Mathematics • Year 7 • Unit 4 • Lesson 18
Sample Space and Tree Diagrams
Build fluency with three ways to list every outcome of a multi-stage experiment: tree diagrams, the grid/array method, and the multiplication principle (options × options). Each complete path = one outcome.
1. I do — fully worked example
Read every line. The same experiment is solved by tree diagram, grid, and multiplication.
Problem. A die is rolled and a coin is flipped. List the sample space and find the total number of outcomes.
Step 1 — Tree diagram (paths read left to right).
Stage 1 (die): 1, 2, 3, 4, 5, 6 → 6 branches.
From each die branch, Stage 2 (coin): H or T → 2 branches.
Reason: every full left-to-right path is one outcome of the combined experiment.
Step 2 — Multiplication principle.
Total outcomes = Stage 1 × Stage 2 = 6 × 2 = 12
Reason: for each die result, there are 2 coin choices — so multiply, never add.
Step 3 — Listed sample space (check the count is 12).
{1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T} ✓ 12 outcomes.
Answer: 12 equally likely outcomes.
2. We do — fill in the missing steps
An outfit is made from 3 shirts (red, blue, white) and 4 pairs of pants (black, grey, navy, beige). Fill the blanks. 4 marks
Step 1 — Stage 1: shirt choices.
Shirt options = _______ List: ____, ____, ____
Step 2 — Stage 2: pant choices.
Pants options = _______ List: ____, ____, ____, ____
Step 3 — Apply the multiplication principle.
Total outfits = ___ × ___ = _______
Step 4 — Verify by listing for one shirt.
Red shirt: Red-black, Red-grey, Red-navy, Red-beige = ___ outfits.
Three shirts × 4 = _______ total. ✓
3. You do — independent practice
Show your method (tree, grid, or multiplication) and list the sample space when asked.
Foundation — count the outcomes
3.1 Two coins are flipped. How many outcomes are in the sample space? List them. 2 marks
3.2 Three coins are flipped at once. How many outcomes? List the sample space. 2 marks
3.3 A spinner with 4 colours is spun, then a coin is flipped. Use the multiplication principle to find the total outcomes. 1 mark
3.4 A class canteen has 3 mains and 2 desserts. How many lunch combinations are possible? 1 mark
Standard — list and find probability
3.5 A die is rolled and a coin is flipped. (i) Use the multiplication principle to find the total. (ii) Find P(rolling a 5 AND getting heads). 2 marks
3.6 Letters {A, B} are chosen, then digits {1, 2, 3}. (i) Draw a tree diagram (sketch is fine). (ii) List all outcomes. (iii) Find P(A3). 3 marks
Extension — push your thinking
3.7 A pizza has 4 base choices, 3 sauce choices, and 6 topping choices (one of each). How many different pizzas are possible? Show your multiplication. 2 marks
3.8 Two dice are rolled (one red, one blue). (i) How many outcomes are in the sample space? (ii) Set up a 6 × 6 grid that lists each outcome as an ordered pair (red, blue). (iii) Find P(both showing the same number). 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — Outfit combinations (We do)
Step 1: Shirts = 3 (red, blue, white). Step 2: Pants = 4 (black, grey, navy, beige). Step 3: Total = 3 × 4 = 12. Step 4: Red shirt gives 4 outfits; three shirts × 4 = 12. ✓
3.1 — Two coins
2 × 2 = 4 outcomes: {HH, HT, TH, TT}.
3.2 — Three coins
2 × 2 × 2 = 8 outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
3.3 — Spinner (4 colours) then coin
4 × 2 = 8 outcomes.
3.4 — Canteen lunch
3 × 2 = 6 combinations.
3.5 — Die and coin
(i) 6 × 2 = 12 outcomes. (ii) Only {5H} is favourable, so P(5 AND heads) = 1/12 ≈ 0.083.
3.6 — {A, B} × {1, 2, 3}
(i) Tree: from Start, A and B; from A: A1, A2, A3; from B: B1, B2, B3. (ii) Sample space {A1, A2, A3, B1, B2, B3} = 6 outcomes. (iii) P(A3) = 1/6.
3.7 — Pizza combinations
4 × 3 × 6 = 72 different pizzas.
3.8 — Two dice
(i) 6 × 6 = 36 outcomes.
(ii) Grid: red dice down rows (1–6), blue dice across columns (1–6); each cell is (red, blue) — for example, row 3 column 4 = (3, 4).
(iii) "Same number" outcomes are the diagonal: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 favourable. P(same) = 6/36 = 1/6.