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.