Mathematics • Year 7 • Unit 4 • Lesson 18

Sample Space and Tree Diagrams — Mixed Challenge

Bring together tree diagrams, the grid method and the multiplication principle. Spot a common counting mistake and design your own multi-stage experiment.

Master · Mixed Challenge

1. Mixed problems

Show your method. 2 marks each

1.1 A coin is flipped 4 times in a row. How many outcomes are in the sample space?

1.2 Two dice (red and blue) are rolled. (i) How many outcomes in total? (ii) Find P(sum = 7) by counting favourables in a 6 × 6 grid.

1.3 Three coins are flipped at once. (i) Sketch a tree diagram. (ii) Find P(exactly 2 heads).

1.4 A 4-digit PIN where digits 0–9 are allowed AND repeats are allowed: how many PINs? Then compare to a 4-digit PIN where digits CANNOT repeat: how many PINs? Briefly explain the difference.

1.5 A breakfast at a café is one cereal (3 choices) + one yoghurt (2 choices) + one fruit (4 choices) + one drink (3 choices). (i) How many breakfasts in total? (ii) Find P("oats AND apple") if cereal and fruit are chosen at random (yoghurt and drink can be anything).

1.6 A bus route has 4 stops (A, B, C, D). A driver writes down (start, end) — start and end must be different stops. (i) Use the multiplication principle to find total outcomes. (ii) Find P(start = A).

Stuck on 1.6? Stage 1 = 4 stops; Stage 2 = 3 remaining stops. Total = 4 × 3 = 12.

2. Find the mistake

A Year 7 student tries to count outcomes for "a die rolled AND a coin flipped". Exactly one line contains a serious error. Spot it, explain why it's wrong, then redo the calculation correctly. 3 marks

Student's working:

Line 1:   Stage 1 (die) has 6 outcomes: 1, 2, 3, 4, 5, 6.

Line 2:   Stage 2 (coin) has 2 outcomes: H, T.

Line 3:   Total outcomes = 6 + 2 = 8.

Line 4:   P(rolling a 5 AND getting H) = 1/8.

(a) Which line contains the mistake?

(b) Explain in one or two sentences why that line is wrong.

(c) Write the correct total and the correct probability.

Stuck? Revisit lesson § "Spot the Trap" — for independent stages, MULTIPLY, never add.

3. Open-ended challenge — design your own multi-stage experiment

This question has many correct answers. Show your work clearly. 4 marks

3.1 Design a three-stage experiment for a younger student. You must provide:

  • a scenario describing the three independent stages (e.g. spinner, coin, die),
  • the multiplication-principle calculation of the total outcomes,
  • a partial sample space of at least 6 outcomes written using your own ordered notation (e.g. R-H-3),
  • one probability question about your experiment and its worked solution.
Stuck? Try: spinner (4 colours), coin, die. Total = 4 × 2 × 6 = 48 outcomes. Question: P("Red AND H AND 6").

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Coin flipped 4 times

2 × 2 × 2 × 2 = 16 outcomes.

1.2 — Two dice

(i) 6 × 6 = 36 outcomes. (ii) Sum = 7 pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 favourable. P(sum = 7) = 6/36 = 1/6.

1.3 — Three coins

(i) Tree: 2 × 2 × 2 = 8 paths giving {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
(ii) Exactly 2 heads = {HHT, HTH, THH} = 3 favourable. P(exactly 2 H) = 3/8.

1.4 — 4-digit PIN

Repeats allowed: 10 × 10 × 10 × 10 = 10 000 PINs. No repeats: 10 × 9 × 8 × 7 = 5040 PINs. The no-repeats count is smaller because once a digit is used, the pool shrinks by 1 for each later stage.

1.5 — Café breakfast

(i) Total = 3 × 2 × 4 × 3 = 72 breakfasts.
(ii) Cereal-and-fruit pairs = 3 × 4 = 12. "Oats AND apple" = 1 favourable pair. P = 1/12.

1.6 — Bus route (start, end)

(i) Total = 4 × 3 = 12 outcomes.
(ii) Start = A gives 1 × 3 = 3 favourable. P(start = A) = 3/12 = 1/4.

2 — Find the mistake

(a) The mistake is on Line 3.
(b) For two independent stages we must multiply the options at each stage, not add them. Each of the 6 die outcomes can be paired with each of the 2 coin outcomes, giving 6 × 2 combinations.
(c) Correct total = 6 × 2 = 12 outcomes. P(5 AND H) = 1/12 ≈ 0.083.

3 — Design your own experiment (sample answer)

Scenario: Stage 1 = spinner with 4 colours (R, B, G, Y). Stage 2 = coin (H, T). Stage 3 = die (1–6).
Total outcomes: 4 × 2 × 6 = 48.
Partial sample space: R-H-1, R-H-2, R-T-1, R-T-6, B-H-3, B-T-4 (6 of 48).
Question: Find P("Red AND H AND 6"). Solution: only 1 favourable outcome (R-H-6) out of 48 equally likely. P = 1/48 ≈ 0.021.

Marking: 1 mark each for scenario, multiplication, partial sample space, and worked probability question.