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O Level Additional Mathematics Statistics Probability Quiz

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Questions

O-Level Additional Mathematics Quiz - Statistics Probability

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 45

Duration: 60 Minutes
Total Marks: 45

Instructions:

Answer all questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
You are expected to use an approved scientific calculator where appropriate.
Show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Section A: Permutations and Combinations (Questions 1-5)

1. A committee of 4 people is to be chosen from a group of 6 men and 5 women. (a) Find the number of different committees that can be formed if there are no restrictions. [2]

(b) Find the number of different committees that can be formed if the committee must contain exactly 2 men and 2 women. [2]

2. Seven distinct books are to be arranged on a shelf. (a) Find the number of different arrangements if there are no restrictions. [1]

(b) Find the number of different arrangements if two particular books must always be together. [2]

3. How many different 4-digit numbers greater than 5000 can be formed using the digits 1, 2, 3, 4, 5, 6, 7 if: (a) Repetition of digits is allowed? [2]

(b) Repetition of digits is not allowed? [1]

4. In how many ways can the letters of the word "SINGAPORE" be arranged if: (a) There are no restrictions? [1]

(b) The vowels (I, A, O, E) must always be together? [2]

5. A class consists of 8 boys and 7 girls. A team of 5 students is to be selected. (a) Find the number of ways to select the team if there are no restrictions. [1]

(b) Find the number of ways to select the team if it must contain at least 3 boys. [3]

Section B: Probability Basics and Conditional Probability (Questions 6-10)

6. Events $A$ and $B$ are such that $P(A) = 0.4$ , $P(B) = 0.5$ , and $P(A \cap B) = 0.2$ . (a) Find $P(A \cup B)$ . [2]

(b) Find $P(A' \cap B)$ . [2]

7. A bag contains 5 red balls, 3 blue balls, and 2 green balls. Two balls are drawn from the bag one after the other without replacement. (a) Draw a tree diagram to represent the possible outcomes and their probabilities. [2]

(Space for rough work/diagram)

(b) Find the probability that both balls are red. [2]

8. In a certain school, 60% of the students study Additional Mathematics ( $A$ ) and 40% study Physics ( $P$ ). It is known that 20% of the students study both subjects. (a) Given that a student studies Additional Mathematics, find the probability that they also study Physics. [2]

(b) Given that a student does not study Physics, find the probability that they study Additional Mathematics. [2]

9. A biased coin is such that the probability of getting a Head is $0.6$ . The coin is tossed 3 times. (a) Find the probability of getting exactly 2 Heads. [2]

(b) Find the probability of getting at least one Tail. [2]

10. Two fair six-sided dice are thrown. Let $E$ be the event that the sum of the scores is 7, and $F$ be the event that the first die shows a 4. (a) Find $P(E)$ . [2]

(b) Find $P(F)$ . [1]

Section C: Discrete Random Variables (Questions 11-15)

11. The discrete random variable $X$ has the following probability distribution:

$x$	1	2	3	4
$P(X=x)$	$k$	$2k$	$3k$	$4k$

(a) Find the value of $k$ . [2]

(b) Find $E(X)$ , the expected value of $X$ . [2]

12. A fair six-sided die is thrown. Let the random variable $Y$ be the square of the score obtained. (a) Write down the probability distribution of $Y$ . [2]

(b) Calculate $E(Y)$ . [2]

13. The random variable $Z$ is defined by $Z = 3X - 2$ , where $X$ is the random variable defined in Question 11. (a) Find $E(Z)$ . [1]

(b) Find $\text{Var}(Z)$ . [1]

14. The probability distribution of a discrete random variable $W$ is given by $P(W=w) = \frac{w}{10}$ for $w = 1, 2, 3, 4$ . (a) Verify that this is a valid probability distribution. [1]

(b) Find $E(W)$ . [2]

15. A game involves spinning a spinner with sectors numbered 1, 2, 3, and 4. The probabilities of landing on 1, 2, 3, and 4 are $0.1, 0.2, 0.3,$ and $0.4$ respectively. Let $S$ be the score. (a) Find the expected score $E(S)$ . [2]

(b) If the payout is $\$ (S^2)$, find the expected payout. [3]

Section D: Binomial Distribution and Applications (Questions 16-20)

16. The random variable $X$ follows a binomial distribution $B(10, 0.3)$ . (a) Find $P(X = 4)$ . [2]

(b) Find $P(X \le 2)$ . [3]

17. In a large population, 15% of people are left-handed. A random sample of 8 people is selected. (a) State the distribution of the number of left-handed people in the sample. [1]

(b) Find the probability that exactly 2 people are left-handed. [2]

18. A multiple-choice test has 10 questions. Each question has 4 options, only one of which is correct. A student guesses the answer to every question. (a) Find the probability that the student gets exactly 3 questions correct. [2]

(b) Find the probability that the student gets more than 1 question correct. [3]

19. The probability that a machine produces a defective item is 0.05. Items are produced independently. (a) In a batch of 20 items, find the probability that exactly 1 item is defective. [2]

(b) In a batch of 20 items, find the probability that at most 2 items are defective. [3]

20. A fair coin is tossed 12 times. Let $H$ be the number of heads obtained. (a) Find $P(H = 6)$ . [2]

(b) Find $P(H \ge 10)$ . [3]

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Answers

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

1. Committee Selection (a) Total people = $6 + 5 = 11$ . Choose 4. $\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$ Answer: 330 [2]

(b) Choose 2 men from 6 and 2 women from 5. $\binom{6}{2} \times \binom{5}{2} = 15 \times 10 = 150$ Answer: 150 [2]

(c) At least 3 women means (3 women, 1 man) or (4 women, 0 men). Case 1: $\binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60$ Case 2: $\binom{5}{4} \times \binom{6}{0} = 5 \times 1 = 5$ Total = $60 + 5 = 65$ Answer: 65 [3]

2. Book Arrangements (a) 7 distinct books. $7! = 5040$ Answer: 5040 [1]

(b) Treat the 2 particular books as 1 unit. Now arranging 6 units. $6! \times 2! = 720 \times 2 = 1440$ Answer: 1440 [2]

3. 4-Digit Numbers > 5000 Digits available: $\{1, 2, 3, 4, 5, 6, 7\}$ . First digit must be 5, 6, or 7.

(a) Repetition allowed. 1st digit: 3 choices (5, 6, 7). 2nd, 3rd, 4th digits: 7 choices each. $3 \times 7 \times 7 \times 7 = 1029$ Answer: 1029 [2]

(b) Repetition not allowed. 1st digit: 3 choices. 2nd digit: 6 choices. 3rd digit: 5 choices. 4th digit: 4 choices. $3 \times 6 \times 5 \times 4 = 360$ Answer: 360 [1]

4. Word "SINGAPORE" Letters: S, I, N, G, A, P, O, R, E (9 distinct letters). Vowels: I, A, O, E (4). Consonants: S, N, G, P, R (5).

(a) No restrictions. $9! = 362,880$ Answer: 362,880 [1]

(b) Vowels together. Treat {IAOE} as 1 unit. Total units = 5 consonants + 1 vowel unit = 6 units. Arrange units: $6!$ . Arrange vowels within unit: $4!$ . $6! \times 4! = 720 \times 24 = 17,280$ Answer: 17,280 [2]

(c) Start with consonant, end with vowel. 1st pos: 5 choices (consonants). Last pos: 4 choices (vowels). Middle 7 positions: Arrange remaining 7 letters in $7!$ ways. $5 \times 4 \times 7! = 20 \times 5040 = 100,800$ Answer: 100,800 [2]

5. Team Selection (8 Boys, 7 Girls) (a) No restrictions. Choose 5 from 15. $\binom{15}{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{120} = 3003$ Answer: 3003 [1]

(b) At least 3 boys. Cases: 3B/2G, 4B/1G, 5B/0G. 3B/2G: $\binom{8}{3}\binom{7}{2} = 56 \times 21 = 1176$ 4B/1G: $\binom{8}{4}\binom{7}{1} = 70 \times 7 = 490$ 5B/0G: $\binom{8}{5}\binom{7}{0} = 56 \times 1 = 56$ Total = $1176 + 490 + 56 = 1722$ Answer: 1722 [3]

(c) Specific boy and girl included. Need to choose 3 more from remaining 13 students. $\binom{13}{3} = \frac{13 \times 12 \times 11}{6} = 286$ Answer: 286 [1]

6. Probability of Events A and B Given: $P(A)=0.4, P(B)=0.5, P(A \cap B)=0.2$ .

(a) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.2 = 0.7$ Answer: 0.7 [2]

(b) $P(A' \cap B) = P(B) - P(A \cap B) = 0.5 - 0.2 = 0.3$ Answer: 0.3 [2]

(c) Check independence: $P(A)P(B) = 0.4 \times 0.5 = 0.2$ . Since $P(A \cap B) = 0.2$ , they are independent. Answer: Yes, independent [2]

7. Balls in a Bag (5R, 3B, 2G) Total 10 balls. Without replacement.

(a) Tree Diagram: First branch R(0.5), B(0.3), G(0.2). Second branches adjust denominators to 9. [2]

(b) P(Red, Red) = $\frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9}$ Answer: $\frac{2}{9}$ [2]

(c) P(Different) = 1 - P(Same). P(R,R) = 20/90. P(B,B) = $\frac{3}{10} \times \frac{2}{9} = \frac{6}{90}$ . P(G,G) = $\frac{2}{10} \times \frac{1}{9} = \frac{2}{90}$ . P(Same) = $\frac{28}{90}$ . P(Different) = $1 - \frac{28}{90} = \frac{62}{90} = \frac{31}{45}$ Answer: $\frac{31}{45}$ [3]

8. Conditional Probability (School Subjects) $P(A)=0.6, P(P)=0.4, P(A \cap P)=0.2$ .

(a) $P(P | A) = \frac{P(A \cap P)}{P(A)} = \frac{0.2}{0.6} = \frac{1}{3}$ Answer: $\frac{1}{3}$ [2]

(b) $P(A | P') = \frac{P(A \cap P')}{P(P')}$ . $P(P') = 1 - 0.4 = 0.6$ . $P(A \cap P') = P(A) - P(A \cap P) = 0.6 - 0.2 = 0.4$ . $P(A | P') = \frac{0.4}{0.6} = \frac{2}{3}$ Answer: $\frac{2}{3}$ [2]

(c) Check independence of A and P'. $P(A)P(P') = 0.6 \times 0.6 = 0.36$ . $P(A \cap P') = 0.4$ . $0.36 \neq 0.4$ , so not independent. Answer: No, not independent [2]

9. Biased Coin (P(H)=0.6) 3 tosses.

(a) Exactly 2 Heads. Outcomes: HHT, HTH, THH. $P(HHT) = 0.6 \times 0.6 \times 0.4 = 0.144$ . Total = $3 \times 0.144 = 0.432$ . Answer: 0.432 [2]

(b) At least one Tail = 1 - P(No Tails) = 1 - P(HHH). $P(HHH) = 0.6^3 = 0.216$ . $1 - 0.216 = 0.784$ . Answer: 0.784 [2]

(c) P(1st H | Exactly 2 H). Let E = Exactly 2 H. Let F = 1st is H. Outcomes in E: {HHT, HTH, THH}. All equally likely? No, probabilities are same for each sequence ( $0.144$ ). F $\cap$ E = {HHT, HTH}. $P(F \cap E) = 0.144 + 0.144 = 0.288$ . $P(E) = 0.432$ . $P(F|E) = \frac{0.288}{0.432} = \frac{2}{3}$ . Answer: $\frac{2}{3}$ [2]

10. Two Dice Total outcomes = 36. E: Sum is 7. {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. 6 outcomes. F: First die is 4. {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)}. 6 outcomes.

(a) $P(E) = \frac{6}{36} = \frac{1}{6}$ Answer: $\frac{1}{6}$ [2]

(b) $P(F) = \frac{6}{36} = \frac{1}{6}$ Answer: $\frac{1}{6}$ [1]

(c) $E \cap F$ : Sum 7 AND First 4. Only (4,3). 1 outcome. $P(E \cap F) = \frac{1}{36}$ . $P(E | F) = \frac{P(E \cap F)}{P(F)} = \frac{1/36}{6/36} = \frac{1}{6}$ . Answer: $\frac{1}{6}$ [2]

11. Discrete Random Variable X Table: $x \in \{1,2,3,4\}$ , $P(x) \in \{k, 2k, 3k, 4k\}$ .

(a) $\sum P(x) = 10k = 1 \Rightarrow k = 0.1$ . Answer: $k = 0.1$ [2]

(b) $E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1+0.4+0.9+1.6 = 3$ . Answer: 3 [2]

(c) $E(X^2) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 0.1+0.8+2.7+6.4 = 10$ . $\text{Var}(X) = 10 - 3^2 = 1$ . Answer: 1 [3]

12. Die Square Variable Y $Y \in \{1, 4, 9, 16, 25, 36\}$ . $P(Y=y) = 1/6$ .

(a) Distribution table with values 1,4,9,16,25,36 each prob 1/6. [2]

(b) $E(Y) = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$ . Answer: $\frac{91}{6}$ [2]

(c) $E(Y^2) = \frac{1+16+81+256+625+1296}{6} = \frac{2275}{6}$ . $\text{Var}(Y) = \frac{2275}{6} - (\frac{91}{6})^2 = \frac{2275}{6} - \frac{8281}{36} = \frac{13650-8281}{36} = \frac{5369}{36}$ . Answer: $\frac{5369}{36}$ [2]

13. Linear Transformation Z = 3X - 2 Using Q11 results: $E(X)=3, \text{Var}(X)=1$ .

(a) $E(Z) = 3E(X) - 2 = 3(3) - 2 = 7$ . Answer: 7 [1]

(b) $\text{Var}(Z) = 3^2 \text{Var}(X) = 9(1) = 9$ . Answer: 9 [1]

(c) $Z > 4 \Rightarrow 3X - 2 > 4 \Rightarrow 3X > 6 \Rightarrow X > 2$ . $X \in \{3, 4\}$ . $P(X>2) = P(X=3) + P(X=4) = 3k + 4k = 7k = 0.7$ . Answer: 0.7 [2]

14. Random Variable W $P(W=w) = w/10$ for $w=1,2,3,4$ .

(a) Sum = $1/10 + 2/10 + 3/10 + 4/10 = 10/10 = 1$ . Valid. [1]

(b) $E(W) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1+0.4+0.9+1.6 = 3$ . Answer: 3 [2]

(c) $E(W^2) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 10$ . $\text{Var}(W) = 10 - 3^2 = 1$ . SD = $\sqrt{1} = 1$ . Answer: 1 [3]

15. Spinner Game $S \in \{1,2,3,4\}$ . $P(1)=0.1, P(2)=0.2, P(3)=0.3, P(4)=0.4$ .

(a) $E(S) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1+0.4+0.9+1.6 = 3$ . Answer: 3 [2]

(b) Payout $S^2$ . $E(S^2) = 1^2(0.1) + 2^2(0.2) + 3^2(0.3) + 4^2(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10$ . Expected payout is $\$ 10$. Answer: 10 [3]

16. Binomial Distribution B(10, 0.3) $n=10, p=0.3$ .

(a) $P(X=4) = \binom{10}{4} (0.3)^4 (0.7)^6$ . $\binom{10}{4} = 210$ . $210 \times 0.0081 \times 0.117649 \approx 0.2001$ . Answer: 0.200 [2]

(b) $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$ . $P(X=0) = (0.7)^{10} \approx 0.0282$ . $P(X=1) = 10(0.3)(0.7)^9 \approx 0.1211$ . $P(X=2) = 45(0.3)^2(0.7)^8 \approx 0.2335$ . Sum $\approx 0.0282 + 0.1211 + 0.2335 = 0.3828$ . Answer: 0.383 [3]

(c) Mean $E(X) = np = 10 \times 0.3 = 3$ . Variance $\text{Var}(X) = np(1-p) = 10 \times 0.3 \times 0.7 = 2.1$ . Answer: Mean = 3, Variance = 2.1 [2]

17. Left-Handed Sample $n=8, p=0.15$ . Let $L$ be number of left-handed people. $L \sim B(8, 0.15)$ .

(a) $B(8, 0.15)$ . Answer: $B(8, 0.15)$ [1]

(b) $P(L=2) = \binom{8}{2} (0.15)^2 (0.85)^6$ . $\binom{8}{2} = 28$ . $28 \times 0.0225 \times 0.3771 \approx 0.2376$ . Answer: 0.238 [2]

18. Multiple Choice Test $n=10, p=0.25$ (1 correct out of 4). Let $C$ be correct answers. $C \sim B(10, 0.25)$ .

(a) $P(C=3) = \binom{10}{3} (0.25)^3 (0.75)^7$ . $\binom{10}{3} = 120$ . $120 \times 0.015625 \times 0.1335 \approx 0.2503$ . Answer: 0.250 [2]

(b) $P(C > 1) = 1 - P(C \le 1) = 1 - [P(C=0) + P(C=1)]$ . $P(C=0) = (0.75)^{10} \approx 0.0563$ . $P(C=1) = 10(0.25)(0.75)^9 \approx 0.1877$ . $P(C \le 1) \approx 0.2440$ . $P(C > 1) = 1 - 0.2440 = 0.7560$ . Answer: 0.756 [3]

19. Defective Items $p=0.05$ .

(a) $n=20$ . $P(D=1) = \binom{20}{1} (0.05)^1 (0.95)^{19}$ . $20 \times 0.05 \times 0.3774 \approx 0.3774$ . Answer: 0.377 [2]

(b) $P(D \le 2) = P(0) + P(1) + P(2)$ . $P(0) = (0.95)^{20} \approx 0.3585$ . $P(1) \approx 0.3774$ . $P(2) = \binom{20}{2} (0.05)^2 (0.95)^{18} = 190 \times 0.0025 \times 0.3972 \approx 0.1887$ . Sum $= 0.3585 + 0.3774 + 0.1887 = 0.9246$ . Answer: 0.925 [3]

20. Coin Tosses $n=12, p=0.5$ . $H \sim B(12, 0.5)$ .

(a) $P(H=6) = \binom{12}{6} (0.5)^{12}$ . $\binom{12}{6} = 924$ . $924 \times (0.5)^{12} = \frac{924}{4096} \approx 0.2256$ . Answer: 0.226 [2]

(b) $P(H \ge 10) = P(10) + P(11) + P(12)$ . $P(10) = \binom{12}{10} (0.5)^{12} = 66 \times (0.5)^{12}$ . $P(11) = \binom{12}{11} (0.5)^{12} = 12 \times (0.5)^{12}$ . $P(12) = \binom{12}{12} (0.5)^{12} = 1 \times (0.5)^{12}$ . Sum $= (66+12+1)(0.5)^{12} = 79 \times (0.5)^{12} = \frac{79}{4096} \approx 0.0193$ . Answer: 0.0193 [3]

(c) $P(H=11 | H \ge 10) = \frac{P(H=11)}{P(H \ge 10)}$ . $P(H=11) = 12 \times (0.5)^{12}$ . $P(H \ge 10) = 79 \times (0.5)^{12}$ . Ratio $= \frac{12}{79} \approx 0.1519$ . Answer: $\frac{12}{79}$ or 0.152 [2]