Mathematics • Year 9 • Unit 4 • Lesson 13
Probability — Mixed Challenge
Combine the probability formula, sample spaces and the complement rule from Lesson 13. Choose the right approach for each problem, fix a classmate's error, and design your own probability puzzle.
1. Mixed problems — choose the right tool
Decide which rule or sample space to use before you start writing. Show working. 3 marks each
1.1 A bag has 5 red, 3 blue and 2 green marbles. Find P(red), P(blue) and P(not green).
1.2 A card is drawn from a standard 52-card deck. Find P(ace) and P(not a face card). Face cards = J, Q, K (12 in total).
1.3 Two fair dice are rolled and the numbers are added. Find P(sum = 7).
1.4 Two fair dice are rolled. Find P(sum > 9) by listing favourable outcomes.
1.5 A biased spinner has P(red) = 0.4, P(blue) = 0.35 and one other colour, green. Find P(green) and P(not red).
1.6 A spinner shows numbers 1-12 in equal sectors. Find P(prime) and P(divisible by 3). Verify that for these two events P(A or B) ≠ P(A) + P(B) by checking the overlap.
2. Find the mistake
Another student has tried to find P(rolling a 5 or a 6 on a fair die) and P(not rolling a 5 or 6). Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — fair 6-sided die:
Line 1: Sample space = {1, 2, 3, 4, 5, 6}, total = 6.
Line 2: P(5) = 1/6 and P(6) = 1/6.
Line 3: P(5 or 6) = P(5) × P(6) = 1/6 × 1/6 = 1/36.
Line 4: P(not 5 or 6) = 1 − 1/36 = 35/36.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answers for both P(5 or 6) and P(not 5 or 6).
Stuck? Revisit lesson § "Misconceptions" — P(A) × P(B) is for AND of independent events, not OR. For "5 or 6" on a single roll, the favourable outcomes are 5 and 6, so add or count directly.3. Open-ended challenge — design a bag
This question has more than one valid answer — there are many bags that work. 4 marks
3.1 Design a bag containing only red, blue and green marbles (whole numbers of each, at least one of each colour) with all three of these properties at the same time:
- The total number of marbles is between 15 and 30.
- P(red) = 1/3 exactly.
- P(not blue) = 3/4 exactly.
For your bag:
(i) Write the number of each colour.
(ii) Compute P(red), P(blue) and P(green) and confirm both required properties.
(iii) State one different bag (still meeting the conditions) and explain in one sentence why the ratios are the same even though the totals are different.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Bag of 5R, 3B, 2G
Total = 10. P(red) = 5/10 = 1/2. P(blue) = 3/10. P(not green) = 1 − 2/10 = 4/5.
1.2 — Card
P(ace) = 4/52 = 1/13. Face cards = 12. P(not face) = (52 − 12) ÷ 52 = 40/52 = 10/13.
1.3 — Two dice, sum = 7
Total outcomes = 36. Favourable pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes. P(sum = 7) = 6/36 = 1/6.
1.4 — Two dice, sum > 9
Sums > 9 mean 10, 11 or 12. Pairs: (4,6), (5,5), (6,4) for 10 = 3; (5,6), (6,5) for 11 = 2; (6,6) for 12 = 1. Total favourable = 6. P(sum > 9) = 6/36 = 1/6.
1.5 — Biased spinner
P(red) + P(blue) + P(green) = 1, so P(green) = 1 − 0.4 − 0.35 = 0.25. P(not red) = 1 − 0.4 = 0.6.
1.6 — 1-12 spinner
Primes 1-12: {2, 3, 5, 7, 11}: P(prime) = 5/12. Multiples of 3 in 1-12: {3, 6, 9, 12}: P(div by 3) = 4/12 = 1/3.
Overlap = {3} (prime AND multiple of 3). So P(A) + P(B) = 5/12 + 4/12 = 9/12, but the actual P(A or B) = 5/12 + 4/12 − 1/12 = 8/12 = 2/3 — the sum overcounts by the one overlap.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student multiplied probabilities, which is the rule for "A AND B" with independent events. The question asked for "5 OR 6" on a single roll, which uses the addition rule (or direct counting). On one die you can't get both 5 and 6 at the same time, so they are mutually exclusive — just add them.
(c) Corrected working: favourable outcomes for "5 or 6" = {5, 6} → 2 outcomes. P(5 or 6) = 2/6 = 1/3. By the complement rule, P(not 5 or 6) = 1 − 1/3 = 2/3.
3 — Open-ended challenge (sample solution)
P(blue) = 1 − 3/4 = 1/4. P(red) = 1/3. So total = LCM-style multiple of 3 and 4, between 15 and 30 → total = 24.
Bag A: red = 24 × 1/3 = 8; blue = 24 × 1/4 = 6; green = 24 − 8 − 6 = 10.
Check: P(red) = 8/24 = 1/3 ✓. P(blue) = 6/24 = 1/4. P(not blue) = 1 − 1/4 = 3/4 ✓. P(green) = 10/24 = 5/12. All three sum to 1 ✓.
Bag B (different total, same ratios — but must still fit 15-30): only total = 24 lies in 15-30 as a common multiple of 3 and 4. If we relax the bound, total = 12 works too (red 4, blue 3, green 5), or total = 36 (red 12, blue 9, green 15). The ratios are the same because probability depends on ratios, not absolute counts — doubling every count leaves every fraction unchanged.
Marking: 1 mark for total 24 (or other valid total inside the range); 1 mark for correct count of each colour; 1 mark for verifying both required probabilities; 1 mark for the second-bag explanation referencing ratios.