Mathematics • Year 7 • Unit 4 • Lesson 17
Theoretical Probability — Mixed Challenge
Bring together P = f/n, equally-likely checks, sample space listing, and the complement rule. Spot a classic mistake and design your own probability question.
1. Mixed problems
Show working. 2 marks each
1.1 Cards numbered 1–20 are placed in a hat. One card is drawn. Find P(even card) and P(card > 15).
1.2 A standard 52-card deck. Find P(diamond OR spade) and P(red picture card).
1.3 A spinner with 12 equal sectors numbered 1–12. Find P(multiple of 3) and P(prime).
1.4 Use the complement rule to find P(not a 6) on a single die roll AND P(not heart) from a 52-card deck.
1.5 A bag holds 2 red, 5 blue, 3 green marbles. Find P(red OR green) two ways: (i) direct count, (ii) complement of blue.
1.6 Three fair coins are flipped. (i) How many outcomes are in the sample space? (ii) Use the complement rule to find P(at least one head).
2. Find the mistake
A Year 7 student tries to find P(even number) on a roll of a standard die. Exactly one line contains a serious error. Spot it, explain why it's wrong, then write the correct working. 3 marks
Student's working:
Line 1: Even numbers on a die: {2, 4, 6} → 3 outcomes.
Line 2: Sample space = {2, 4, 6} → 3 outcomes total.
Line 3: P(even) = 3 ÷ 3 = 1.
Line 4: So the die is certain to land on an even number.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write the correct sample space, the correct calculation, and the correct answer.
Stuck? Revisit lesson § "Common Pitfalls" — the denominator MUST be all outcomes, not just the favourables.3. Open-ended challenge — invent your own probability question
This question has many correct answers. Show your work clearly. 4 marks
3.1 Design a one-page probability puzzle for a Year 6 classmate. You must give:
- a scenario describing the equally-likely random device (a bag, spinner, deck, etc.) in 1–2 sentences,
- three questions: one direct P = f/n, one "find P(not …)" using the complement, and one "at least one" using the complement,
- worked solutions for all three with the formula line and the final answer as a fraction in lowest terms.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cards 1–20
Evens {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} → 10. P(even) = 10/20 = 1/2. Cards > 15 = {16, 17, 18, 19, 20} → 5. P(> 15) = 5/20 = 1/4.
1.2 — 52-card deck
P(diamond or spade) = 26/52 = 1/2. Red picture cards = 6 (J, Q, K of hearts + J, Q, K of diamonds). P(red picture) = 6/52 = 3/26 ≈ 0.115.
1.3 — Spinner 1–12
Multiples of 3 = {3, 6, 9, 12} → 4. P(multiple of 3) = 4/12 = 1/3. Primes = {2, 3, 5, 7, 11} → 5. P(prime) = 5/12 ≈ 0.417.
1.4 — Complement rule
P(not 6) = 1 − 1/6 = 5/6. P(not heart) = 1 − 13/52 = 1 − 1/4 = 3/4.
1.5 — Bag of marbles
Total = 2 + 5 + 3 = 10. (i) Direct: P(red or green) = (2 + 3)/10 = 5/10 = 1/2. (ii) Complement: 1 − P(blue) = 1 − 5/10 = 1/2. ✓
1.6 — Three coin flips
(i) Sample space = 2 × 2 × 2 = 8 outcomes {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
(ii) Complement of "at least one head" = "all tails" = {TTT} → 1 outcome. P(at least one head) = 1 − 1/8 = 7/8 = 0.875.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The sample space is all possible outcomes of rolling a die — that is {1, 2, 3, 4, 5, 6}, not just the even ones. The student has confused "favourable outcomes" with "sample space".
(c) Sample space {1, 2, 3, 4, 5, 6}, n = 6. Favourables {2, 4, 6}, f = 3. P(even) = 3 ÷ 6 = 1/2 = 0.5.
3 — Invent your own question (sample answer)
Scenario: A bag holds 3 red, 4 blue and 5 yellow lollies. One lolly is drawn at random.
Q1 (direct): Find P(yellow). Solution: total = 12, favourables = 5. P(yellow) = 5/12.
Q2 (complement): Find P(not blue). Solution: P(blue) = 4/12 = 1/3. P(not blue) = 1 − 1/3 = 2/3.
Q3 (at least one): Two lollies are drawn one at a time with replacement. Find P(at least one yellow). Solution: P(no yellow on a single draw) = 7/12. With replacement, P(no yellow twice) = (7/12) × (7/12) = 49/144. P(at least one yellow) = 1 − 49/144 = 95/144 ≈ 0.660.
Marking: 1 mark for scenario, 1 for each well-formed question with worked solution.