Mathematics • Year 10 • Unit 4 • Lesson 16
Compound Events — Mixed Challenge
Pull together Lesson 16: multiply along a branch, add for mutually exclusive events, subtract the overlap when events can both happen, and tell "with replacement" (independent) from "without replacement" (dependent). Then spot a Year 10 mistake and design your own three-stage tree-diagram puzzle.
1. Mixed problems — pick the right rule
Each question uses a different idea from Lesson 16. Show working. 3 marks each
1.1 A fair die is rolled and a fair coin is flipped. Find P(an even number AND a tail).
1.2 A bag has 5 white and 3 black counters. Two counters are drawn without replacement. Find P(both black) and explain why your calculation uses 3/8 × 2/7 (not 3/8 × 3/8).
1.3 Events A and B have P(A) = 0.5, P(B) = 0.3, P(A and B) = 0.1. Find P(A or B). Is the answer the same as if you assumed A and B were mutually exclusive?
1.4 A fair coin is tossed four times. Use the tree-diagram pattern from Lesson 16 to find P(exactly three heads). (You may list the favourable outcomes instead of drawing the whole tree.)
1.5 A spinner has 8 equal sectors: 3 red, 2 blue, 3 green. The spinner is spun twice. Are the spins independent? Find P(red on first AND green on second).
1.6 A box contains 2 red, 3 blue and 5 green pens. Two pens are drawn one after another without replacement. Find P(at least one red), giving a one-sentence explanation of why the "1 − P(none)" shortcut is faster than listing every favourable branch.
2. Find the mistake
A Year 10 student has tried this problem: "A bag has 3 red and 2 blue marbles. Two marbles are drawn without replacement. Find P(both red)." Their working is below. Exactly one line contains a mistake. Identify it, explain why it is wrong, and re-do the working correctly. 3 marks
Student's working:
Line 1: "Without replacement" means the second draw depends on the first.
Line 2: P(first red) = 3/5.
Line 3: After taking a red, the bag has 2 red and 2 blue, total 5 marbles.
Line 4: So P(second red | first red) = 2/5.
Line 5: P(both red) = 3/5 × 2/5 = 6/25.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working and give the correct P(both red).
Stuck? After one marble is removed, how many marbles are left? Don't forget to reduce the total.3. Open-ended challenge — design a three-stage scenario
Many valid answers. Follow every rule. 4 marks
3.1 Design a real-world scenario that involves exactly three stages, where:
- the three stages are independent (with-replacement, or genuinely unrelated like a coin / die / spinner combination),
- each stage has at least two possible outcomes,
- the question you ask requires both the multiplication rule AND the addition rule (because two different paths down the tree count as "successful").
Write up:
(i) the scenario in one sentence,
(ii) a labelled tree diagram (rough sketch is fine),
(iii) the probability you are computing in words,
(iv) the full calculation and final answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Die and coin
P(even) = 3/6 = 1/2. P(tail) = 1/2. Independent.
P(even and tail) = 1/2 × 1/2 = 1/4.
1.2 — Two black counters, no replacement
P(first black) = 3/8. After one black is removed, 2 black remain out of 7. So P(second black | first black) = 2/7.
P(both black) = 3/8 × 2/7 = 6/56 = 3/28.
Using 3/8 × 3/8 would treat the second draw as independent, but the lesson is clear: without replacement makes the draws dependent — the bag has changed.
1.3 — Overlap matters
P(A or B) = P(A) + P(B) − P(A and B) = 0.5 + 0.3 − 0.1 = 0.7.
If you assumed mutually exclusive, you would get 0.5 + 0.3 = 0.8 — that's wrong: it double-counts the 0.1 overlap.
1.4 — Three heads in four tosses
Favourable outcomes (three H, one T): HHHT, HHTH, HTHH, THHH → 4 outcomes.
Each has probability (1/2)⁴ = 1/16.
P(exactly three heads) = 4 × 1/16 = 4/16 = 1/4.
1.5 — Two spins of an 8-sector spinner
The spinner is reset each spin, so the spins are independent.
P(red) = 3/8. P(green) = 3/8.
P(red then green) = 3/8 × 3/8 = 9/64.
1.6 — At least one red (without replacement)
P(no reds) = P(first not red) × P(second not red | first not red) = 8/10 × 7/9 = 56/90 = 28/45.
P(at least one red) = 1 − 28/45 = 17/45.
Shortcut: "at least one" has many favourable branches (RR, RB, RG, BR, GR), but "none" has only one type. Computing 1 − P(none) is one calculation instead of five.
2 — Find the mistake (without-replacement marbles)
(a) The mistake is on Line 3.
(b) After one red is removed, the bag has 2 red and 2 blue, for a total of 4 marbles (not 5). The student forgot to reduce the total.
(c) Corrected: P(first red) = 3/5. P(second red | first red) = 2/4 = 1/2. P(both red) = 3/5 × 1/2 = 3/10.
3 — Open-ended challenge (sample solution)
Scenario: Toss a fair coin three times. Find P(exactly two heads).
Tree (sketch): three levels, each branching H/T, eight outcomes at the leaves, each probability (1/2)³ = 1/8.
Probability in words: The probability of two heads and one tail in any order.
Calculation: Favourable paths HHT, HTH, THH. Each path uses the multiplication rule along the tree to give 1/8. The addition rule combines the three mutually exclusive paths: P = 1/8 + 1/8 + 1/8 = 3/8.
Marking: 1 mark — three-stage scenario with at least two outcomes per stage; 1 — labelled tree; 1 — correct use of multiplication along a path; 1 — correct addition across mutually exclusive paths with a final answer.