Mathematics • Year 8 • Unit 4 • Lesson 16
Sample Space
Build fluency with listing sample spaces, applying the counting principle (m × n), and calculating probabilities from systematic outcome lists. One worked example, one guided example, then eight independent problems graded from foundation to extension.
1. I do — fully worked example
Watch how we list a sample space systematically, then count favourable outcomes.
Problem. A die is rolled and a coin is flipped. (a) List the sample space. (b) Find P(even number AND tails).
Step 1 — Count outcomes for each stage.
Die: 6 outcomes Coin: 2 outcomes
Reason: each stage has its own outcome count — we count these first.
Step 2 — Apply the counting principle.
|S| = 6 × 2 = 12 outcomes
Reason: multiply the stage counts (m × n) so we know how many ordered pairs to expect.
Step 3 — List the 12 ordered pairs systematically.
(1,H), (1,T), (2,H), (2,T), (3,H), (3,T),
(4,H), (4,T), (5,H), (5,T), (6,H), (6,T)
Reason: fix the die face, list both coin outcomes, then move on — nothing is missed.
Step 4 — Count favourable outcomes (even AND T).
Even: {2, 4, 6}. With T: (2,T), (4,T), (6,T) → 3 favourable
Step 5 — Apply the probability formula and simplify.
P(even and T) = 3 / 12 = 1 / 4
Answer: P(even and T) = 1/4.
2. We do — fill in the missing steps
Same shape as Section 1, but with the working faded. Fill in each blank. 4 marks
Problem. Two coins are flipped. (a) List the sample space. (b) Find P(exactly one head).
Step 1 — Stage outcome counts:
Coin 1: ______ outcomes Coin 2: ______ outcomes
Step 2 — Counting principle:
|S| = ______ × ______ = ______ outcomes
Step 3 — List all pairs:
(H, ___), (H, ___), (T, ___), (T, ___)
Step 4 — Count favourable (exactly one H):
Favourable: ______ and ______ → ______ outcomes
Step 5 — Apply formula:
P(exactly one H) = ______ / ______ = ______
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation, the middle two are standard, and the last two are extension.
Foundation — counting and listing
3.1 Write the sample space S for rolling a single standard die. State |S|. 1 mark
3.2 A coin is flipped and a 4-sector spinner (sectors A, B, C, D) is spun. How many outcomes are in the sample space? Show the calculation. 1 mark
3.3 Two dice are rolled. State |S|. Then explain in one sentence why (1, 2) and (2, 1) count as different outcomes. 1 mark
3.4 Three coins are flipped. Use the counting principle to state |S|, then list all the outcomes systematically. 1 mark
Standard — probability from sample space
3.5 Two dice are rolled. Find P(sum = 7) by listing all favourable pairs (d₁, d₂). Simplify your fraction. 2 marks
3.6 A coin is flipped three times. Find P(exactly 2 heads). List the favourable outcomes from the sample space {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. 2 marks
Extension — combine ideas
3.7 A bag has 1 Red, 1 Blue, and 1 Green marble. A marble is drawn AND a die is rolled. (a) How many outcomes are in the sample space? (b) Find P(Blue marble AND a prime number on the die). 2 marks
3.8 Two dice are rolled. Find P(sum ≤ 4) by listing all favourable ordered pairs systematically. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (two coins)
Step 1: Coin 1 = 2, Coin 2 = 2. Step 2: |S| = 2 × 2 = 4. Step 3: (H,H), (H,T), (T,H), (T,T). Step 4: Favourable = (H,T) and (T,H) → 2 outcomes. Step 5: P(exactly one H) = 2 / 4 = 1/2.
3.1 — Sample space for one die
S = {1, 2, 3, 4, 5, 6}. |S| = 6.
3.2 — Coin and 4-sector spinner
|S| = 2 × 4 = 8 outcomes.
3.3 — Two dice
|S| = 6 × 6 = 36. (1, 2) means die 1 = 1 and die 2 = 2; (2, 1) means die 1 = 2 and die 2 = 1 — different ordered outcomes, so both are counted separately.
3.4 — Three coins
|S| = 2 × 2 × 2 = 8. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
3.5 — P(sum = 7) on two dice
Favourable: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes. P(sum = 7) = 6 / 36 = 1/6.
3.6 — P(exactly 2 heads) on three coins
Favourable: HHT, HTH, THH → 3 outcomes. P(exactly 2 H) = 3/8.
3.7 — Marble and die
(a) |S| = 3 × 6 = 18. (b) Primes on die: {2, 3, 5}. Favourable: (B, 2), (B, 3), (B, 5) → 3 outcomes. P(Blue and prime) = 3 / 18 = 1/6.
3.8 — P(sum ≤ 4)
Sum 2: (1,1). Sum 3: (1,2), (2,1). Sum 4: (1,3), (2,2), (3,1). Total favourable = 6. P(sum ≤ 4) = 6 / 36 = 1/6.