Mathematics • Year 10 • Unit 4 • Lesson 15
Probability — Mixed Challenge
Combine every Lesson 15 idea: sample space, favourable outcomes, complement rule, simple tree diagrams, and the two big misconceptions ("probability can exceed 1" and "HH, HT, TT are the only outcomes for two flips").
1. Mixed problems — choose the right tool
Each question uses a different idea from Lesson 15. Decide whether the question is about sample space, complement, mutually exclusive events or a misconception before you start writing. 3 marks each
1.1 A standard 52-card deck. Find: (a) P(king), (b) P(red card), (c) P(king or queen).
1.2 A bag has 8 marbles: 3 red, 2 blue, 2 green, 1 yellow. (a) Find P(blue or yellow). (b) Use the complement rule to find P(not red).
1.3 A spinner is split into three UNEQUAL sections: half red, one-quarter blue, one-quarter green. (a) State each probability as a fraction. (b) Check that the three probabilities sum to 1.
1.4 Two fair dice are rolled and the totals are added. (a) Write the size of the sample space (consider both dice as distinguishable). (b) Find P(total = 7). (c) Find P(total ≠ 7) using the complement rule.
1.5 Decide whether each statement is True or False, with a one-sentence reason: (a) "Any probability is between 0 and 1 inclusive." (b) "If P(A) = 0.4 and P(B) = 0.5 and A and B are mutually exclusive, then P(A or B) = 0.9." (c) "Probability can be expressed as a fraction, decimal or percentage."
1.6 A coin is flipped 3 times. (a) Write the full sample space. (b) Find P(all three the same). (c) Find P(at least one head) using the complement rule.
2. Find the mistake
Another Year 10 student has answered three probability questions. Their reasoning is below. Exactly one line contains a Lesson 15 misconception. Spot it, explain why it's wrong, and re-do that part correctly. 3 marks
Student's working — bag has 4 red, 4 blue and 2 green marbles:
Line 1: Total = 10. P(red) = 4/10 = 2/5.
Line 2: P(red or blue) = 4/10 + 4/10 = 8/10 = 4/5.
Line 3: P(not green) = 1 + P(green) = 1 + 2/10 = 1.2.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that claim is wrong, quoting the complement rule from Lesson 15.
(c) Re-do the calculation correctly.
Stuck? Complement rule: P(A′) = 1 − P(A), NOT 1 + P(A). And any probability must be ≤ 1.3. Open-ended challenge — design a probability game
This question has many valid answers. Be creative but follow every rule. 4 marks
3.1 Design a simple chance game for a school fair. Your write-up must include all of the following:
- a clear description of the equipment (cards, dice, spinner, marbles, etc.) and how to play,
- the full sample space of one play,
- the probability of winning a prize, as a fraction AND a decimal,
- the probability of not winning, found using the complement rule,
- a "fairness check": does the probability of winning lie between 0 and 1? Is the game roughly fair for the player, or stacked in the school's favour?
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 52-card deck
(a) P(king) = 4/52 = 1/13.
(b) P(red) = 26/52 = 1/2.
(c) P(king or queen) = 4/52 + 4/52 = 8/52 = 2/13. (Kings and queens are mutually exclusive.)
1.2 — Marbles
Total = 8.
(a) P(blue or yellow) = (2 + 1)/8 = 3/8.
(b) P(red) = 3/8 → P(not red) = 1 − 3/8 = 5/8.
1.3 — Unequal spinner
(a) P(red) = 1/2, P(blue) = 1/4, P(green) = 1/4.
(b) Sum = 1/2 + 1/4 + 1/4 = 4/4 = 1 ✓ (Lesson 15: probabilities of all outcomes in a sample space sum to 1.)
1.4 — Two dice
(a) Sample space size = 6 × 6 = 36.
(b) Total = 7 outcomes: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 favourable. P = 6/36 = 1/6.
(c) P(total ≠ 7) = 1 − 1/6 = 5/6.
1.5 — True/False
(a) TRUE — probability is between 0 and 1 inclusive (Lesson 15 definition).
(b) TRUE — for mutually exclusive A and B, P(A or B) = P(A) + P(B) = 0.4 + 0.5 = 0.9.
(c) TRUE — probability can be a fraction (1/2), decimal (0.5) or percentage (50 %); all three are equivalent representations of the same value.
1.6 — Three coin flips
(a) S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, size 8.
(b) "All same" = {HHH, TTT}, count 2. P = 2/8 = 1/4.
(c) P(no heads) = P(TTT) = 1/8. P(at least one head) = 1 − 1/8 = 7/8.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The complement rule is P(A′) = 1 − P(A), NOT 1 + P(A). The student added instead of subtracting, getting 1.2 — which Lesson 15 says is impossible because probability is always between 0 and 1.
(c) Correct: P(green) = 2/10 = 1/5. P(not green) = 1 − 2/10 = 8/10 = 4/5 = 0.8. Sanity check: 0 ≤ 0.8 ≤ 1 ✓.
3 — Open-ended challenge (sample solution)
Game: Heart of the Deck. Pay $1 to draw a card from a shuffled standard 52-card deck; if it is a heart, win a $3 prize.
Sample space: 52 cards.
P(win) = 13/52 = 1/4 = 0.25. P(not win) = 1 − 1/4 = 3/4 = 0.75 (using the complement rule).
Fairness check: 0 ≤ 0.25 ≤ 1 ✓. The game is stacked in the school's favour: on average, every 4 plays the school takes $4 in entry fees and pays out $3 once, netting $1 per 4 plays. To make it fair, the prize would need to be $4 (so expected return matches the $1 entry).
Marking: 1 mark for clear game description + sample space, 1 for correct P(win) as fraction and decimal, 1 for correct use of the complement rule, 1 for a thoughtful fairness check.