1.
Your design must include:
(i) Define the winning event clearly (e.g., "sum is more than 9", "doubles", "product is even").
(ii) List all favourable outcomes from the 36-outcome sample space.
(iii) Calculate the probability of winning. Simplify your fraction.
(iv) Decide whether the game is fair, profitable for the player, or profitable for the stallholder. Justify: out of 36 plays the stallholder expects to take in $36 and pay out $3 × (number of wins) = $___ . Net = $___ .
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Three-stage counting
|S| = 2 × 6 × 4 = 48 outcomes.
1.2 — P(both dice the same)
Doubles: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) → 6 outcomes. P(same) = 6 / 36 = 1/6.
1.3 — Marbles with replacement
|S| = 4 × 4 = 16. Favourable (both Red) = (R, R) → 1 outcome. P(both Red) = 1/16.
1.4 — Three coin flips, at least one T
|S| = 8. Only HHH has no tails, so P(no tails) = 1/8. P(at least one T) = 1 − 1/8 = 7/8.
1.5 — P(sum is a multiple of 4)
Multiples of 4 in range 2–12: 4, 8, 12.
Sum 4: (1,3), (2,2), (3,1) → 3 pairs.
Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 pairs.
Sum 12: (6,6) → 1 pair.
Total favourable = 3 + 5 + 1 = 9. P(sum mult of 4) = 9 / 36 = 1/4.
1.6 — 4-sided and 5-sided dice
(a) |S| = 4 × 5 = 20.
(b) Same number: (1,1), (2,2), (3,3), (4,4) → 4 outcomes. (Note: the 4-sided die has no 5, so there is no (5,5).) P(same) = 4 / 20 = 1/5.
2 — Find the mistake
(a) The mistake is on Line 1.
(b) The counting principle says we multiply the stage outcomes, not add them: |S| = 6 × 6 = 36, not 12. The student added when they should have multiplied.
(c) Corrected working:
|S| = 6 × 6 = 36.
Pairs summing to 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes.
P(sum = 6) = 5 / 36 ≈ 0.139 (about 13.9%). The student's listing in Line 2 was correct — only the denominator was wrong.
3 — Open-ended challenge (sample solution)
(i) Event: the sum of the two dice is greater than 9 (i.e., 10, 11, or 12).
(ii) Favourable outcomes:
Sum 10: (4,6), (5,5), (6,4) → 3 pairs.
Sum 11: (5,6), (6,5) → 2 pairs.
Sum 12: (6,6) → 1 pair.
Total = 6 outcomes.
(iii) P(win) = 6 / 36 = 1/6.
(iv) Fairness: In 36 expected plays, the stallholder collects 36 × $1 = $36. The stallholder pays out $3 × 6 = $18. Net profit = $36 − $18 = +$18 to the stallholder. The game is profitable for the stallholder: on average a player loses $18/36 = $0.50 per play. A "fair" game would need P(win) = 1/3 (so the $3 prize × 1/3 chance = $1 expected value to match the $1 stake).
Marking: 1 mark for clearly defined event; 1 mark for full list of favourable outcomes; 1 mark for correct simplified probability; 1 mark for fairness analysis with stallholder profit calculation.