O Level Elementary Mathematics Statistics Probability Quiz

O-Level Elementary Mathematics Quiz - Statistics Probability (Answer Key)

Total Marks: 40
Topic: Statistics and Probability (syllabus-first generated content, not past-year derived)

Section A: Data Handling and Representation

Q1. [2 marks]
Total books = ( (0 \times 4) + (1 \times 9) + (2 \times 8) + (3 \times 6) + (4 \times 3) )
= ( 0 + 9 + 16 + 18 + 12 = 55 )
Answer: 55 books.
Teaching note: Multiply each value by its frequency then sum. Common mistake: adding frequencies only (30) instead of weighted books.

Q2. [2 marks]
Mean = total books ÷ number of students = ( 55 \div 30 = 1.833... )
= 1.83 (2 d.p.)
Answer: 1.83.
Teaching note: Mean = sum of all data ÷ number of data points. Use total from Q1.

Q3. [1 mark]
Ordered masses: 2.0, 2.1, 2.2, 2.3, 2.3, 2.4, 2.5. Middle (4th) value = 2.3.
Answer: 2.3 kg.
Teaching note: Median is the middle number after sorting. For 7 values, the 4th is median.

Q4. [1 mark]
7 appears most often (three times).
Answer: 7.
Teaching note: Mode is the most frequent value.

Q5. [1 mark]
Range = largest – smallest = 24 – 12 = 12.
Answer: 12.
Teaching note: Range shows spread; subtract min from max.

Q6. [2 marks]
Bus students = 60 out of 120.
Angle = ( \frac{60}{120} \times 360^\circ = 180^\circ ).
Answer: 180°.
Teaching note: Pie angle = (frequency ÷ total) × 360°. Check: MRT 108°, Walk 54°, Cycle 18°, sum = 360°.

Q7. [1 mark]
Highest frequency is 8 at mark 30.
Answer: 30.
Teaching note: Modal mark is the mark with highest frequency.

Section B: Probability Basics

Q8. [2 marks]
Total marbles = 5 + 3 + 2 = 10.
P(blue) = ( \frac{3}{10} ).
Answer: ( \frac{3}{10} ).
Teaching note: Probability = favourable ÷ total outcomes.

Q9. [1 mark]
Numbers >4 on die: 5, 6 → 2 outcomes.
P = ( \frac{2}{6} = \frac{1}{3} ).
Answer: ( \frac{1}{3} ).
Teaching note: Fair die has 6 equal outcomes.

Q10. [2 marks]
Outcomes: HH, HT, TH, TT.
Answer: HH, HT, TH, TT.
Teaching note: Two coins, each H or T; list systematically.

Q11. [1 mark]
Exactly one head: HT, TH → 2 of 4.
P = ( \frac{2}{4} = \frac{1}{2} ).
Answer: ( \frac{1}{2} ).

Q12. [2 marks]
"ELEMENTARY" has 10 letters; E appears at positions 1,4,7 → 3 times.
P(E) = ( \frac{3}{10} ).
Answer: ( \frac{3}{10} ).
Teaching note: Count repeated letters carefully.

Q13. [1 mark]
Red appears in 2 of 4 equal sections.
P(red) = ( \frac{2}{4} = \frac{1}{2} ).
Answer: ( \frac{1}{2} ).

Q14. [1 mark]
P(no rain) = 1 – 0.3 = 0.7.
Answer: 0.7.
Teaching note: Total probability = 1.

Section C: Combined and Extended Probability

Q15. [3 marks]
P(first white) = ( \frac{6}{10} ). After drawing one white, 5 white left of 9 total.
P(second white) = ( \frac{5}{9} ).
P(both) = ( \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} ).
Answer: ( \frac{1}{3} ).
Marking: 1m for first prob, 1m second, 1m final. Without replacement reduces total.

Q16. [3 marks]
Use inclusion: n(Phys ∪ Chem) = 12 + 8 – 5 = 15.
P(at least one) = ( \frac{15}{20} = \frac{3}{4} ).
Answer: ( \frac{3}{4} ).
Teaching note: Avoid double-counting overlap.

Q17. [3 marks]
From diagram: n(A∪B) = 7 + 5 + 3 = 15.
Total students = 15 + 10 = 25.
P(A∪B) = ( \frac{15}{25} = \frac{3}{5} ).
Answer: n(A∪B)=15, P=( \frac{3}{5} ).
Teaching note: Union includes all inside circles.

Q18. [2 marks]
P(red) = ( \frac{3}{5} ). With replacement, independent.
P(both red) = ( \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} ).
Answer: ( \frac{9}{25} ).

Q19. [2 marks]
Total students = 10+8+6+4+7+9+5+3 = 52.
Girls liking Badminton = 9.
P = ( \frac{9}{52} ).
Answer: ( \frac{9}{52} ).

Q20. [3 marks]
Hearts = 13, Kings = 4, King of Hearts counted twice.
Favourable = 13 + 4 – 1 = 16.
P = ( \frac{16}{52} = \frac{4}{13} ).
Answer: ( \frac{4}{13} ).
Teaching note: Use OR rule: P(A∪B)=P(A)+P(B)–P(A∩B).