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O Level Additional Mathematics Statistics Probability Quiz

Free O Level A Maths Statistics quiz, Gemma31B Exam version, with questions, answers, and O Level-style practice for Singapore students.

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O Level Additional Mathematics From Real Exams Generated by Gemma 4 31B Updated 2026-08-17
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O-Level Additional Mathematics Quiz Answers - Statistics Probability

Section A: Probability and Discrete Distributions

  1. Working: P(Same)=P(RR)+P(BB)=(58×47)+(38×27)=2056+656=2656=1328P(\text{Same}) = P(RR) + P(BB) = (\frac{5}{8} \times \frac{4}{7}) + (\frac{3}{8} \times \frac{2}{7}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28} Answer: 0.464 (or 13/28) [3 marks]

  2. Working: P(At least one)=1−P(None)=1−(1−0.7)(1−0.6)=1−(0.3×0.4)=1−0.12=0.88P(\text{At least one}) = 1 - P(\text{None}) = 1 - (1-0.7)(1-0.6) = 1 - (0.3 \times 0.4) = 1 - 0.12 = 0.88 Answer: 0.88 [3 marks]

  3. Working: P(C∪P)=P(C)+P(P)−P(C∩P)=0.6+0.5−0.3=0.8P(C \cup P) = P(C) + P(P) - P(C \cap P) = 0.6 + 0.5 - 0.3 = 0.8. P(Neither)=1−0.8=0.2P(\text{Neither}) = 1 - 0.8 = 0.2 Answer: 0.2 [3 marks]

  4. Working: (53)(0.5)3(0.5)2=10×0.125×0.25=0.3125\binom{5}{3} (0.5)^3 (0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125 Answer: 0.313 [3 marks]

  5. Working: P(X≤1)=P(X=0)+P(X=1)=(100)(0.05)0(0.95)10+(101)(0.05)1(0.95)9≈0.5987+0.3151=0.9138P(X \le 1) = P(X=0) + P(X=1) = \binom{10}{0}(0.05)^0(0.95)^{10} + \binom{10}{1}(0.05)^1(0.95)^9 \approx 0.5987 + 0.3151 = 0.9138 Answer: 0.914 [4 marks]

  6. Working: P(A∩B)=P(A)+P(B)−P(A∪B)=0.4+0.5−0.7=0.2P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.7 = 0.2. Check independence: P(A)×P(B)=0.4×0.5=0.2P(A) \times P(B) = 0.4 \times 0.5 = 0.2. Since P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B), they are independent. Answer: Independent [4 marks]

  7. Working: (32)×(51)(83)=3×556=1556\frac{\binom{3}{2} \times \binom{5}{1}}{\binom{8}{3}} = \frac{3 \times 5}{56} = \frac{15}{56} Answer: 0.268 [4 marks]

  8. Working: P(X>4)=P(X=5)+P(X=6)=(65)(0.8)5(0.2)1+(66)(0.8)6(0.2)0=0.3932+0.2621=0.6553P(X > 4) = P(X=5) + P(X=6) = \binom{6}{5}(0.8)^5(0.2)^1 + \binom{6}{6}(0.8)^6(0.2)^0 = 0.3932 + 0.2621 = 0.6553 Answer: 0.655 [4 marks]

  9. Working: np=4np = 4 and np(1−p)=3np(1-p) = 3. Divide: (1−p)=3/4  ⟹  p=0.25(1-p) = 3/4 \implies p = 0.25. n(0.25)=4  ⟹  n=16n(0.25) = 4 \implies n = 16. Answer: n=16,p=0.25n=16, p=0.25 [4 marks]

  10. Working: P(H∪K)=P(H)+P(K)−P(H∩K)=1352+452−152=1652=413P(H \cup K) = P(H) + P(K) - P(H \cap K) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} Answer: 0.308 [3 marks]

Section B: Statistics and Linear Regression

  1. Working: Total of 5 = 5×12=605 \times 12 = 60. Total of 6 = 6×15=906 \times 15 = 90. 6th number = 90−60=3090 - 60 = 30. Answer: 30 [3 marks]

  2. Working: Mean xˉ=4+7+7+8+12+15+207=737≈10.4\bar{x} = \frac{4+7+7+8+12+15+20}{7} = \frac{73}{7} \approx 10.4. σ2=∑(x−xˉ)2n=(4−10.4)2+⋯+(20−10.4)27≈211.77≈30.2\sigma^2 = \frac{\sum(x-\bar{x})^2}{n} = \frac{(4-10.4)^2 + \dots + (20-10.4)^2}{7} \approx \frac{211.7}{7} \approx 30.2. σ≈5.5\sigma \approx 5.5. Answer: Mean = 10.4, SD = 5.5 [4 marks]

  3. Working: New Variance = k2×Old Variance=32×25=9×25=225k^2 \times \text{Old Variance} = 3^2 \times 25 = 9 \times 25 = 225. Answer: 225 [3 marks]

  4. Working: xˉ=120/10=12\bar{x} = 120/10 = 12. Var=∑x2n−xˉ2=300010−122=300−144=156\text{Var} = \frac{\sum x^2}{n} - \bar{x}^2 = \frac{3000}{10} - 12^2 = 300 - 144 = 156. σ=156≈12.5\sigma = \sqrt{156} \approx 12.5. Answer: 12.5 [4 marks]

  5. Working: New Mean = 50+5=5550 + 5 = 55. Standard deviation is invariant under translation. New SD = 10. Answer: Mean = 55, SD = 10 [3 marks]

  6. Working: xˉ=2+4+6+8+105=6\bar{x} = \frac{2+4+6+8+10}{5} = 6. yˉ=500+420+350+280+2105=17605=352\bar{y} = \frac{500+420+350+280+210}{5} = \frac{1760}{5} = 352. Answer: xˉ=6,yˉ=352\bar{x} = 6, \bar{y} = 352 [3 marks]

  7. Working: ∑(x−xˉ)2=16+4+0+4+16=40\sum(x-\bar{x})^2 = 16+4+0+4+16 = 40. ∑(x−xˉ)(y−yˉ)=(−4)(148)+(−2)(68)+(0)(−2)+(2)(−72)+(4)(−142)=−592−136+0−144−568=−1440\sum(x-\bar{x})(y-\bar{y}) = (-4)(148) + (-2)(68) + (0)(-2) + (2)(-72) + (4)(-142) = -592 - 136 + 0 - 144 - 568 = -1440. r=−144040×∑(y−yˉ)2r = \frac{-1440}{\sqrt{40 \times \sum(y-\bar{y})^2}}. ∑(y−yˉ)2=1482+682+(−2)2+(−72)2+(−142)2=21904+4624+4+5184+20164=51880\sum(y-\bar{y})^2 = 148^2 + 68^2 + (-2)^2 + (-72)^2 + (-142)^2 = 21904 + 4624 + 4 + 5184 + 20164 = 51880. r=−144040×51880=−14401444≈−0.997r = \frac{-1440}{\sqrt{40 \times 51880}} = \frac{-1440}{1444} \approx -0.997. Answer: -0.997 [5 marks]

  8. Working: b=∑(x−xˉ)(y−yˉ)∑(x−xˉ)2=−144040=−36b = \frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2} = \frac{-1440}{40} = -36. a=yˉ−bxˉ=352−(−36)(6)=352+216=568a = \bar{y} - b\bar{x} = 352 - (-36)(6) = 352 + 216 = 568. Answer: y=568−36xy = 568 - 36x [5 marks]

  9. Working: y=568−36(7)=568−252=316y = 568 - 36(7) = 568 - 252 = 316. Answer: 316 thousand dollars [3 marks]

  10. Working: r≈−0.997r \approx -0.997. Since ∣r∣|r| is very close to 1, it is a strong relationship. Since rr is negative, it is a negative correlation. Answer: Strong negative correlation [3 marks]