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Secondary 4 Additional Mathematics Statistics Probability Quiz
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Secondary 4 Additional Mathematics Quiz – Statistics Probability
Answer Key
Section A: Basic Probability (Questions 1–5)
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Total balls = 5 + 3 + 2 = 10.
(a) P(red) = 5/10 = 1/2.
(b) P(not blue) = 1 – P(blue) = 1 – 3/10 = 7/10. -
Sample space = {1,2,3,4,5,6}.
(a) Prime numbers: 2,3,5 → P(prime) = 3/6 = 1/2.
(b) Numbers > 4: 5,6 → P(>4) = 2/6 = 1/3. -
(a) P(King) = 4/52 = 1/13.
(b) P(red or Queen) = P(red) + P(Queen) – P(red Queen) = 26/52 + 4/52 – 2/52 = 28/52 = 7/13. -
Sample space = {HH, HT, TH, TT}.
(a) Exactly one head: {HT, TH} → P = 2/4 = 1/2.
(b) At least one head: {HH, HT, TH} → P = 3/4. -
Let R = rain, U = umbrella.
P(U) = P(U|R)P(R) + P(U|R')P(R') = (0.8)(0.3) + (0.2)(0.7) = 0.24 + 0.14 = 0.38.
Section B: Combined Events (Questions 6–10)
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(a) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.5 + 0.4 – 0.2 = 0.7.
(b) P(A' ∩ B) = P(B) – P(A ∩ B) = 0.4 – 0.2 = 0.2. -
(a) Since independent, P(A ∩ B) = P(A)P(B) = 0.6 × 0.5 = 0.3.
(b) P(A ∪ B) = 0.6 + 0.5 – 0.3 = 0.8. -
Let M = Math, P = Physics.
(a) P(M ∪ P) = P(M) + P(P) – P(M ∩ P) = 18/30 + 15/30 – 10/30 = 23/30.
(b) P(neither) = 1 – P(M ∪ P) = 1 – 23/30 = 7/30. -
Mutually exclusive → P(X ∩ Y) = 0.
(a) P(X ∪ Y) = 0.35 + 0.45 = 0.80.
(b) P(X' ∩ Y') = 1 – P(X ∪ Y) = 1 – 0.80 = 0.20. -
(a) P(C ∩ D) = P(C) + P(D) – P(C ∪ D) = 0.7 + 0.6 – 0.9 = 0.4.
(b) P(C | D) = P(C ∩ D) / P(D) = 0.4 / 0.6 = 2/3.
Section C: Conditional Probability (Questions 11–15)
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(a) P(both white) = (4/10) × (3/9) = 12/90 = 2/15.
(b) P(2nd black | 1st white) = 6/9 = 2/3. -
Let T = tea, C = coffee.
(a) P(T | C) = P(T ∩ C) / P(C) = 0.30 / 0.50 = 3/5 = 0.6.
(b) P(C' | T) = P(C' ∩ T) / P(T). P(C' ∩ T) = P(T) – P(T ∩ C) = 0.6 – 0.3 = 0.3. So P = 0.3 / 0.6 = 1/2 = 0.5. -
P(red) = P(P) × P(red|P) + P(Q) × P(red|Q) = (1/2)(3/8) + (1/2)(4/6) = 3/16 + 4/12 = 3/16 + 1/3 = (9 + 16)/48 = 25/48.
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Let A = pass AMath, P = pass Physics.
P(A ∪ P) = 0.88, P(A) = 0.75, P(P) = 0.65.
(a) P(A ∩ P) = P(A) + P(P) – P(A ∪ P) = 0.75 + 0.65 – 0.88 = 0.52.
(b) P(P | A) = P(A ∩ P) / P(A) = 0.52 / 0.75 = 52/75. -
Let D = defective.
P(B|D) = [P(D|B)P(B)] / [P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)]
= (0.03 × 0.45) / (0.02×0.30 + 0.03×0.45 + 0.04×0.25)
= 0.0135 / (0.006 + 0.0135 + 0.01) = 0.0135 / 0.0295 = 135/295 = 27/59.
Section D: Probability Distributions and Expectation (Questions 16–20)
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(a) P(X > 2) = P(X=3) + P(X=4) = 0.4 + 0.2 = 0.6.
(b) E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7. -
(a) Σ P(Y=y) = k(0+1) + k(1+1) + k(2+1) + k(3+1) = k(1+2+3+4) = 10k = 1 → k = 0.1.
(b) E(Y) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.4) = 0 + 0.2 + 0.6 + 1.2 = 2.0. -
Sum S ranges from 2 to 12. Distribution:
S=2:1/36, S=3:2/36, S=4:3/36, S=5:4/36, S=6:5/36, S=7:6/36, S=8:5/36, S=9:4/36, S=10:3/36, S=11:2/36, S=12:1/36.
(a) P(S ≥ 10) = P(10)+P(11)+P(12) = 3/36 + 2/36 + 1/36 = 6/36 = 1/6.
(b) E(S) = Σ s·P(S=s) = 7 (by symmetry of two dice). -
(a) P(Ace) = 4/52 = 1/13, net gain = 5.
P(Other) = 36/52 = 9/13, net gain = –0.538. -
H ~ Binomial(n=3, p=0.4).
(a) P(H=0) = (0.6)^3 = 0.216; P(H=1) = 3C1(0.4)(0.6)^2 = 3×0.4×0.36 = 0.432;
P(H=2) = 3C2(0.4)^2(0.6) = 3×0.16×0.6 = 0.288; P(H=3) = (0.4)^3 = 0.064.
(b) E(H) = np = 3×0.4 = 1.2.
Var(H) = np(1–p) = 3×0.4×0.6 = 0.72.