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S5 Mathematics Practice Paper: Probability, Permutations and Statistics (12 questions, 100 marks, 75 minutes)
S5 Mathematics: Probability, Permutations and Statistics
S5 maths practice paper: probability without replacement, dice, combinations, conditional probability, mean and standard deviation, normal z-scores, expectation,
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Paper details Level: S5 | Subject: Mathematics | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Short Questions (25 marks)
Answer all questions and show your working.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A bag contains 5 red balls and 3 white balls. Two balls are drawn at random without replacement. Find the probability that both balls are red, giving the answer as a fraction in lowest terms and as a decimal. | P(first red) = 5/8. After a red ball is removed the bag holds 4 red and 3 white, so P(second red) = 4/7. Probability = 5/8 x 4/7 = 20/56 = 5/14 = 0.357. Marks: 2 + 2 + 2. |
| 2 | Two fair dice are thrown. Find the probability that the sum of the scores is 8. | The pairs summing to 8 are (2,6), (3,5), (4,4), (5,3), (6,2): 5 outcomes out of 36. Probability = 5/36 = 0.1389. Marks: list 3, sample space 2, answer 1. |
| 3 | Three students are chosen from a group of 6 to form a committee. (a) How many different committees are possible? (b) How many of them include a particular student, Ming? | (a) C(6,3) = 20 committees. (b) With Ming fixed, choose 2 from the remaining 5: C(5,2) = 10. Marks: (a) 4; (b) 3. |
| 4 | In a class of 40 students, 22 are boys and 18 are girls. 20 students wear glasses, 12 of them boys. A student who wears glasses is chosen at random. Find the probability that this student is a boy. | Glasses wearers total 20, of whom 12 are boys, so the conditional probability is 12/20 = 0.6. Marks: numerator 3, denominator 2, answer 1. |
Section B: Statistics and Expectation (30 marks)
State the formula used and give answers to the accuracy required.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | The scores of 10 students in a test are 4, 6, 7, 8, 8, 9, 9, 9, 10, 10. Find the mean, the median, the mode and the range. | Mean = 80/10 = 8.0; median = (8+9)/2 = 8.5; mode = 9 (three times); range = 10 – 4 = 6. Marks: 2 each. |
| 6 | A data set is 2, 4, 4, 4, 5, 5, 7, 9. Find its mean and its population standard deviation (3 significant figures). | Mean = 40/8 = 5; sum of squared deviations = 9+1+1+1+0+0+4+16 = 32.0; variance = 32.0/8 = 4; population standard deviation = 2.00. Marks: mean 2, deviations 3, variance 2, answer 1. |
| 7 | The scores of a test are normally distributed with mean 65 and standard deviation 10. (a) Find the z-score of a score of 80. (b) Using the normal table, estimate the percentage of students scoring above 80. | (a) z = (80 – 65)/10 = 1.5. (b) The table gives P(Z below 1.5) = 0.9332, so the proportion above 80 is about 1 – 0.9332 = 0.0668, that is 6.7%. Marks: (a) 4; (b) 4. |
| 8 | A raffle sells 1,000 tickets at 5 dollars each. One ticket wins 500 dollars, 10 tickets win 50 dollars each, and the rest win nothing. Find the expected prize per ticket and the expected net gain per ticket. | Expected prize = (1 x 500 + 10 x 50)/1000 = 1,000/1,000 = 1 dollar; expected net gain = 1 – 5 = -4 dollars, a loss of 4 dollars per ticket. Marks: expected prize 4, net gain 3. |
Section C: Long Questions (45 marks)
Answer in full steps or with a table.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A basketball player scores free throws with probability 0.7 and each attempt is independent (binomial model). He takes 10 attempts. (a) Find the probability of exactly 8 successes. (b) Find the probability of at least 9 successes. | (a) C(10,8) x 0.7^8 x 0.3^2 = 45 x 0.05764801 x 0.09 = 0.2335. (b) P(at least 9) = C(10,9) x 0.7^9 x 0.3 + 0.7^10 = 0.1211 + 0.0282 = 0.1493. Marks: (a) 6; (b) 6. |
| 10 | Events A and B are independent with P(A) = 0.5 and P(B) = 0.4. (a) Find P(A and B). (b) Find P(A or B). (c) Find P(A given B). (d) Explain why A and B cannot also be mutually exclusive. | (a) Independence gives P(A and B) = 0.5 x 0.4 = 0.2. (b) P(A or B) = 0.5 + 0.4 – 0.2 = 0.7. (c) P(A given B) = 0.2/0.4 = 0.5, the same as P(A), which confirms independence. (d) Mutually exclusive means P(A and B) = 0, but independence requires P(A and B) = 0.2; both cannot hold unless one event has probability zero. Marks: (a) 2; (b) 3; (c) 3; (d) 3. |
| 11 | Three letters are chosen from A, B, C, D, E and arranged in a row without repetition. (a) How many different strings are possible? (b) How many of them start with A? (c) How many combinations of 3 letters are possible if order is ignored? | (a) P(5,3) = 5 x 4 x 3 = 60 strings. (b) With A first, arrange 2 of the remaining 4 letters: 4 x 3 = 12. (c) C(5,3) = 10 combinations. Marks: (a) 4; (b) 4; (c) 4. |
| 12 | A school has 1,200 students: 300 in S4, 420 in S5 and 480 in S6. A stratified sample of 120 students is taken. (a) Find the number sampled from each level. (b) Give one reason why stratified sampling is better than simple random sampling here. | (a) Sampling fraction = 120/1200 = 0.1, so take 30 from S4, 42 from S5 and 48 from S6, totalling 120. (b) Stratified sampling keeps the level structure of the sample the same as the school, so no level is over- or under-represented and the estimate is more precise. Marks: (a) 6; (b) 4. |
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