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HKDSE Mathematics Practice Paper: Probability and Statistics
HKDSE Mathematics: Probability and Statistics
HKDSE Mathematics practice paper: 12 original questions, 100 marks, 75 minutes on counting, binomial and Poisson distributions, normal distribution and regression.
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Paper details Level: S6 | Subject: Mathematics (Compulsory) | Time: 75 minutes | Marks: 100 | Questions: 12
Section A (Q1–Q4): Counting, Binomial and Normal Distribution (32 marks)
Each question carries 8 marks. Show your working.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A bag contains 6 red, 4 blue and 5 green balls, 15 in total. Three balls are drawn at random without replacement. (a) Find the number of possible combinations of 3 balls. (2 marks) (b) Find the probability that the three balls are of the same colour. (3 marks) (c) Find the probability that exactly two of the balls are red. (3 marks) |
(a) Number of combinations = C(15,3) = 455. [2 marks] (b) Same colour: C(6,3) + C(4,3) + C(5,3) = 20 + 4 + 10 = 34; probability = 34/455 = 0.0747. [3 marks] (c) Exactly two red: C(6,2) x C(9,1) = 15 x 9 = 135; probability = 135/455 = 0.2967. [3 marks] |
| 2 | A game has probability of success 0.3 on each independent trial and is played 10 times. Let X be the number of successes. (a) Find P(X = 3). (3 marks) (b) Find the mean and standard deviation of X (4 decimal places). (2 marks) (c) Find the probability of at least one success. (3 marks) |
(a) P(X = 3) = C(10,3)(0.3)^3(0.7)^7 = 120 x 0.027 x 0.0824 = 0.2668. [3 marks] (b) Mean = np = 3; variance = np(1 – p) = 2.1; standard deviation = 1.4491. [2 marks] (c) P(X >= 1) = 1 – 0.7^10 = 1 – 0.0282 = 0.9718. [3 marks] |
| 3 | Test scores in a subject are normally distributed with mean 52 and standard deviation 8. (a) Find the probability that a score exceeds 60. (3 marks) (b) Find the probability that a score lies between 44 and 60. (2 marks) (c) If only the top 10% gain a distinction, find the minimum distinction score (1 decimal place). (3 marks) |
(a) z = (60 – 52)/8 = 1; P(Z > 1) = 0.1587. [3 marks] (b) P(44 < X < 60) = P(-1 < Z < 1) = 0.6827. [2 marks] (c) For the top 10%, z = 1.2816; minimum score = 52 + 1.2816 x 8 = 62.3. [3 marks] |
| 4 | Factory A makes 60% of the output and factory B the remaining 40%. The defect rates are 2% for A and 5% for B. (a) Find the probability that a randomly chosen item is defective. (3 marks) (b) Given a defective item, find the probability that it came from A. (3 marks) (c) Three items are chosen independently. Find the probability that at least one is defective. (2 marks) |
(a) P(defective) = 0.6 x 0.02 + 0.4 x 0.05 = 0.012 + 0.020 = 0.032. [3 marks] (b) P(A | defective) = 0.012/0.032 = 0.375. [3 marks] (c) 1 – 0.968^3 = 1 – 0.9070 = 0.0929. [2 marks] |
Section B (Q5–Q8): Descriptive Statistics, Regression and Independence (36 marks)
Each question carries 9 marks. Show calculations and explanations.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | The test scores of 8 students are: [12, 15, 15, 18, 20, 22, 25, 33]. (a) Find the mean, median and mode. (3 marks) (b) Find the population standard deviation (3 decimal places). (3 marks) (c) A ninth student scoring 60 joins the class. Find the new mean and median. (3 marks) |
(a) Mean = 160/8 = 20; median = (18 + 20)/2 = 19; mode = 15. [3 marks] (b) Sum of squared deviations = 316; variance = 316/8 = 39.5; standard deviation = 6.285. [3 marks] (c) New mean = 220/9 = 24.44; new median = 20 (the fifth value). [3 marks] |
| 6 | Advertising spend x (thousand dollars) and sales y (ten thousand dollars) for 5 branches are: x = [1, 2, 3, 4, 5], y = [2, 4, 5, 8, 11]. (a) Find the regression line of y on x (slope and intercept to 1 decimal place). (4 marks) (b) Estimate sales when advertising spend is 7 thousand dollars. (2 marks) (c) Find the correlation coefficient r and comment on the reliability of predicting at x = 7. (3 marks) |
(a) xbar = 3 and ybar = 6; Sxy = 22 and Sxx = 10; slope = 22/10 = 2.2, intercept = 6 – 2.2 x 3 = -0.6; so y = -0.6 + 2.2x. [4 marks] (b) At x = 7, y = -0.6 + 2.2 x 7 = 14.8. [2 marks] (c) r = 22/sqrt(10 x 50) = 0.984, a very strong positive correlation; but x = 7 is outside the data range 1 to 5, so the prediction is an extrapolation and may be unreliable. [3 marks] |
| 7 | A committee of 4 is chosen from 6 boys and 5 girls. (a) Find the number of possible committees. (2 marks) (b) Find the probability that exactly 2 members are girls (4 decimal places). (3 marks) (c) Find the probability that at least one member is a girl (4 decimal places). (2 marks) (d) If one girl must chair the committee and the other 3 members are unrestricted, find the number of committees. (2 marks) |
(a) C(11,4) = 330. [2 marks] (b) C(5,2) x C(6,2) = 10 x 15 = 150; probability = 150/330 = 0.4545. [3 marks] (c) All-boy committees: C(6,4) = 15; P(at least one girl) = 1 – 15/330 = 0.9545. [2 marks] (d) 5 choices for the chair, then 3 members from the remaining 10: 5 x C(10,3) = 5 x 120 = 600. [2 marks] |
| 8 | Of 200 students, 120 take Mathematics, 90 take Physics and 60 take both. One student is chosen at random. (a) Find the probability that the student takes Mathematics or Physics. (2 marks) (b) Given that the student takes Mathematics, find the probability of taking Physics. (3 marks) (c) Test whether the two choices are independent, with calculation. (2 marks) (d) Find the number of students taking exactly one of the two subjects. (2 marks) |
(a) (120 + 90 – 60)/200 = 0.75. [2 marks] (b) 60/120 = 0.5. [3 marks] (c) P(M) x P(P) = 0.6 x 0.45 = 0.27 while P(both) = 0.3; since 0.27 is not 0.3, the two choices are not independent. [2 marks] (d) (120 – 60) + (90 – 60) = 60 + 30 = 90 students. [2 marks] |
Section C (Q9–Q12): Poisson, Expectation and Sampling (32 marks)
Each question carries 8 marks. Give answers to a suitable number of decimal places.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A helpline receives on average 2.5 calls per hour and the hourly number of calls follows a Poisson distribution. (a) Find the probability of no calls in an hour (4 decimal places). (2 marks) (b) Find the probability of at least 3 calls in an hour (4 decimal places). (3 marks) (c) State the mean, variance and standard deviation of the hourly number of calls (4 decimal places). (3 marks) |
(a) P(X = 0) = e^(-2.5) = 0.0821. [2 marks] (b) P(X <= 2) = 0.0821 + 0.2052 + 0.2565 = 0.5438; P(X >= 3) = 1 – 0.5438 = 0.4562. [3 marks] (c) For a Poisson variable, mean = variance = 2.5; standard deviation = sqrt(2.5) = 1.5811. [3 marks] |
| 10 | A raffle ticket costs $6. The prize X has distribution P(X = 0) = 0.7, P(X = 10) = 0.25, P(X = 50) = 0.05. (a) Find E(X). (3 marks) (b) Find the expected net gain per ticket and advise whether it is worth playing in the long run. (3 marks) (c) Find the fair price of a ticket. (2 marks) |
(a) E(X) = 0 x 0.7 + 10 x 0.25 + 50 x 0.05 = 0 + 2.5 + 2.5 = 5. [3 marks] (b) Expected net gain = 5 – 6 = -1, a loss of $1 per ticket on average, so it is not worth playing in the long run. [3 marks] (c) Fair price = E(X) = $5. [2 marks] |
| 11 | 100 students are randomly sampled from a district; they study 68 minutes per day on average with standard deviation 12 minutes. (a) Find the standard error of the mean. (2 marks) (b) Find the 95% confidence interval for the mean study time (2 decimal places, z = 1.96). (3 marks) (c) If the sample size rises to 400, how do the standard error and the interval width change? State the new standard error. (3 marks) |
(a) Standard error = 12/sqrt(100) = 1.2. [2 marks] (b) 68 +/- 1.96 x 1.2 = 68 +/- 2.352, i.e. (65.65, 70.35). [3 marks] (c) With n = 400 the standard error becomes 12/20 = 0.6; the interval width falls from 4.70 to 2.35, so the estimate is more precise. [3 marks] |
| 12 | Student numbers by form: S1 300, S2 300, S3 250, S4 200, S5 100, S6 50, total 1200. A stratified sample of 120 students is drawn. (a) Find the sampling fraction. (3 marks) (b) Find the number sampled from each form. (3 marks) (c) Explain one advantage of stratified sampling over simple random sampling. (2 marks) |
(a) Sampling fraction = 120/1200 = 0.1, i.e. 10%. [3 marks] (b) S1 30, S2 30, S3 25, S4 20, S5 10 and S6 5, totalling 120. [3 marks] (c) Stratified sampling guarantees that every form is represented, which reduces sampling error and reflects the population structure better than simple random sampling. [2 marks] |
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