Secondary 1 Mathematics Statistics Probability Quiz

Secondary 1 Mathematics Quiz - Statistics Probability (Answer Key)

Total Marks: 40

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Section A: Data Collection and Organisation (Questions 1–5, 10 marks)

1. (a) [1 mark]

The frequency table is already complete as given.
Answer: Table completed (no missing values).
Marking note: Award 1 mark if student confirms table is complete or fills any missing entry correctly.

1. (b) [1 mark]

The mode is the value with the highest frequency.
Frequency: 0→4, 1→6, 2→8, 3→7, 4→3, 5→2.
Highest frequency is 8 for "2 books".
Answer: 2 books
Marking note: Accept "2" or "2 books".

2. [2 marks]

Class intervals: 150–154, 155–159, 160–164.
Count data:
150–154: 152, 154 → 2
155–159: 158, 155, 157, 159, 156, 158, 157, 158, 155, 159 → 10
160–164: 160, 162, 161, 160, 163, 160, 161 → 7
Total = 2+10+7 = 19? Wait, 20 students. Recount:
Data: 152, 158, 160, 155, 162, 157, 159, 161, 156, 158, 160, 154, 159, 163, 157, 158, 160, 155, 159, 161
150–154: 152, 154 → 2
155–159: 155, 157, 159, 156, 158, 158, 157, 158, 155, 159, 159 → 11? Let's list: 155 (positions 4,18), 157 (6,15), 159 (7,13,19), 156 (9), 158 (2,10,16) → that's 2+2+3+1+3 = 11.
160–164: 160 (3,11,17), 162 (5), 161 (8,20), 163 (14) → 3+1+2+1 = 7.
Total = 2+11+7 = 20. Correct.

Height (cm)	Frequency
150–154	2
155–159	11
160–164	7

Answer: Table as above.
Marking: 1 mark for correct intervals, 1 mark for correct frequencies.
Common mistake: Miscounting data points; ensure tallying is done carefully.

3. (a) [1 mark]

Modal class = class with highest frequency.
Frequencies: 0–2:3, 3–5:7, 6–8:9, 9–11:4, 12–14:2.
Highest is 9 for class 6–8.
Answer: 6–8 hours
Marking note: Accept "6–8" or "6–8 hours".

3. (b) [2 marks]

Estimated mean = Σ(f × x) / Σf, where x = class midpoint.
Midpoints: 0–2→1, 3–5→4, 6–8→7, 9–11→10, 12–14→13.
Σf = 25.
Σ(fx) = 3(1) + 7(4) + 9(7) + 4(10) + 2(13) = 3 + 28 + 63 + 40 + 26 = 160.
Mean = 160 / 25 = 6.4 hours.
Answer: 6.4 hours
Working marks: 1 mark for correct midpoints and Σ(fx), 1 mark for correct division and answer.
Common mistake: Using class boundaries instead of midpoints; forgetting to divide by total frequency.

4. (a) [1 mark]

Relative frequency = frequency of outcome / total trials = 9 / 60 = 3/20 = 0.15.
Answer: 0.15 or 3/20 or 15%
Marking note: Accept fraction, decimal, or percentage.

4. (b) [1 mark]

For a fair die, each outcome should have relative frequency ≈ 1/6 ≈ 0.1667.
Observed relative frequencies: 1:12/60=0.2, 2:8/60≈0.133, 3:10/60≈0.167, 4:9/60=0.15, 5:11/60≈0.183, 6:10/60≈0.167.
These vary around 1/6. With only 60 trials, some variation is expected. The die appears fair (no strong evidence of bias).
Answer: Yes, the die appears fair because the relative frequencies are close to 1/6 and variation is expected in 60 trials.
Marking note: Accept reasonable explanation referencing expected probability 1/6 and sample size.

5. (a) [1 mark]

Stem 3 | leaves 1 2 4 6 8 9 → 6 teachers in their 30s (31,32,34,36,38,39).
Answer: 6
Marking note: Count leaves in the 3 stem.

5. (b) [1 mark]

Total 15 teachers. Median is the 8th value (since (15+1)/2 = 8).
Ordered data from diagram:
23, 25, 27, 28, 29, 31, 32, 34, 36, 38, 39, 40, 42, 45. Wait, 15 values:
Stem 2: 23,25,27,28,29 (5 values)
Stem 3: 31,32,34,36,38,39 (6 values) → positions 6 to 11
Stem 4: 40,42,45 (3 values) → positions 12,13,14? Wait 5+6+3=14. Missing one.
Stem 4 has 0,2,5 → 3 values. Total 5+6+3=14. But says 15 teachers. Recount:
2 | 3 5 7 8 9 → 5
3 | 1 2 4 6 8 9 → 6
4 | 0 2 5 → 3
Total 14. Diagram error? Assume 15 teachers, maybe stem 4 has 4 leaves? But given as 0,2,5.
For median of 14, it's average of 7th and 8th.
7th = 32, 8th = 34 → median = 33.
But question says 15 teachers. Likely a typo in diagram. We'll proceed with 14 data points, median = (32+34)/2 = 33.
Or if 15, maybe stem 4 has 4 leaves? Let's assume the diagram has 15 leaves: perhaps 4 | 0 2 5 and one more? Not specified.
Better: Use the given diagram as is (14 values) but question says 15. I'll assume stem 4 has 4 leaves: 0,2,5, and say 7? But not given.
Resolution: In answer key, note the discrepancy. For 14 values, median = 33. For 15 values (if one more in 40s), median = 8th value = 34.
Answer: 33 years (based on 14 data points shown) or 34 years (if 15th value in 40s).
Marking: Accept 33 or 34 with correct reasoning.
Teaching note: Always count total frequency first. Median position = (n+1)/2 for odd n.

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Section B: Data Representation and Interpretation (Questions 6–12, 14 marks)

6. (a) [1 mark]

From bar chart: Sports has highest bar (45).
Answer: Sports
Marking: 1 mark for correct identification.

6. (b) [2 marks]

Total students = 45+30+25+20+15 = 135.
Uniformed Groups = 25.
Percentage = (25/135) × 100% = 18.518...% ≈ 18.5% (or 500/27 %).
Answer: 18.5% (or 500/27 %)
Working: 1 mark for correct total, 1 mark for correct calculation and answer.
Common mistake: Using wrong total or misreading bar height.

7. (a) [2 marks]

Total angle = 360°. Oranges sector = 108°.
Number of students = (108/360) × 200 = 0.3 × 200 = 60.
Answer: 60 students
Working: 1 mark for correct fraction (108/360), 1 mark for correct multiplication and answer.

7. (b) [1 mark]

Apples sector = 90°. Fraction = 90/360 = 1/4.
Answer: 1/4
Marking: 1 mark for simplest form.

8. (a) [1 mark]

Highest temperature = 34°C at 2 p.m.
Answer: 2 p.m.
Marking: 1 mark.

8. (b) [1 mark]

Temperature at 8 a.m. = 28°C, at 12 p.m. = 33°C.
Increase = 33 – 28 = 5°C.
Answer: 5°C
Marking: 1 mark.

8. (c) [1 mark]

At 11 a.m., between 10 a.m. (31°C) and 12 p.m. (33°C). Linear interpolation: halfway → 32°C.
Answer: 32°C (accept 31–33°C with reasoning)
Marking: 1 mark for reasonable estimate.

9. (a) [1 mark]

Mode = value with most dots = 1 sibling (6 dots).
Answer: 1 sibling
Marking: 1 mark.

9. (b) [2 marks]

Mean = Σ(x × f) / Σf.
x: 0,1,2,3,4,5; f: 3,6,5,4,2,0.
Σf = 20.
Σ(xf) = 0(3)+1(6)+2(5)+3(4)+4(2)+5(0) = 0+6+10+12+8+0 = 36.
Mean = 36/20 = 1.8.
Answer: 1.8 siblings
Working: 1 mark for correct Σ(xf), 1 mark for division and answer.

10. (a) [1 mark]

Class 4–6 kg has frequency 12 (height of bar).
Answer: 12 parcels
Marking: 1 mark.

10. (b) [2 marks]

Estimate total mass using midpoints × frequencies.
Class width = 2 kg. Midpoints: 1, 3, 5, 7, 9.
Frequencies: 5, 8, 12, 10, 5.
Total mass ≈ 5(1) + 8(3) + 12(5) + 10(7) + 5(9) = 5 + 24 + 60 + 70 + 45 = 204 kg.
Answer: 204 kg
Working: 1 mark for correct midpoints and products, 1 mark for sum and answer.
Note: This is an estimate because exact masses within intervals are unknown.

11. [2 marks]

Line graph: Points at (Jan,120), (Feb,150), (Mar,180), (Apr,160), (May,200), (Jun,190). Connected by straight lines. Axes labelled.
Marking: 1 mark for correct plotting of all 6 points, 1 mark for connecting with lines and labelled axes.
Common mistake: Not labelling axes, incorrect scale, joining points with curves instead of straight segments.

12. (a) [1 mark]

Class A median = 70, Class B median = 65. Class A higher.
Answer: Class A
Marking: 1 mark.

12. (b) [1 mark]

IQR = Q3 – Q1.
Class A: 85 – 55 = 30.
Class B: 80 – 50 = 30.
Both have same IQR = 30.
Answer: Both classes have the same interquartile range of 30.
Marking: 1 mark for correct calculation and comparison.

12. (c) [1 mark]

Other comparisons:

• Class A has higher minimum (40 vs 30), higher maximum (95 vs 90).
• Class A has higher Q1 (55 vs 50) and higher Q3 (85 vs 80).
• Class A's distribution is shifted higher overall.
• Range: Class A = 55, Class B = 60.
  Answer: Any valid comparison, e.g., "Class A has a higher minimum score" or "Class A's scores are generally higher."
  Marking: 1 mark for any correct comparative statement.

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Section C: Probability (Questions 13–20, 16 marks)

13. (a) [1 mark]

Total balls = 5+3+2 = 10.
P(red) = 5/10 = 1/2.
Answer: 1/2 or 0.5 or 50%
Marking: 1 mark.

13. (b) [1 mark]

P(not blue) = 1 – P(blue) = 1 – 3/10 = 7/10.
Or: red + green = 5+2 = 7 out of 10.
Answer: 7/10 or 0.7 or 70%
Marking: 1 mark.

14. (a) [1 mark]

Sample space = {1, 2, 3, 4, 5, 6}.
Answer: {1, 2, 3, 4, 5, 6}
Marking: 1 mark for complete set.

14. (b) [1 mark]

Prime numbers on die: 2, 3, 5. (1 is not prime).
Favourable outcomes = 3. Total = 6.
P(prime) = 3/6 = 1/2.
Answer: 1/2
Marking: 1 mark.
Common mistake: Including 1 as prime.

14. (c) [1 mark]

Numbers > 4: 5, 6.
P(>4) = 2/6 = 1/3.
Answer: 1/3
Marking: 1 mark.

15. (a) [1 mark]

Word "MATHEMATICS" has 11 letters.
Answer: 11
Marking: 1 mark.

15. (b) [1 mark]

Letters: M,A,T,H,E,M,A,T,I,C,S. 'A' appears 2 times.
P(A) = 2/11.
Answer: 2/11
Marking: 1 mark.

15. (c) [1 mark]

Vowels in word: A, E, A, I → 4 vowels (A appears twice).
P(vowel) = 4/11.
Answer: 4/11
Marking: 1 mark.
Common mistake: Counting distinct vowels only (A,E,I = 3) instead of occurrences.

16. (a) [1 mark]

Even numbers on spinner: 2,4,6,8 → 4 out of 8.
P(even) = 4/8 = 1/2.
Answer: 1/2
Marking: 1 mark.

16. (b) [1 mark]

Multiples of 3: 3, 6 → 2 out of 8.
P(multiple of 3) = 2/8 = 1/4.
Answer: 1/4
Marking: 1 mark.

16. (c) [1 mark]

Numbers < 3: 1, 2 → 2 out of 8.
P(<3) = 2/8 = 1/4.
Answer: 1/4
Marking: 1 mark.

17. (a) [1 mark]

Multiples of 4 from 1 to 20: 4,8,12,16,20 → 5 numbers.
P(multiple of 4) = 5/20 = 1/4.
Answer: 1/4
Marking: 1 mark.

17. (b) [1 mark]

Factors of 12: 1,2,3,4,6,12 → 6 numbers.
P(factor of 12) = 6/20 = 3/10.
Answer: 3/10
Marking: 1 mark.

17. (c) [1 mark]

Multiples of 3 and 5 = multiples of 15. From 1 to 20: 15 only.
P = 1/20.
Answer: 1/20
Marking: 1 mark.

18. (a) [2 marks]

Venn diagram:
Universal set = 40.
Basketball = 25, Football = 18, Both = 10.
Only Basketball = 25 – 10 = 15.
Only Football = 18 – 10 = 8.
Neither = 40 – (15+10+8) = 7.
Regions: B only=15, F only=8, Both=10, Neither=7.
Marking: 1 mark for correct values in overlapping region and only regions, 1 mark for correct neither and total 40.
Common mistake: Putting 25 and 18 in the circles without subtracting overlap.

18. (b) [1 mark]

P(neither) = 7/40.
Answer: 7/40
Marking: 1 mark.

19. (a) [1 mark]

Two coins: outcomes = {HH, HT, TH, TT}.
Answer: HH, HT, TH, TT (or set notation)
Marking: 1 mark for all 4 outcomes.
Common mistake: Listing only 3 outcomes (e.g., treating HT and TH as same).

19. (b) [1 mark]

Exactly one head: HT, TH → 2 out of 4.
P = 2/4 = 1/2.
Answer: 1/2
Marking: 1 mark.

19. (c) [1 mark]

At least one tail: HT, TH, TT → 3 out of 4.
Or 1 – P(no tail) = 1 – P(HH) = 1 – 1/4 = 3/4.
Answer: 3/4
Marking: 1 mark.

20. (a) [2 marks]

Tree diagram:
First draw: W (4/10), B (6/10).
If first W: remaining 3W, 6B → P(W)=3/9, P(B)=6/9.
If first B: remaining 4W, 5B → P(W)=4/9, P(B)=5/9.
Marking: 1 mark for correct first-level probabilities, 1 mark for correct second-level conditional probabilities.
Common mistake: Using same probabilities for second draw (with replacement) instead of without replacement.

20. (b) [1 mark]

P(both white) = P(W then W) = (4/1

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Secondary 1 Mathematics Quiz - Statistics Probability (Answer Key)

Total Marks: 40

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Section A: Data Collection and Organisation (Questions 1–5, 10 marks)

1. (a) [1 mark]

The frequency table is already complete as given.
Answer: Table completed (no missing values).
Marking note: Award 1 mark if student confirms table is complete or fills any missing entry correctly.

1. (b) [1 mark]

The mode is the value with the highest frequency.
Frequency: 0→4, 1→6, 2→8, 3→7, 4→3, 5→2.
Highest frequency is 8 for "2 books".
Answer: 2 books
Marking note: Accept "2" or "2 books".

2. [2 marks]

Class intervals: 150–154, 155–159, 160–164.
Count data:
150–154: 152, 154 → 2
155–159: 158, 155, 157, 159, 156, 158, 157, 158, 155, 159, 159 → 11
160–164: 160, 162, 161, 163, 160, 160, 161 → 7
Total = 2+11+7 = 20.

Height (cm)	Frequency
150–154	2
155–159	11
160–164	7

Answer: Table as above.
Marking: 1 mark for correct intervals, 1 mark for correct frequencies.
Common mistake: Miscounting data points; ensure tallying is done carefully.

3. (a) [1 mark]

Modal class = class with highest frequency.
Frequencies: 0–2:3, 3–5:7, 6–8:9, 9–11:4, 12–14:2.
Highest is 9 for class 6–8.
Answer: 6–8 hours
Marking note: Accept "6–8" or "6–8 hours".

3. (b) [2 marks]

Estimated mean = Σ(f × x) / Σf, where x = class midpoint.
Midpoints: 0–2→1, 3–5→4, 6–8→7, 9–11→10, 12–14→13.
Σf = 25.
Σ(fx) = 3(1) + 7(4) + 9(7) + 4(10) + 2(13) = 3 + 28 + 63 + 40 + 26 = 160.
Mean = 160 / 25 = 6.4 hours.
Answer: 6.4 hours
Working marks: 1 mark for correct midpoints and Σ(fx), 1 mark for correct division and answer.
Common mistake: Using class boundaries instead of midpoints; forgetting to divide by total frequency.

4. (a) [1 mark]

Relative frequency = frequency of outcome / total trials = 9 / 60 = 3/20 = 0.15.
Answer: 0.15 or 3/20 or 15%
Marking note: Accept fraction, decimal, or percentage.

4. (b) [1 mark]

For a fair die, each outcome should have relative frequency ≈ 1/6 ≈ 0.1667.
Observed relative frequencies: 1:12/60=0.2, 2:8/60≈0.133, 3:10/60≈0.167, 4:9/60=0.15, 5:11/60≈0.183, 6:10/60≈0.167.
These vary around 1/6. With only 60 trials, some variation is expected. The die appears fair (no strong evidence of bias).
Answer: Yes, the die appears fair because the relative frequencies are close to 1/6 and variation is expected in 60 trials.
Marking note: Accept reasonable explanation referencing expected probability 1/6 and sample size.

5. (a) [1 mark]

Stem 3 | leaves 1 2 4 6 8 9 → 6 teachers in their 30s (31,32,34,36,38,39).
Answer: 6
Marking note: Count leaves in the 3 stem.

5. (b) [1 mark]

Total 15 teachers. Median is the 8th value (since (15+1)/2 = 8).
Ordered data from diagram:
23, 25, 27, 28, 29, 31, 32, 34, 36, 38, 39, 40, 42, 45. (Note: diagram shows 14 values; assuming 15th value in 40s or accepting 14 values gives median 33).
For 15 values, 8th value = 34.
Answer: 34 years
Marking: Accept 33 or 34 with correct reasoning.
Teaching note: Always count total frequency first. Median position = (n+1)/2 for odd n.

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Section B: Data Representation and Interpretation (Questions 6–12, 14 marks)

6. (a) [1 mark]

Answer: Sports
Marking note: Direct reading from bar chart.

6. (b) [2 marks]

Total students = 45 + 30 + 25 + 20 + 15 = 135.
Uniformed Groups = 25.
Percentage = (25 / 135) × 100% = 18.518...% ≈ 18.5% (or 500/27 %).
Answer: 18.5% (or 500/27 %)
Working marks: 1 mark for correct total, 1 mark for correct calculation and answer.

7. (a) [2 marks]

Total angle = 360°. Oranges sector = 108°.
Number of students = (108° / 360°) × 200 = 0.3 × 200 = 60.
Answer: 60 students
Working marks: 1 mark for correct fraction, 1 mark for correct answer.

7. (b) [1 mark]

Apples sector = 90°. Fraction = 90° / 360° = 1/4.
Answer: 1/4
Marking note: Must be in simplest form.

8. (a) [1 mark]

Answer: 2 p.m. (or 14:00)
Marking note: Highest point on graph is 34°C at 2 p.m.

8. (b) [1 mark]

Temperature at 8 a.m. = 28°C, at 12 p.m. = 33°C.
Increase = 33 – 28 = 5°C.
Answer: 5°C

8. (c) [1 mark]

At 11 a.m., halfway between 10 a.m. (31°C) and 12 p.m. (33°C).
Estimate = (31 + 33) / 2 = 32°C.
Answer: 32°C
Marking note: Accept 31.5°C to 32.5°C with valid interpolation.

9. (a) [1 mark]

Mode = value with most dots = 1 sibling (6 dots).
Answer: 1 sibling

9. (b) [2 marks]

Mean = Σ(x × f) / Σf = (0×3 + 1×6 + 2×5 + 3×4 + 4×2 + 5×0) / 20 = (0 + 6 + 10 + 12 + 8 + 0) / 20 = 36 / 20 = 1.8.
Answer: 1.8 siblings
Working marks: 1 mark for correct Σ(xf) and Σf, 1 mark for correct division and answer.

10. (a) [1 mark]

Class interval 4–6 kg has frequency 12 (bar height = 12, class width = 2, so frequency = 12).
Answer: 12 parcels

10. (b) [2 marks]

Estimated total mass = Σ(f × midpoint).
Midpoints: 0–2→1, 2–4→3, 4–6→5, 6–8→7, 8–10→9.
Frequencies: 5, 8, 12, 10, 5.
Total mass = 5(1) + 8(3) + 12(5) + 10(7) + 5(9) = 5 + 24 + 60 + 70 + 45 = 204 kg.
Answer: 204 kg
Working marks: 1 mark for correct midpoints and Σ(fx), 1 mark for correct answer.

11. [2 marks]

Line graph with points: (Jan,120), (Feb,150), (Mar,180), (Apr,160), (May,200), (Jun,190). Points connected by straight line segments. Axes labelled, scales correct.
Marking: 1 mark for correct plotting of all 6 points, 1 mark for correct line segments and labels.

12. (a) [1 mark]

Class A median = 70, Class B median = 65.
Answer: Class A

12. (b) [1 mark]

Class A IQR = Q3 – Q1 = 85 – 55 = 30.
Class B IQR = 80 – 50 = 30.
Answer: Both classes have the same interquartile range of 30.

12. (c) [1 mark]

Possible comparisons:

• Class A has a higher minimum (40 vs 30) and higher maximum (95 vs 90).
• Class A has a higher median (70 vs 65).
• Class A has a smaller range (55 vs 60).
• The distribution of Class A is shifted higher overall.
  Answer: Any valid comparison, e.g., "Class A has a higher minimum and maximum score" or "Class A has a smaller range (55) compared to Class B (60)."

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Section C: Probability (Questions 13–20, 16 marks)

13. (a) [1 mark]

Total balls = 5 + 3 + 2 = 10.
P(red) = 5/10 = 1/2.
Answer: 1/2

13. (b) [1 mark]

P(not blue) = 1 – P(blue) = 1 – 3/10 = 7/10.
Or: red + green = 5 + 2 = 7, so 7/10.
Answer: 7/10

14. (a) [1 mark]

Sample space = {1, 2, 3, 4, 5, 6}.
Answer: {1, 2, 3, 4, 5, 6}

14. (b) [1 mark]

Prime numbers on die: 2, 3, 5 → 3 outcomes.
P(prime) = 3/6 = 1/2.
Answer: 1/2

14. (c) [1 mark]

Numbers > 4: 5, 6 → 2 outcomes.
P(>4) = 2/6 = 1/3.
Answer: 1/3

15. (a) [1 mark]

Word "MATHEMATICS" has 11 letters.
Answer: 11

15. (b) [1 mark]

Letter 'A' appears 2 times.
P(A) = 2/11.
Answer: 2/11

15. (c) [1 mark]

Vowels in "MATHEMATICS": A, E, A, I → 4 vowels (A appears twice).
P(vowel) = 4/11.
Answer: 4/11

16. (a) [1 mark]

Even numbers: 2, 4, 6, 8 → 4 outcomes.
P(even) = 4/8 = 1/2.
Answer: 1/2

16. (b) [1 mark]

Multiples of 3: 3, 6 → 2 outcomes.
P(multiple of 3) = 2/8 = 1/4.
Answer: 1/4

16. (c) [1 mark]

Numbers < 3: 1, 2 → 2 outcomes.
P(<3) = 2/8 = 1/4.
Answer: 1/4

17. (a) [1 mark]

Multiples of 4 from 1 to 20: 4, 8, 12, 16, 20 → 5 numbers.
P(multiple of 4) = 5/20 = 1/4.
Answer: 1/4

17. (b) [1 mark]

Factors of 12: 1, 2, 3, 4, 6, 12 → 6 numbers.
P(factor of 12) = 6/20 = 3/10.
Answer: 3/10

17. (c) [1 mark]

Multiples of 3 and 5 = multiples of 15. From 1 to 20: 15 only → 1 number.
P(both) = 1/20.
Answer: 1/20

18. (a) [2 marks]

Venn diagram:

• Basketball only = 25 – 10 = 15
• Football only = 18 – 10 = 8
• Both = 10
• Neither = 40 – (15 + 10 + 8) = 7
  Marking: 1 mark for correct overlapping circles with labels, 1 mark for correct numbers in all four regions (including neither).

18. (b) [1 mark]

P(neither) = 7/40.
Answer: 7/40

19. (a) [1 mark]

Sample space: {HH, HT, TH, TT}.
Answer: HH, HT, TH, TT (or set notation)

19. (b) [1 mark]

Exactly one head: HT, TH → 2 outcomes.
P(exactly one head) = 2/4 = 1/2.
Answer: 1/2

19. (c) [1 mark]

At least one tail: HT, TH, TT → 3 outcomes.
P(at least one tail) = 3/4.
Answer: 3/4

20. (a) [2 marks]

Tree diagram:
First draw: W (4/10), B (6/10).
From W: W (3/9), B (6/9).
From B: W (4/9), B (5/9).
All branches labelled with probabilities.
Marking: 1 mark for correct structure and first-level probabilities, 1 mark for correct second-level conditional probabilities.

20. (b) [1 mark]

P(both white) = (4/10) × (3/9) = 12/90 = 2/15.
Answer: 2/15

20. (c) [1 mark]

P(different colours) = P(W then B) + P(B then W) = (4/10)(6/9) + (6/10)(4/9) = 24/90 + 24/90 = 48/90 = 8/15.
Answer: 8/15

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End of Answer Key