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).
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 from 15 to 48 inclusive.
- 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. Check that at least two distinct bags satisfy the total range.
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, and P(red) = 1/3. Whole-number counts require the total to be a multiple of 3 and 4. In the revised range 15–48, totals 24 and 36 are both possible.
Bag A (total 24): red = 24 × 1/3 = 8; blue = 24 × 1/4 = 6; green = 24 − 8 − 6 = 10.
Bag B (total 36): red = 36 × 1/3 = 12; blue = 36 × 1/4 = 9; green = 36 − 12 − 9 = 15.
For Bag A, P(red) = 8/24 = 1/3 and P(not blue) = (8+10)/24 = 18/24 = 3/4. For Bag B, P(red) = 12/36 = 1/3 and P(not blue) = (12+15)/36 = 27/36 = 3/4. Each colour count in Bag B is 1.5 times its count in Bag A, so the ratios and probabilities are unchanged.
Marking: 1 mark for a valid total; 1 mark for correct whole-number counts of each colour; 1 mark for verifying both required probabilities; 1 mark for a second distinct valid bag and an explanation of the unchanged ratios.