Mathematics • Year 8 • Unit 4 • Lesson 14
Probability — Mixed Challenge
Pull together the probability formula, sample-space listing, complementary events, and expected outcomes. Six mixed problems, one "find the mistake", and one open-ended challenge.
1. Mixed problems — choose the right move
Each question uses a different idea from Lesson 14. Show working. 3 marks each
1.1 A spinner has 8 equal sectors numbered 1-8. Find: (a) P(odd number), (b) P(prime number), (c) P(number divisible by 3).
1.2 A standard deck of 52 cards. Find: (a) P(black card), (b) P(face card — Jack, Queen, or King), (c) P(NOT a face card) using the complement rule.
1.3 Two coins are flipped together. (a) List the complete sample space. (b) Find P(exactly one head). (c) Find P(at least one head) using the complement rule.
1.4 A bag contains some red and blue marbles. The probability of drawing a red marble is 3/7. (a) If there are 14 marbles in total, how many are red? (b) If 21 draws are made (with replacement), how many should be expected to be red?
1.5 A class of 30 students elects one captain by drawing names from a hat. (a) Find P(a specific student is chosen). (b) The probabilities of selecting any one student are equally likely. What is the sum of all 30 individual probabilities? Explain why.
1.6 A standard die is rolled twice. Find: (a) P(both rolls are even), (b) P(sum equals 7). (Hint: list the (1st, 2nd) pairs that give a sum of 7 from the 36-outcome sample space.)
2. Find the mistake
Another student tried this problem. Exactly one line contains a mistake. Spot it, explain why, and correct it. 3 marks
Problem: A bag contains 6 red, 4 blue, and 2 green marbles. Find P(red) and P(NOT red).
Line 1: Total marbles = 6 + 4 + 2 = 12.
Line 2: Favourable for red = 6.
Line 3: P(red) = 6 ÷ 12 = 1/2.
Line 4: P(NOT red) = 1 + 1/2 = 1.5.
Line 5: Answer: P(red) = 1/2 and P(NOT red) = 1.5.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong, and what should have been done.
(c) Write the corrected working and the correct answer.
Stuck? Probability must always be between 0 and 1. P = 1.5 is impossible — it signals a calculation error. The complement rule uses SUBTRACTION: P(not A) = 1 − P(A).3. Open-ended challenge — design your own game
This question has many valid answers. 4 marks
3.1 Your job: design a simple game of chance using a spinner, a bag of marbles, or a die. The game should have ONE winning outcome where the probability of winning is exactly 1/4.
(i) Describe the game equipment (e.g. spinner with sectors, bag with marbles, etc.).
(ii) Define what counts as "winning" and what counts as "losing".
(iii) Calculate P(win) and confirm it equals 1/4. Show the favourable outcomes and the total outcomes.
(iv) Use the complement rule to calculate P(lose).
(v) If your game is played 80 times, how many wins are expected? Show the working.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Spinner 1-8
(a) Odd = {1, 3, 5, 7}, 4 favourable. P(odd) = 4/8 = 1/2.
(b) Prime = {2, 3, 5, 7}, 4 favourable. P(prime) = 4/8 = 1/2.
(c) Divisible by 3 = {3, 6}, 2 favourable. P(÷3) = 2/8 = 1/4.
1.2 — Standard deck
(a) Black cards (clubs + spades) = 26. P(black) = 26/52 = 1/2.
(b) Face cards = J, Q, K in each of 4 suits = 12. P(face) = 12/52 = 3/13.
(c) P(NOT face) = 1 − 3/13 = 10/13.
1.3 — Two coins
(a) Sample space = {HH, HT, TH, TT}, 4 outcomes.
(b) Exactly one head = {HT, TH}, 2 outcomes. P = 2/4 = 1/2.
(c) P(at least one head) = 1 − P(no heads) = 1 − P(TT) = 1 − 1/4 = 3/4.
1.4 — Red marbles
(a) P(red) = 3/7. Red count = 14 × 3/7 = 6 red marbles (and 8 blue).
(b) Expected red in 21 draws = 21 × 3/7 = 9 reds.
1.5 — Class election
(a) P(specific student chosen) = 1/30 ≈ 3.33%.
(b) Sum of all 30 probabilities = 30 × 1/30 = 1. The sum is 1 because exactly one student WILL be chosen — the probabilities cover all possible outcomes.
1.6 — Two dice
(a) P(both even) = (3/6) × (3/6) = 9/36 = 1/4.
(b) Pairs summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 favourable. P(sum = 7) = 6/36 = 1/6.
2 — Find the mistake
(a) The mistake is on Line 4.
(b) The complement rule uses SUBTRACTION: P(not A) = 1 − P(A), NOT 1 + P(A). The student added instead of subtracting, producing P = 1.5 which is impossible (probabilities must be between 0 and 1 inclusive).
(c) Corrected: P(NOT red) = 1 − 1/2 = 1/2. (Alternatively: count NOT red = 4 + 2 = 6 favourable, P = 6/12 = 1/2 — same answer.)
3 — Open-ended challenge (sample solution)
(i) Equipment: A spinner divided into 12 equal sectors. 3 sectors are coloured red ("WIN"); 9 sectors are coloured white ("LOSE").
(ii) Game: Player spins once. WIN if the arrow lands on a red sector, LOSE if it lands on white.
(iii) P(win): Favourable = 3 (red sectors). Total = 12. P(win) = 3 ÷ 12 = 1/4 ✓.
(iv) P(lose): Using complement rule, P(lose) = 1 − 1/4 = 3/4.
(v) Expected wins in 80 plays: 80 × 1/4 = 20 wins. (And 60 losses.)
Marking: 1 mark for a sensible equipment description. 1 mark for clear win/lose rules. 1 mark for correctly computing P(win) = 1/4 with favourable and total shown. 1 mark for using the complement for P(lose) AND showing the expected-wins calculation.