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.