Grade 8 Probability — Answer Key

Victorian Curriculum | Year 8 Mathematics | Tutor Reference Copy

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🎯 Success Criteria

By the end of this unit, students should be able to:

✅ Calculate theoretical probability using P(event) = favourable outcomes ÷ total outcomes

✅ Construct and interpret tree diagrams to list sample spaces for multi-step experiments

✅ Complete and interpret two-way tables and calculate associated probabilities

✅ Draw and interpret Venn diagrams including union (∪), intersection (∩), and complement (A′)

✅ Distinguish between experimental and theoretical probability

✅ Solve multi-step probability problems using correct mathematical notation

📌 Differentiation Note: Core questions = Q1–Q20 | Extension = Q21–Q30. Scaffold lower-level learners with worked examples before independent practice.

📋 Part 1: Basic Probability (Q1–Q10) — Answers

Q1. A bag contains 4 red, 3 blue, and 5 green marbles. What is the probability of picking a red marble?

Answer: Total marbles = 4 + 3 + 5 = 12
P(red) = 4/12 = 1/3 ≈ 0.333

Q2. What is the probability of NOT picking a green marble from the same bag?

Answer: P(green) = 5/12, so P(not green) = 1 − 5/12 = 7/12 ≈ 0.583

Q3. A standard die is rolled. Find: (a) P(even number) (b) P(number greater than 4) (c) P(1 or 6)

(a) Even numbers: {2, 4, 6} → P(even) = 3/6 = 1/2
(b) Numbers > 4: {5, 6} → P(>4) = 2/6 = 1/3
(c) {1, 6} → P(1 or 6) = 2/6 = 1/3

Q4. A card is drawn from a standard 52-card deck. Find: (a) P(heart) (b) P(king) (c) P(red card) (d) P(king of hearts)

(a) P(heart) = 13/52 = 1/4
(b) P(king) = 4/52 = 1/13
(c) P(red) = 26/52 = 1/2
(d) P(king of hearts) = 1/52 ≈ 0.019

Q5. In a class of 30 students, 18 play sport and 12 do not. A student is chosen at random. Find P(plays sport).

Answer: P(plays sport) = 18/30 = 3/5 = 0.6

Q6. A spinner has 8 equal sections numbered 1–8. Find: (a) P(prime number) (b) P(multiple of 3)

(a) Primes in 1–8: {2, 3, 5, 7} → P(prime) = 4/8 = 1/2
(b) Multiples of 3: {3, 6} → P(multiple of 3) = 2/8 = 1/4

Q7. Two events A and B are mutually exclusive. P(A) = 0.35 and P(B) = 0.45. Find P(A or B).

Answer: Since mutually exclusive: P(A ∪ B) = P(A) + P(B) = 0.35 + 0.45 = 0.80

Q8. A letter is chosen randomly from the word PROBABILITY. Find P(vowel).

Answer: Letters: P-R-O-B-A-B-I-L-I-T-Y = 11 letters
Vowels: O, A, I, I → 4 vowels
P(vowel) = 4/11 ≈ 0.364

Q9. If P(event) = 0.72, what is the probability of the complementary event?

Answer: P(complement) = 1 − 0.72 = 0.28

Q10. A coin is flipped 200 times and lands on heads 86 times. (a) What is the experimental probability of heads? (b) How does this compare to theoretical probability?

(a) P(heads) = 86/200 = 0.43
(b) Theoretical P(heads) = 0.5. The experimental probability (0.43) is slightly less than theoretical. With more trials, experimental probability tends to approach the theoretical value. (Law of Large Numbers)

🌳 Part 2: Tree Diagrams & Sample Spaces (Q11–Q18) — Answers

Q11. A coin is flipped and a die is rolled. (a) How many outcomes are in the sample space? (b) List all outcomes. (c) Find P(heads and even number).

(a) 2 × 6 = 12 outcomes
(b) {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
(c) Favourable: {H2, H4, H6} → P = 3/12 = 1/4
Tree diagram: Coin branches (H/T) → each branches into 6 die outcomes

Q12. A bag has 1 red (R) and 1 blue (B) ball. A ball is drawn, replaced, then drawn again. (a) Draw the tree diagram outcomes. (b) Find P(same colour both times). (c) Find P(at least one red).

(a) Outcomes: {RR, RB, BR, BB} — 4 equally likely outcomes
Tree: R → {RR, RB} | B → {BR, BB}
(b) Same colour: {RR, BB} → P = 2/4 = 1/2
(c) At least one red: {RR, RB, BR} → P = 3/4 = 0.75

Q13. A bag has 2 red and 1 blue ball. A ball is drawn WITHOUT replacement, then a second ball is drawn. (a) List the sample space. (b) Find P(both red). (c) Find P(one of each colour).

Label balls: R1, R2, B
(a) Outcomes: {R1R2, R1B, R2R1, R2B, BR1, BR2} — 6 outcomes
(b) Both red: {R1R2, R2R1} → P = 2/6 = 1/3
(c) One of each: {R1B, R2B, BR1, BR2} → P = 4/6 = 2/3

Q14. A family has 3 children. Assuming equal probability of boy (B) or girl (G): (a) How many outcomes in the sample space? (b) Find P(exactly 2 girls). (c) Find P(at least 1 boy).

(a) 2³ = 8 outcomes: {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}
(b) Exactly 2 girls: {BGG, GBG, GGB} → P = 3/8 = 0.375
(c) At least 1 boy: all except {GGG} → P = 7/8 = 0.875

Q15. Two dice are rolled. Find: (a) P(sum = 7) (b) P(sum ≥ 10) (c) P(doubles)

Total outcomes: 6 × 6 = 36
(a) Sum = 7: {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} → P = 6/36 = 1/6
(b) Sum ≥ 10: {(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)} → P = 6/36 = 1/6
(c) Doubles: {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)} → P = 6/36 = 1/6

Q16. A restaurant offers 3 mains (chicken, fish, pasta) and 2 desserts (cake, ice cream). (a) How many meal combinations are possible? (b) Find P(choosing pasta and cake).

(a) 3 × 2 = 6 combinations: {chicken-cake, chicken-ice cream, fish-cake, fish-ice cream, pasta-cake, pasta-ice cream}
(b) P(pasta and cake) = 1/6 ≈ 0.167

Q17. Using a tree diagram, find the probability of getting exactly 2 heads when 3 coins are flipped.

Sample space (8 outcomes): {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Exactly 2 heads: {HHT, HTH, THH} → 3 outcomes
P(exactly 2 heads) = 3/8 = 0.375

Q18. A bag has 3 red and 2 blue balls. Two balls are drawn WITH replacement. Find P(different colours).

P(red) = 3/5, P(blue) = 2/5
P(red then blue) = 3/5 × 2/5 = 6/25
P(blue then red) = 2/5 × 3/5 = 6/25
P(different colours) = 6/25 + 6/25 = 12/25 = 0.48

📊 Part 3: Two-Way Tables & Venn Diagrams (Q19–Q25) — Answers

Q19. Complete the two-way table and answer the questions.
Survey: 80 students — play sport or play music (or both/neither).
Play sport: 45 | Play music: 30 | Both: 12 | Neither: ?

Finding "Neither": Using inclusion-exclusion:
|Sport ∪ Music| = 45 + 30 − 12 = 63
Neither = 80 − 63 = 17 students

Completed table:
                | Sport | No Sport | Total
Music       | 12    |   18     | 30
No Music | 33    |   17     | 50
Total        | 45    |   35     | 80

Q20. Using the table from Q19, find: (a) P(plays sport only) (b) P(plays music | plays sport) (c) P(neither)

(a) Sport only (not music) = 33 → P = 33/80 ≈ 0.4125
(b) P(music | sport) = 12/45 = 4/15 ≈ 0.267 (conditional probability)
(c) P(neither) = 17/80 ≈ 0.2125

Q21. In a class of 25 students: 14 like maths (M), 11 like science (S), 6 like both, and some like neither. (a) Draw and complete the Venn diagram values. (b) Find how many like neither. (c) Find P(likes maths only).

(a) Venn diagram values:
Maths only = 14 − 6 = 8
Science only = 11 − 6 = 5
Both = 6
(b) Neither = 25 − (8 + 6 + 5) = 25 − 19 = 6 students
(c) P(maths only) = 8/25 = 0.32

Q22. Using the Venn diagram from Q21, find: (a) P(M ∪ S) (b) P(M ∩ S) (c) P(M′) — complement of maths

(a) P(M ∪ S) = (8 + 6 + 5)/25 = 19/25 = 0.76
(b) P(M ∩ S) = 6/25 = 0.24
(c) P(M′) = 1 − 14/25 = 11/25 = 0.44

Q23. A survey of 50 people asked about tea (T) and coffee (C) preference: 28 drink tea, 20 drink coffee, 8 drink both. Find P(drinks tea or coffee but not both).

Tea only = 28 − 8 = 20
Coffee only = 20 − 8 = 12
Tea or coffee but NOT both = 20 + 12 = 32
P = 32/50 = 16/25 = 0.64

Q24. Two-way table — 100 people surveyed about exercise habits and diet quality.
Good diet + exercises: 35 | Good diet + no exercise: 15 | Poor diet + exercises: 20 | Poor diet + no exercise: 30
Find: (a) P(good diet) (b) P(exercises | good diet) (c) P(poor diet and no exercise)

Total = 100 ✓
(a) Good diet = 35 + 15 = 50 → P(good diet) = 50/100 = 0.5
(b) P(exercises | good diet) = 35/50 = 7/10 = 0.7 (conditional)
(c) P(poor diet ∩ no exercise) = 30/100 = 0.3

Q25. Are "good diet" and "exercises" independent in Q24? Show working to justify.

Check independence: P(A ∩ B) = P(A) × P(B) if independent
P(good diet) = 0.5 | P(exercises) = (35+20)/100 = 55/100 = 0.55
P(A) × P(B) = 0.5 × 0.55 = 0.275
P(good diet ∩ exercises) = 35/100 = 0.35 ≠ 0.275
Conclusion: NOT independent — diet and exercise are associated.

🚀 Part 4: Extension Activities (Q26–Q30) — Answers

📌 For advanced learners preparing for Maths Olympiad, ICAS, or Selective School entry.

Q26. (Extension) A box contains 5 red, 3 blue, and 2 green balls. Two balls are drawn WITHOUT replacement. Find P(both the same colour).

Total balls = 10
P(both red) = (5/10) × (4/9) = 20/90
P(both blue) = (3/10) × (2/9) = 6/90
P(both green) = (2/10) × (1/9) = 2/90
P(same colour) = (20 + 6 + 2)/90 = 28/90 = 14/45 ≈ 0.311

Q27. (Extension) A game show has 3 doors. Behind one is a car; behind the others are goats. You pick Door 1. The host opens Door 3 (revealing a goat). Should you switch to Door 2? Justify using probability. (Monty Hall Problem)

Initial pick: P(car behind Door 1) = 1/3, P(car behind Door 2 or 3) = 2/3
Host reveals goat behind Door 3.
The probability the car is behind Door 2 = 2/3 (the host's action concentrates the 2/3 probability onto Door 2).
P(win by staying) = 1/3
P(win by switching) = 2/3
Conclusion: YES — always switch. Switching doubles your chance of winning.

Q28. (Extension) In a group of 30 students, what is the probability that at least 2 students share the same birthday? (Use the complement method — assume 365 days, ignore leap years.)

Use complement: P(at least 2 share) = 1 − P(all different)
P(all 30 have different birthdays) = (365/365) × (364/365) × (363/365) × ... × (336/365)
= 365!/(335! × 365³⁰)
≈ 1 − 0.294 = 0.706 (approximately)
Answer: ≈ 70.6% chance at least 2 students share a birthday.
This is the famous Birthday Problem — a surprising result in probability!

Q29. (Extension) A test for a disease is 95% accurate. The disease affects 1% of the population. If a person tests positive, what is the probability they actually have the disease? (Bayes' Theorem)

Let D = has disease, T = tests positive
P(D) = 0.01 | P(D′) = 0.99
P(T|D) = 0.95 (true positive) | P(T|D′) = 0.05 (false positive)

Bayes' Theorem:
P(D|T) = P(T|D) × P(D) / P(T)
P(T) = (0.95 × 0.01) + (0.05 × 0.99) = 0.0095 + 0.0495 = 0.059
P(D|T) = 0.0095 / 0.059 ≈ 0.161 = 16.1%
Despite 95% accuracy, only ~16% of positive tests indicate actual disease — highlighting the importance of base rates!

Q30. (Extension) A card is drawn from a 52-card deck. Event A = drawing a red card. Event B = drawing a face card (J, Q, K). (a) Are A and B independent? (b) Find P(A ∪ B). (c) Find P(A | B).

(a) Independence check:
P(A) = 26/52 = 1/2 | P(B) = 12/52 = 3/13
P(A ∩ B) = 6/52 = 3/26 (red face cards: J♥, Q♥, K♥, J♦, Q♦, K♦)
P(A) × P(B) = 1/2 × 3/13 = 3/26 = P(A ∩ B) ✓
YES — A and B are independent events.
(b) P(A ∪ B) = 1/2 + 3/13 − 3/26 = 13/26 + 6/26 − 3/26 = 16/26 = 8/13 ≈ 0.615
(c) P(A|B) = P(A ∩ B)/P(B) = (6/52)/(12/52) = 6/12 = 1/2 (same as P(A) — confirms independence)

📌 Differentiation Strategies for Diverse Learners

🟢 Foundation (Year 7–8 level): Focus on Q1–Q10. Use concrete materials (dice, cards, spinners). Provide probability scales (0 to 1) and vocabulary cards.

🟡 Core (Year 8 on-level): Complete Q1–Q25. Emphasise correct notation (P(A), ∩, ∪, A′). Use systematic listing before tree diagrams.

🔴 Extension (Advanced / Olympiad): Complete Q26–Q30. Introduce Bayes' Theorem, conditional probability, and independence proofs. Encourage investigation of the Birthday Problem and Monty Hall Problem.

🔵 NAPLAN / ICAS Prep: Focus on multi-step problems, two-way tables (Q19–Q24), and problems requiring justification. Practice writing clear mathematical reasoning.