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Secondary 4 Additional Mathematics Statistics Probability Quiz

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Secondary 4 Additional Mathematics Quiz - Statistics Probability

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 60

Duration: 60 Minutes
Total Marks: 60

Instructions:

Answer all questions.
Show all necessary working clearly. No marks will be given for correct answers without working.
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.
The use of an approved graphing calculator is expected.

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

[15 Marks]

1. A committee of 4 people is to be selected 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. [1] (b) Find the number of different committees that can be formed if the committee must contain exactly 2 men and 2 women. [2]

2. How many distinct arrangements can be made using all the letters of the word STATISTICS? [3]

3. Seven students are to stand in a row for a photograph. (a) Find the number of different arrangements if two specific students, Alice and Bob, must stand next to each other. [2] (b) Find the number of different arrangements if Alice and Bob must not stand next to each other. [2]

4. A code consists of 3 distinct digits chosen from $\{1, 2, 3, 4, 5, 6\}$ followed by 2 distinct letters chosen from $\{A, B, C, D\}$ . (a) Find the total number of possible codes. [2] (b) Find the number of codes that start with an even digit. [1]

5. From a group of 10 students, a President, a Vice-President, and a Secretary are to be chosen. (a) Find the number of ways these positions can be filled. [1] (b) If two specific students refuse to work together in any capacity, find the number of valid ways to fill the positions. [1]

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

[15 Marks]

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 $P(A \cup B)$ . [1] (b) Determine whether events $A$ and $B$ are independent, showing your reasoning. [2]

7. A bag contains 5 red balls, 3 blue balls, and 2 green balls. Two balls are drawn at random without replacement. (a) Draw a tree diagram to represent the possible outcomes. [1] (b) Find the probability that both balls are the same color. [2]

8. In a certain school, 60% of the students study Additional Mathematics (Amath) and 40% do not. Among those who study Amath, 80% pass the subject. Among those who do not study Amath, 10% pass a different elective math subject. (a) Find the probability that a randomly selected student studies Amath and passes. [1] (b) Given that a student passed their math elective, find the probability that they studied Amath. [3]

9. The probability that it rains on any given day in April is 0.3. Assume that the weather on consecutive days is independent. (a) Find the probability that it rains on exactly 2 out of 3 consecutive days. [2] (b) Find the probability that it rains on at least one of the 3 days. [2]

10. Events $X$ and $Y$ are mutually exclusive. Given that $P(X) = 0.3$ and $P(Y) = 0.4$ : (a) Find $P(X \cup Y)$ . [1] (b) Find $P(X | Y)$ . [1]

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

[15 Marks]

11. A 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 $k$ . [1] (b) Calculate $E(X)$ , the expected value of $X$ . [2]

12. The random variable $Y$ represents the score obtained when throwing a biased die. The probability of scoring a 6 is twice the probability of scoring any other number (1 to 5), which are equally likely. (a) Find $P(Y=6)$ . [1] (b) Find the variance of $Y$ , denoted as $\text{Var}(Y)$ . [3]

13. A game costs $\$ 5 $to play. A player rolls a fair die. If the score is 6, the player wins$ $20 $. If the score is 4 or 5, the player wins$ $8 $. Otherwise, the player wins nothing. (a) Construct the probability distribution table for the net profit$ W$. [2] (b) Calculate the expected net profit per game. Is the game fair? Explain. [2]

14. Let $X$ be a random variable with $E(X) = 4$ and $\text{Var}(X) = 9$ . Find the mean and variance of the random variable $Y$ defined by: (a) $Y = 2X + 3$ [2] (b) $Y = X^2$ (Hint: Use $\text{Var}(X) = E(X^2) - [E(X)]^2$ to find $E(X^2)$ first, then note this part asks for mean/var of transformed variable? No, standard question: Find $E(3X - 1)$ and $\text{Var}(3X - 1)$ ). Correction for clarity: Find $E(3X - 1)$ and $\text{Var}(3X - 1)$ . [2]

15. Two fair dice are thrown. Let $S$ be the sum of the scores. (a) Find $P(S = 7)$ . [1] (b) Find $P(S > 10)$ . [1]

Section D: Binomial Distribution (Questions 16–20)

[15 Marks]

16. A fair coin is tossed 10 times. Let $X$ be the number of heads obtained. (a) State the distribution of $X$ , including parameters. [1] (b) Find $P(X = 6)$ . [2]

17. In a factory, 5% of the light bulbs produced are defective. A random sample of 20 bulbs is selected. (a) Find the probability that exactly 2 bulbs are defective. [2] (b) Find the probability that at least 1 bulb is defective. [2]

18. A multiple-choice quiz has 8 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 most likely number of correct answers (the mode). [1]

19. The probability that a basketball player scores a free throw is 0.8. He takes 5 free throws. (a) Find the probability that he scores at least 4 free throws. [2] (b) Find the expected number of successful free throws. [1]

20. A random variable $X \sim B(n, p)$ . Given that $E(X) = 6$ and $\text{Var}(X) = 2.4$ . (a) Find the values of $n$ and $p$ . [3]

*** End of Quiz ***

Answers

Secondary 4 Additional Mathematics Quiz - Statistics Probability (Answer Key)

Total Marks: 60

Section A: Permutations and Combinations

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

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

2. Word: STATISTICS (10 letters). Counts: S=3, T=3, A=1, I=2, C=1. $\frac{10!}{3! \, 3! \, 2! \, 1! \, 1!} = \frac{3,628,800}{6 \times 6 \times 2} = \frac{3,628,800}{72} = 50,400$ Answer: 50,400 [3]

3. (a) Treat Alice and Bob as 1 unit. Total entities = 6 (AB, S3, S4, S5, S6, S7). Arrangements of entities = $6!$ . Internal arrangement of AB = $2!$ . Total = $6! \times 2! = 720 \times 2 = 1440$ . Answer: 1440 [2]

(b) Total arrangements without restriction = $7! = 5040$ . Subtract arrangements where they are together (from part a). $5040 - 1440 = 3600$ . Answer: 3600 [2]

4. (a) Digits: $P(6,3) = 6 \times 5 \times 4 = 120$ . Letters: $P(4,2) = 4 \times 3 = 12$ . Total codes = $120 \times 12 = 1440$ . Answer: 1440 [2]

(b) Even digits from $\{1..6\}$ are $\{2,4,6\}$ (3 choices). First digit: 3 choices. Second digit: 5 remaining choices. Third digit: 4 remaining choices. Digits part = $3 \times 5 \times 4 = 60$ . Letters part = 12. Total = $60 \times 12 = 720$ . Answer: 720 [1]

5. (a) Permutation of 3 from 10. $P(10,3) = 10 \times 9 \times 8 = 720$ Answer: 720 [1]

(b) Total ways = 720. Ways where specific students (A and B) work together: Case 1: A and B are selected. The third person is chosen from remaining 8. Positions for A and B: $P(3,2) = 6$ ways to place them. Third person: 8 choices. Position for third person: 1 remaining spot. Actually, simpler: Select 3 people including A and B. Choose 3rd person: $\binom{8}{1} = 8$ . Arrange A, B, C in 3 positions: $3! = 6$ . Invalid ways = $8 \times 6 = 48$ . Valid ways = $720 - 48 = 672$ . Answer: 672 [1]

Section B: Probability Basics

6. (a) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $= 0.4 + 0.5 - 0.2 = 0.7$ Answer: 0.7 [1]

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

7. Total balls = 10. (a) Tree diagram branches: R (5/10), B (3/10), G (2/10). Second draw depends on first. [1]

(b) P(Same Color) = P(RR) + P(BB) + P(GG) $P(RR) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90}$ $P(BB) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90}$ $P(GG) = \frac{2}{10} \times \frac{1}{9} = \frac{2}{90}$ $\text{Total} = \frac{20+6+2}{90} = \frac{28}{90} = \frac{14}{45}$ Answer: $\frac{14}{45}$ (or 0.311) [2]

8. Let $A$ = Study Amath, $A'$ = Not Amath. $P$ = Pass. $P(A) = 0.6, P(A') = 0.4$ . $P(P|A) = 0.8, P(P|A') = 0.1$ .

(a) $P(A \cap P) = P(A) \times P(P|A) = 0.6 \times 0.8 = 0.48$ . Answer: 0.48 [1]

(b) $P(A|P) = \frac{P(A \cap P)}{P(P)}$ . $P(P) = P(A \cap P) + P(A' \cap P) = 0.48 + (0.4 \times 0.1) = 0.48 + 0.04 = 0.52$ . $P(A|P) = \frac{0.48}{0.52} = \frac{48}{52} = \frac{12}{13}$ Answer: $\frac{12}{13}$ (or 0.923) [3]

9. $P(\text{Rain}) = 0.3, P(\text{No Rain}) = 0.7$ .

(a) Exactly 2 rains in 3 days. Patterns: RRN, RNR, NRR. $3 \times (0.3)^2 \times (0.7)^1 = 3 \times 0.09 \times 0.7 = 0.189$ Answer: 0.189 [2]

(b) At least one rain = $1 - P(\text{No Rain in 3 days})$ . $P(\text{None}) = (0.7)^3 = 0.343$ $1 - 0.343 = 0.657$ Answer: 0.657 [2]

10. Mutually exclusive means $P(X \cap Y) = 0$ .

(a) $P(X \cup Y) = P(X) + P(Y) = 0.3 + 0.4 = 0.7$ . Answer: 0.7 [1]

(b) $P(X|Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.4} = 0$ . Answer: 0 [1]

Section C: Discrete Random Variables

11. (a) Sum of probabilities = 1. $k + 2k + 3k + 4k = 10k = 1 \Rightarrow k = 0.1$ Answer: $k = 0.1$ [1]

(b) $E(X) = \sum x P(X=x)$ $= 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4)$ $= 0.1 + 0.4 + 0.9 + 1.6 = 3.0$ Answer: 3 [2]

12. Let $P(Y=1) = ... = P(Y=5) = p$ . Then $P(Y=6) = 2p$ . Sum = $5p + 2p = 7p = 1 \Rightarrow p = 1/7$ . $P(Y=6) = 2/7$ .

(a) Answer: $2/7$ [1]

(b) $E(Y) = 1(\frac{1}{7}) + 2(\frac{1}{7}) + 3(\frac{1}{7}) + 4(\frac{1}{7}) + 5(\frac{1}{7}) + 6(\frac{2}{7})$ $= \frac{1+2+3+4+5+12}{7} = \frac{27}{7}$ $E(Y^2) = 1^2(\frac{1}{7}) + ... + 5^2(\frac{1}{7}) + 6^2(\frac{2}{7})$ $= \frac{1+4+9+16+25+72}{7} = \frac{127}{7}$ $\text{Var}(Y) = E(Y^2) - [E(Y)]^2 = \frac{127}{7} - (\frac{27}{7})^2$ $= \frac{127}{7} - \frac{729}{49} = \frac{889 - 729}{49} = \frac{160}{49} \approx 3.27$ Answer: $\frac{160}{49}$ or 3.27 [3]

13. Cost = $5. Outcomes:

Score 6 (Prob 1/6): Win $20. Net Profit $W = 20 - 5 = 15$ .
Score 4,5 (Prob 2/6): Win $8. Net Profit $W = 8 - 5 = 3$ .
Score 1,2,3 (Prob 3/6): Win $0. Net Profit $W = 0 - 5 = -5$ .

(a) Table:

$w$	15	3	-5
$P(W=w)$	$1/6$	$1/3$	$1/2$
[2]

(b) $E(W) = 15(\frac{1}{6}) + 3(\frac{2}{6}) - 5(\frac{3}{6})$ $= \frac{15 + 6 - 15}{6} = \frac{6}{6} = 1$ Expected profit is $1. Since $E(W) \neq 0$ , the game is not fair (it favors the player). Answer: $1, Not Fair [2]

14. Given $E(X)=4, \text{Var}(X)=9$ .

(a) $Y = 2X + 3$ . $E(Y) = 2E(X) + 3 = 2(4) + 3 = 11$ . $\text{Var}(Y) = 2^2 \text{Var}(X) = 4(9) = 36$ . Answer: Mean 11, Variance 36 [2]

(b) Question corrected to $Y = 3X - 1$ . $E(Y) = 3E(X) - 1 = 3(4) - 1 = 11$ . $\text{Var}(Y) = 3^2 \text{Var}(X) = 9(9) = 81$ . Answer: Mean 11, Variance 81 [2]

15. Total outcomes = 36.

(a) Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). 6 outcomes. $P(S=7) = 6/36 = 1/6$ . Answer: $1/6$ [1]

(b) Sum > 10: Sum can be 11 or 12. Sum 11: (5,6), (6,5). Sum 12: (6,6). Total 3 outcomes. $P(S>10) = 3/36 = 1/12$ . Answer: $1/12$ [1]

Section D: Binomial Distribution

16. (a) $X \sim B(10, 0.5)$ . [1]

(b) $P(X=6) = \binom{10}{6} (0.5)^6 (0.5)^4 = \binom{10}{6} (0.5)^{10}$ . $\binom{10}{6} = 210$ . $210 \times \frac{1}{1024} = \frac{210}{1024} \approx 0.205$ . Answer: 0.205 [2]

17. $X \sim B(20, 0.05)$ .

(a) $P(X=2) = \binom{20}{2} (0.05)^2 (0.95)^{18}$ . $\binom{20}{2} = 190$ . $190 \times 0.0025 \times 0.3972 \approx 0.1887$ . Answer: 0.189 [2]

(b) $P(X \ge 1) = 1 - P(X=0)$ . $P(X=0) = (0.95)^{20} \approx 0.3585$ . $1 - 0.3585 = 0.6415$ . Answer: 0.642 [2]

18. $X \sim B(8, 0.25)$ .

(a) $P(X=3) = \binom{8}{3} (0.25)^3 (0.75)^5$ . $\binom{8}{3} = 56$ . $56 \times 0.015625 \times 0.2373 \approx 0.2076$ . Answer: 0.208 [2]

(b) Mode is usually floor $((n+1)p)$ . $(8+1)(0.25) = 2.25$ . Floor is 2. Check $P(X=2)$ vs $P(X=3)$ . $P(X=2) = 28(0.25)^2(0.75)^6 \approx 0.311$ . $P(X=3) \approx 0.208$ . Mode is 2. Answer: 2 [1]

19. $X \sim B(5, 0.8)$ .

(a) $P(X \ge 4) = P(X=4) + P(X=5)$ . $P(X=4) = \binom{5}{4}(0.8)^4(0.2)^1 = 5 \times 0.4096 \times 0.2 = 0.4096$ . $P(X=5) = \binom{5}{5}(0.8)^5(0.2)^0 = 1 \times 0.32768 = 0.32768$ . Sum = $0.4096 + 0.32768 = 0.73728$ . Answer: 0.737 [2]

(b) $E(X) = np = 5 \times 0.8 = 4$ . Answer: 4 [1]

20. $E(X) = np = 6$ . $\text{Var}(X) = npq = 2.4$ .

Divide Var by Mean: $\frac{npq}{np} = \frac{2.4}{6} \Rightarrow q = 0.4$ . Since $q = 1-p$ , $p = 1 - 0.4 = 0.6$ . Substitute $p$ into Mean eq: $n(0.6) = 6 \Rightarrow n = 10$ .

Answer: $n=10, p=0.6$ [3]