Grade 8 Probability — Answer Key

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Probability dice and spinner illustration

🎯 Success Criteria

By the end of this topic, students can:

✅ Express probability as a fraction, decimal, and percentage

✅ Calculate theoretical and experimental probability

✅ Use the complementary rule: P(not A) = 1 − P(A)

✅ Construct and interpret two-way tables and tree diagrams

✅ Distinguish between dependent and independent events

📚 Part 1: Multiple Choice — Answers

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

A) 3/10

B) 4/10 = 2/5 ✅

C) 1/4

D) 1/3

Working: P(red) = 4 ÷ (4+3+3) = 4/10 = 2/5 = 0.4 = 40%

2. A fair six-sided die is rolled. What is P(not a 5)?

A) 1/6

B) 4/6

C) 5/6 ✅

D) 1/5

Working: P(not 5) = 1 − P(5) = 1 − 1/6 = 5/6 ≈ 0.833

3. A spinner has 8 equal sections numbered 1–8. What is P(even number)?

A) 3/8

B) 4/8 = 1/2 ✅

C) 5/8

D) 2/8

Working: Even numbers = {2, 4, 6, 8} → 4 outcomes out of 8 → P = 4/8 = 1/2

4. Two coins are flipped. What is the probability of getting exactly one head?

A) 1/4

B) 2/4 = 1/2 ✅

C) 3/4

D) 1/2 only if coins are biased

Working: Sample space = {HH, HT, TH, TT}. Exactly one head: {HT, TH} = 2 outcomes → P = 2/4 = 1/2

✏️ Part 2: Short Answer — Answers & Full Working

5. A class of 25 students contains 10 boys and 15 girls. A student is chosen at random. Find P(girl) as a fraction, decimal, and percentage.

Answer:

P(girl) = 15/25 = 3/5

Decimal: 3 ÷ 5 = 0.6

Percentage: 0.6 × 100 = 60%

6. A student rolled a die 60 times and got a 6 exactly 8 times. (a) What is the experimental probability of rolling a 6? (b) What is the theoretical probability? (c) Explain any difference.

Answer:

(a) Experimental P(6) = 8/60 = 2/15 ≈ 0.133 ≈ 13.3%

(b) Theoretical P(6) = 1/6 ≈ 0.167 ≈ 16.7%

(c) The results differ because experimental probability is based on a limited number of trials. As the number of trials increases, experimental probability approaches theoretical probability (Law of Large Numbers).

7. A card is drawn from a standard 52-card deck. Find the probability that it is a heart OR a king.

Answer:

P(heart) = 13/52, P(king) = 4/52, P(king of hearts) = 1/52

Using the Addition Rule: P(heart OR king) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13 ≈ 0.308 ≈ 30.8%

Key concept: Subtract the overlap to avoid double-counting.

8. A bag has 5 red and 3 blue balls. Two balls are drawn WITHOUT replacement. Complete the tree diagram and find P(both red).

Answer (Tree Diagram Summary):

1st draw: P(R) = 5/8, P(B) = 3/8

2nd draw (given 1st was Red): P(R|R) = 4/7, P(B|R) = 3/7

2nd draw (given 1st was Blue): P(R|B) = 5/7, P(B|B) = 2/7

P(both red) = 5/8 × 4/7 = 20/56 = 5/14 ≈ 0.357 ≈ 35.7%

9. In a survey of 40 students: 22 play football, 18 play basketball, and 7 play both. Complete the two-way table and find P(plays football only).

Answer:

Football only = 22 − 7 = 15

Basketball only = 18 − 7 = 11

Both = 7, Neither = 40 − (15 + 11 + 7) = 7

P(football only) = 15/40 = 3/8 = 0.375 = 37.5%

🚀 Part 3: Extension — Advanced Learners

For students preparing for Maths Olympiad, ICAS, or Selective School exams.

10. EXTENSION: A game show has 3 doors. Behind one is a car; the others hide goats. You pick Door 1. The host (who knows what's behind each door) opens Door 3 to reveal a goat. Should you switch to Door 2? Justify using probability.

Answer (Monty Hall Problem):

If you stay with Door 1: P(win) = 1/3

If you switch to Door 2: P(win) = 2/3

Yes, you should switch. Initially, there is a 2/3 chance the car is behind one of the other two doors. When the host reveals a goat behind Door 3, the full 2/3 probability transfers to Door 2. Switching doubles your chance of winning.

11. EXTENSION: Two independent events A and B have P(A) = 0.4 and P(B) = 0.3. Find: (a) P(A and B) (b) P(A or B) (c) P(neither A nor B)

Answer:

(a) P(A and B) = P(A) × P(B) = 0.4 × 0.3 = 0.12 (independent events → multiply)

(b) P(A or B) = P(A) + P(B) − P(A and B) = 0.4 + 0.3 − 0.12 = 0.58

Alternative for (c): P(not A) × P(not B) = 0.6 × 0.7 = 0.42 ✅

12. CHALLENGE: A fair coin is flipped 4 times. What is the probability of getting at least 2 heads?

Answer:

Total outcomes = 2⁴ = 16

P(0 heads) = 1/16 (TTTT)

P(1 head) = 4/16 (HTTT, THTT, TTHT, TTTH)

P(at least 2 heads) = 1 − P(0 heads) − P(1 head) = 1 − 1/16 − 4/16 = 11/16 ≈ 0.6875 ≈ 68.75%

🌟 Differentiation Strategies

Foundation Learners: Focus on Questions 1–5. Use physical manipulatives (coins, dice, coloured counters). Provide a probability scale (0 to 1) reference card. Practise converting fractions to decimals and percentages separately before combining.

Core Learners: Complete Questions 1–9. Encourage drawing tree diagrams for multi-step problems. Reinforce the complementary rule and addition rule with real-life contexts.

Advanced Learners (Olympiad / ICAS / Selective): Attempt all 12 questions including extensions. Explore conditional probability notation P(A|B). Investigate the Birthday Problem and research the Monty Hall Problem further. Begin exploring combinations (nCr) for counting outcomes efficiently.

NAPLAN Preparation Tip: Practice expressing probability in multiple forms (fraction, decimal, percentage) and interpreting two-way tables — both are common NAPLAN question types.