Secondary 4 Additional Mathematics Quiz - Statistics Probability (Answer Key)
Total Marks: 40
Note: This is syllabus-aligned practice from inferred templates; not past-year exam derived.
Section A (1 mark each)
Q1. Probability = 63=21.
Teaching note: Even numbers on a die are 2, 4, 6 (3 out of 6 equally likely outcomes).
Mark: 1 for 21.
Q2. P(A∪B)=P(A)+P(B)=0.3+0.4=0.7.
Teaching note: Mutually exclusive events cannot occur together, so no overlap to subtract.
Mark: 1 for 0.7.
Q3. P(red)=5+75=125.
Mark: 1 for 125.
Q4. p+q=1.
Teaching note: Exhaustive outcomes cover all possibilities.
Mark: 1 for 1.
Q5. Vowels: A, E, A, I (4 of 11 letters). P=114.
Mark: 1 for 114.
Q6. Independent: P(X∩Y)=P(X)×P(Y)=0.5×0.2=0.1.
Mark: 1 for 0.1.
Q7. 1−0.15=0.85.
Mark: 1 for 0.85.
Q8. P(yellow)=3+2+52=102=51.
Mark: 1 for 51 or 0.2.
Section B (2 marks each)
Q9. Sample space = {HH, HT, TH, TT}. Exactly one head: HT, TH → P=42=21.
Marking: 1 for sample space, 1 for probability.
Q10. P(Q∪H)=P(Q)+P(H)−P(Q∩H)=524+5213−521=5216=134.
Marking: 1 for correct addition/subtraction, 1 for 134.
Q11. P(both white)=106×95=9030=31.
Marking: 1 for method, 1 for answer.
Q12. P(F∪B)=3016+17−11=3022=1511.
Marking: 1 for method using inclusion-exclusion, 1 for answer.
Q13. Let p=P(1)=⋯=P(5). Then 5p+0.25=1⇒5p=0.75⇒p=0.15.
Marking: 1 for equation, 1 for 0.15.
Q14. P(A∪B)=P(A)+P(B)−P(A)P(B) (independent).
0.7=0.4+P(B)−0.4P(B)⇒0.3=0.6P(B)⇒P(B)=0.5.
Marking: 1 for equation, 1 for 0.5.
Section C
Q15. (a) Tree diagram: first branch R(5/10), B(3/10), G(2/10); second branch adjusted for without replacement.
(b) Same colour: RR + BB + GG = 105⋅94+103⋅92+102⋅91=9020+6+2=9028=4514.
Marking: 2 for tree, 2 for probability.
Q16. (a) P(B∪C)=10060+50−30=10080=0.8.
(b) Neither = 1−0.8=0.2.
(c) Since P(B∩C)=30=0, not mutually exclusive.
Marking: 2, 1, 1.
Q17. (a) Sum 5: (1,4),(2,3),(3,2),(4,1) → 164=41.
(b) At least one 4: total − no 4 = 1−43⋅43=1−169=167.
Marking: 2 each.
Q18. X∼B(10,0.08). P(X=2)=(210)(0.08)2(0.92)8=45×0.0064×0.5132≈0.1478.
Marking: 1 for identify binomial, 1 for substitution, 1 for answer.
Q19. (a) 0.1+p+0.3+0.2=1⇒p=0.4.
(b) E(X)=0(0.1)+1(0.4)+2(0.3)+3(0.2)=0+0.4+0.6+0.6=1.6.
(c) E(X2)=0+0.4+1.2+1.8=3.4; Var(X)=3.4−1.62=3.4−2.56=0.84.
Marking: 1, 2, 2.
Q20. (a) P(A)P(B)=0.5×0.4=0.2=P(A∩B) → independent.
(b) P(A′∩B′)=1−P(A∪B)=1−(0.5+0.4−0.2)=1−0.7=0.3.
Marking: 2, 2.