Mathematics • Year 8 • Unit 4 • Lesson 20
Probability — Mixed Challenge (Unit 4 Final)
Bring together every Unit 4 tool: sample spaces, Venn diagrams, tree diagrams, experimental probability, the complement, and the counting principle. Six mixed problems, one "find the mistake", and one open-ended design challenge that finishes Year 8 maths.
1. Mixed problems — choose the right tool
Each question uses a different idea from across the unit. Show working. 3 marks each
1.1 A bag has 7 red, 5 blue, and 3 green marbles. One marble is drawn. (a) Find P(red), P(blue), P(green). (b) Verify they sum to 1. (c) Find P(not green) using the complement.
1.2 n(ξ) = 80. P(A) = 0.45, P(B) = 0.35, P(A ∩ B) = 0.20. (a) Use the addition rule to find P(A ∪ B). (b) Find P(neither A nor B).
1.3 Two dice are rolled. Find P(product of the two dice is even). Hint: easier via complement — find P(both odd) first.
1.4 A bag has 4 R, 2 B (6 total). Two are drawn without replacement. Use a tree to find P(both R) and P(one of each colour).
1.5 A spinner of 8 equal sectors is spun 400 times. Sector A appears 65 times. (a) Find experimental P(A) and theoretical P(A). (b) Comment whether the spinner appears fair.
1.6 A school of 120 students surveyed: 70 own a smartphone (P), 45 own a tablet (T), 25 own both. (a) Fill in the Venn diagram. (b) Find P(owns at least one device). (c) Find P(owns neither).
2. Find the mistake
A student attempted this multi-tool problem. 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
Problem: Two coins are flipped. Find P(at least one head).
Line 1: Sample space: {HH, HT, TH, TT} → |S| = 4.
Line 2: Favourable for "at least one H": HH, HT, TH → 3 outcomes.
Line 3: P(at least one H) = 1 − P(at least one T) = 1 − 3/4 = 1/4.
Line 4: Final answer: P(at least one H) = 1/4.
(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 and state the correct probability.
Stuck? The complement of "at least one H" is NOT "at least one T". The complement is "no H at all" = "TT". The student used the wrong complement.3. Open-ended challenge — design a probability investigation
This question has many valid answers. Year 8 mathematics ends here — make your final task one to be proud of. 4 marks
3.1 Your job: design a complete probability investigation that uses at least TWO Unit 4 tools (e.g., a Venn AND experimental probability, or a sample space AND a tree, or a tree AND a complement rule).
Your design must include:
(i) State the real-world question or scenario.
(ii) Identify the two Unit 4 tools you will use and explain why each is the right tool for that part.
(iii) Show the full working for both tools — diagrams, calculations, and the final probability.
(iv) Write one sentence explaining what your final probability means for the real-world scenario.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Marbles, formula and complement
Total = 15. P(red) = 7/15; P(blue) = 5/15 = 1/3; P(green) = 3/15 = 1/5.
Sum = 7/15 + 5/15 + 3/15 = 15/15 = 1 ✓.
P(not green) = 1 − 3/15 = 12/15 = 4/5.
1.2 — Addition rule with probabilities
(a) P(A ∪ B) = 0.45 + 0.35 − 0.20 = 0.60.
(b) P(neither) = 1 − 0.60 = 0.40.
1.3 — Two dice, product even (complement)
P(odd on a die) = 3/6 = 1/2. P(both odd) = 1/2 × 1/2 = 1/4 (these are independent so multiply; equivalent to 9 ordered pairs of (odd, odd) out of 36 = 9/36 = 1/4).
P(product even) = 1 − P(both odd) = 1 − 1/4 = 3/4.
1.4 — Tree without replacement
Stage 1: P(R) = 4/6 = 2/3; P(B) = 2/6 = 1/3.
Stage 2 if R first: P(R | R) = 3/5, P(B | R) = 2/5.
Stage 2 if B first: P(R | B) = 4/5, P(B | B) = 1/5.
P(both R) = 2/3 × 3/5 = 6/15 = 2/5.
P(one of each) = P(RB) + P(BR) = (2/3 × 2/5) + (1/3 × 4/5) = 4/15 + 4/15 = 8/15.
1.5 — Spinner fairness
Experimental P(A) ≈ 65/400 = 0.1625. Theoretical P(A) = 1/8 = 0.125. Difference is about 0.04 above expected. The expected count would be 50; observed is 65 (15 above). With 400 trials a deviation this large suggests possible bias toward sector A — but not conclusively so without more data. Reasonable to say "slight evidence the spinner may favour A, but not conclusive without further trials".
1.6 — Smartphone and tablet Venn
(a) P only = 70 − 25 = 45; Both = 25; T only = 45 − 25 = 20; Neither = 120 − 45 − 25 − 20 = 30. Check: 45+25+20+30 = 120 ✓.
(b) P(at least one) = (45+25+20)/120 = 90/120 = 3/4.
(c) P(neither) = 30/120 = 1/4.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The complement of "at least one H" is "NO heads at all" — that is, TT — NOT "at least one T". Almost every outcome has at least one T (HT, TH, TT), so "at least one T" is itself 3/4, not the complement. The student confused the two events.
(c) Corrected working:
Sample space: {HH, HT, TH, TT}.
Favourable for "at least one H": HH, HT, TH → 3 outcomes.
P(at least one H) = 3/4. OR using complement: P(at least one H) = 1 − P(no H) = 1 − P(TT) = 1 − 1/4 = 3/4.
3 — Open-ended investigation (sample solution)
(i) Scenario: A school has 200 students. Past data shows 30% catch the school bus (B) and 60% walk (W) — with 10% doing both (live close to a bus stop but sometimes walk). I want to find: (a) how many students do neither, and (b) the probability that on any random day a bus-catching student arrives on time, given that records of the last 50 trips show 42 on-time arrivals.
(ii) Tools:
TOOL 1 — Venn diagram: best for overlapping groups (some students both bus AND walk).
TOOL 2 — Experimental probability: best for "on-time" because bus reliability is not theoretical — it must be estimated from historical data.
(iii) Working:
VENN: n(B) = 60, n(W) = 120, n(B ∩ W) = 20. B only = 40; W only = 100; Both = 20; Neither = 200 − 40 − 20 − 100 = 40 students.
EXPERIMENTAL: P(bus on time) ≈ 42/50 = 21/25 = 0.84.
(iv) Real-world meaning: About 40 students get to school by another method (e.g., car, scooter); of those who catch the bus, the bus is on time about 84% of the time, so we can expect roughly 50 of the 60 bus catchers to arrive on time each day.
Marking: 1 mark for clear scenario; 1 mark for justified choice of two tools; 1 mark for full working using both tools correctly; 1 mark for real-world interpretation sentence.