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Secondary 4 Additional Mathematics Statistics Probability Quiz

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Questions

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Secondary 4 Additional Mathematics Quiz – Statistics Probability

Instructions: Answer all questions. Show your working clearly. Each question carries 5 marks.

Section A: Basic Probability (Questions 1–5)

  1. A bag contains 5 red balls, 3 blue balls, and 2 green balls. One ball is drawn at random. Find the probability that the ball drawn is
    (a) red,
    (b) not blue.

  2. A fair six-sided die is rolled once. Find the probability of obtaining
    (a) a prime number,
    (b) a number greater than 4.

  3. A card is drawn at random from a standard pack of 52 playing cards. Find the probability that the card is
    (a) a King,
    (b) a red card or a Queen.

  4. Two fair coins are tossed. List the sample space and find the probability of getting
    (a) exactly one head,
    (b) at least one head.

  5. The probability that it rains on a given day is 0.3. The probability that a student brings an umbrella on a rainy day is 0.8, and on a non-rainy day is 0.2. Find the probability that on a randomly chosen day, the student brings an umbrella.

Section B: Combined Events (Questions 6–10)

  1. Events A and B are such that P(A) = 0.5, P(B) = 0.4, and P(A ∩ B) = 0.2. Find
    (a) P(A ∪ B),
    (b) P(A' ∩ B).

  2. A and B are independent events with P(A) = 0.6 and P(B) = 0.5. Find
    (a) P(A ∩ B),
    (b) P(A ∪ B).

  3. In a class of 30 students, 18 study Mathematics, 15 study Physics, and 10 study both. A student is chosen at random. Find the probability that the student studies
    (a) Mathematics or Physics,
    (b) neither subject.

  4. Two events X and Y are mutually exclusive. Given that P(X) = 0.35 and P(Y) = 0.45, find
    (a) P(X ∪ Y),
    (b) P(X' ∩ Y').

  5. For two events C and D, P(C) = 0.7, P(D) = 0.6, and P(C ∪ D) = 0.9. Find
    (a) P(C ∩ D),
    (b) P(C | D).

Section C: Conditional Probability (Questions 11–15)

  1. A box contains 4 white and 6 black balls. Two balls are drawn without replacement. Find the probability that
    (a) both balls are white,
    (b) the second ball is black given that the first ball is white.

  2. In a survey, 60% of people like tea, 50% like coffee, and 30% like both. A person is selected at random. Find the probability that the person
    (a) likes tea given that they like coffee,
    (b) does not like coffee given that they like tea.

  3. Bag P contains 3 red and 5 blue marbles. Bag Q contains 4 red and 2 blue marbles. A bag is chosen at random, and then a marble is drawn from it. Find the probability that the marble drawn is red.

  4. The probability that a student passes Additional Mathematics is 0.75. The probability that a student passes Physics is 0.65. The probability that a student passes at least one of these subjects is 0.88. Find the probability that a student
    (a) passes both subjects,
    (b) passes Physics given that the student passes Additional Mathematics.

  5. Three machines A, B, and C produce 30%, 45%, and 25% of the items in a factory, respectively. The percentages of defective items produced are 2%, 3%, and 4% for A, B, and C respectively. An item is selected at random and found to be defective. Find the probability that it was produced by machine B.

Section D: Probability Distributions and Expectation (Questions 16–20)

  1. A discrete random variable X has the following probability distribution.

    x1234
    P(X=x)0.10.30.40.2

    Find
    (a) P(X > 2),
    (b) E(X).

  2. The random variable Y has probability distribution given by P(Y = y) = k(y + 1) for y = 0, 1, 2, 3.
    (a) Find the value of k.
    (b) Hence, find E(Y).

  3. A fair die is rolled twice. Let S be the sum of the two numbers obtained. Construct the probability distribution table for S, and find
    (a) P(S ≥ 10),
    (b) E(S).

  4. In a game, a player draws a card from a standard pack of 52 cards. If it is an Ace, the player wins 10;ifitisaKing,Queen,orJack,theplayerwins10; if it is a King, Queen, or Jack, the player wins 5; otherwise, the player loses $2. Let W be the player's net gain.
    (a) Construct the probability distribution table for W.
    (b) Calculate the expected net gain, E(W).

  5. A biased coin is tossed three times. The probability of getting a head on any toss is 0.4. Let H be the number of heads obtained.
    (a) Find the probability distribution of H.
    (b) Calculate E(H) and Var(H).

Answers

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Secondary 4 Additional Mathematics Quiz – Statistics Probability

Answer Key

Section A: Basic Probability (Questions 1–5)

  1. 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.

  2. 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.

  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.

  4. 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.

  5. 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)

  1. (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.

  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.

  3. 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.

  4. 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.

  5. (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)

  1. (a) P(both white) = (4/10) × (3/9) = 12/90 = 2/15.
    (b) P(2nd black | 1st white) = 6/9 = 2/3.

  2. 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.

  3. 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.

  4. 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.

  5. 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)

  1. (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.

  2. (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.

  3. 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).

  4. (a) P(Ace) = 4/52 = 1/13, net gain = 10.P(King,Queen,Jack)=12/52=3/13,netgain=10. P(King, Queen, Jack) = 12/52 = 3/13, net gain = 5.
    P(Other) = 36/52 = 9/13, net gain = –2.Distribution:W=10withprob1/13;W=5withprob3/13;W=–2withprob9/13.(b)E(W)=10(1/13)+5(3/13)+(–2)(9/13)=(10+15–18)/13=7/13≈2. Distribution: W = 10 with prob 1/13; W = 5 with prob 3/13; W = –2 with prob 9/13. (b) E(W) = 10(1/13) + 5(3/13) + (–2)(9/13) = (10 + 15 – 18)/13 = 7/13 ≈ 0.538.

  5. 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.