A Level H1 Mathematics Practice Paper 5

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TuitionGoWhere Practice Paper - Maths H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Mathematics H1
Level: A-Level
Paper: Practice Paper — Statistics & Probability
Duration: 1 hour 30 minutes
Total Marks: 60
Version: 5 of 5

Name: ___________________________
Class: ___________________________
Date: ___________________________

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Instructions

• Write your answers in the spaces provided.
• Show all working clearly. Marks are awarded for correct method even if the final answer is wrong.
• Give answers correct to 3 significant figures unless otherwise stated.
• A graphing calculator may be used where appropriate.
• The total mark for this paper is 60.
• The number of marks for each question or part-question is shown in brackets [ ].

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Section A: Pure Statistics (30 marks)

Answer all questions in this section.

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Question 1 [2 marks]

A random sample of 8 students recorded the number of hours they spent on revision in a week:

$12,\ 15,\ 10,\ 18,\ 14,\ 11,\ 16,\ 13$

Calculate the unbiased estimate of the population mean.

$\bar{x} = \frac{\sum x_i}{n}$

Answer: _______________

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Question 2 [3 marks]

Using the data from Question 1, calculate the unbiased estimate of the population variance.

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n - 1}$

Answer: _______________

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Question 3 [3 marks]

The heights of a certain species of plant are normally distributed with mean $\mu$ cm and standard deviation $\sigma$ cm. A botanist measures a random sample of 5 plants and obtains the following heights (in cm):

$42,\ 45,\ 38,\ 47,\ 43$

(a) Find the unbiased estimates of $\mu$ and $\sigma^2$. [2 marks]

(b) State one assumption required for these to be unbiased estimates of the population parameters. [1 mark]

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Question 4 [4 marks]

A factory produces light bulbs. The lifetime of the bulbs, $X$ hours, follows a normal distribution with mean 800 hours and standard deviation 50 hours.

(a) Find $\mathrm{P}(X > 860)$. [2 marks]

(b) Find the value of $k$ such that $\mathrm{P}(X < k) = 0.90$. [2 marks]

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Question 5 [3 marks]

A fair six-sided die is rolled 4 times. Let $X$ be the number of times a six appears.

(a) State the distribution of $X$. [1 mark]

(b) Find $\mathrm{P}(X = 2)$. [2 marks]

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Question 6 [4 marks]

The number of emails received by an employee per hour follows a Poisson distribution with mean 3.5.

(a) Find the probability that in a given hour, the employee receives exactly 4 emails. [2 marks]

(b) Find the probability that in a given 2-hour period, the employee receives at least 5 emails. [2 marks]

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Question 7 [3 marks]

A box contains 7 red balls and 5 blue balls. Three balls are selected at random without replacement.

Find the probability that exactly 2 red balls and 1 blue ball are selected.

Answer: _______________

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Question 8 [4 marks]

A random variable $X \sim \mathrm{B}(20, 0.3)$.

(a) Find $\mathrm{E}(X)$ and $\mathrm{Var}(X)$. [2 marks]

(b) Using a suitable approximation, find $\mathrm{P}(X \geq 10)$. [2 marks]

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Question 9 [4 marks]

A continuous random variable $X$ has probability density function given by

$f(x) = \begin{cases} kx(4 - x) & 0 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}$

(a) Show that $k = \frac{3}{32}$. [2 marks]

(b) Find $\mathrm{E}(X)$. [2 marks]

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Section B: Applied Statistics & Data Interpretation (30 marks)

Answer all questions in this section.

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Question 10 [5 marks]

The following table shows the daily sales (in $) of a small café over 6 consecutive days:

Day	Mon	Tue	Wed	Thu	Fri	Sat
Sales ($)	320	280	350	410	390	460

(a) Calculate the mean daily sales. [1 mark]

(b) Calculate the standard deviation of the daily sales. [2 marks]

(c) On Sunday, the café recorded sales of $520. Explain the effect this additional value has on the mean and standard deviation. [2 marks]

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Question 11 [6 marks]

A market researcher investigates the relationship between advertising expenditure (in thousands of dollars) and monthly sales revenue (in thousands of dollars) for 8 small businesses.

Advertising ($x$)	2.0	3.5	4.0	5.5	6.0	7.5	8.0	9.0
Sales ($y$)	15	22	25	30	33	38	40	45

(a) Calculate the equation of the least squares regression line of $y$ on $x$. Give your answer in the form $y = a + bx$, where $a$ and $b$ are correct to 3 significant figures. [3 marks]

(b) Interpret the value of $b$ in context. [1 mark]

(c) Use your regression line to estimate the sales revenue when advertising expenditure is $6,500. Comment on the reliability of this estimate. [2 marks]

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Question 12 [5 marks]

A medical researcher claims that the mean cholesterol level of adult males in a certain city is 5.0 mmol/L. A random sample of 36 adult males from the city is taken. The sample mean cholesterol level is 5.3 mmol/L with a standard deviation of 1.2 mmol/L.

Test, at the 5% significance level, whether there is evidence that the mean cholesterol level is greater than 5.0 mmol/L.

(a) State the null and alternative hypotheses. [1 mark]

(b) Calculate the test statistic. [1 mark]

(c) State the critical value and your conclusion. [3 marks]

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Question 13 [5 marks]

A game involves rolling two fair six-sided dice and summing the scores.

(a) Complete the probability distribution table for the sum $S$ of the two dice. [3 marks]

$s$	2	3	4	5	6	7	8	9	10	11	12
$\mathrm{P}(S = s)$

(b) Find $\mathrm{E}(S)$ and $\mathrm{Var}(S)$. [2 marks]

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Question 14 [4 marks]

The weights of a certain variety of apple are normally distributed with mean 180 g and standard deviation 15 g.

(a) Find the probability that a randomly chosen apple weighs between 165 g and 195 g. [2 marks]

(b) A random sample of 9 apples is selected. Find the probability that the sample mean weight is greater than 185 g. [2 marks]

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Question 15 [5 marks]

A survey was conducted on 200 university students regarding their preferred mode of transport to campus. The results are summarised below:

Transport	Bus	MRT	Car	Walk	Total
Frequency	55	72	38	35	200

(a) Find the probability that a randomly selected student takes the MRT. [1 mark]

(b) Two students are selected at random. Find the probability that both take the bus. [2 marks]

(c) Three students are selected at random. Find the probability that at least one walks. [2 marks]

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End of Paper

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Mark Summary

Section	Marks
Section A: Questions 1–9	30
Section B: Questions 10–15	30
Total	60

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TuitionGoWhere Practice Paper — Maths H1 A-Level

Answer Key & Marking Scheme

Paper: Practice Paper — Statistics & Probability (Version 5 of 5)
Total Marks: 60

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Section A: Pure Statistics (30 marks)

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Question 1 [2 marks]

Answer: $\bar{x} = 13.625$

Working:

$\bar{x} = \frac{12 + 15 + 10 + 18 + 14 + 11 + 16 + 13}{8} = \frac{109}{8} = 13.625$

Marking:

• M1: Correct substitution into the formula for the mean
• A1: Correct answer (13.625 or 13.6 to 3 s.f.)

Teaching Note: The unbiased estimate of the population mean is simply the sample mean. We add all data values and divide by the number of observations $n$. This is the best single-number estimate of the true population mean $\mu$ from sample data.

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Question 2 [3 marks]

Answer: $s^2 = 6.268$ (or 6.27 to 3 s.f.)

Working:

Using $\bar{x} = 13.625$ and $n = 8$:

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
12	−1.625	2.6406
15	1.375	1.8906
10	−3.625	13.1406
18	4.375	19.1406
14	0.375	0.1406
11	−2.625	6.8906
16	2.375	5.6406
13	−0.625	0.3906

$\sum(x_i - \bar{x})^2 = 49.875$

$s^2 = \frac{49.875}{8 - 1} = \frac{49.875}{7} = 7.125$

Correction: Let me recalculate carefully:

$\sum(x_i - \bar{x})^2 = 2.6406 + 1.8906 + 13.1406 + 19.1406 + 0.1406 + 6.8906 + 5.6406 + 0.3906 = 49.875$

$s^2 = \frac{49.875}{7} = 7.125$

Answer: $s^2 = 7.125$ (or 7.13 to 3 s.f.)

Marking:

• M1: Correct calculation of deviations from the mean
• M1: Correct use of $n - 1 = 7$ in the denominator (not $n = 8$)
• A1: Correct final answer

Common Mistake: Using $n = 8$ instead of $n - 1 = 7$ gives $\frac{49.875}{8} = 6.234$, which is the biased sample variance. The unbiased estimate requires dividing by $n - 1$ to correct for the fact that we are estimating the population parameter from sample data.

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Question 3 [3 marks]

(a) [2 marks]

Answer: $\hat{\mu} = 43.0$ cm, $\hat{\sigma}^2 = 11.5$ cm$^2$

Working:

$\bar{x} = \frac{42 + 45 + 38 + 47 + 43}{5} = \frac{215}{5} = 43.0$

$s^2 = \frac{(42-43)^2 + (45-43)^2 + (38-43)^2 + (47-43)^2 + (43-43)^2}{5-1}$

$= \frac{1 + 4 + 25 + 16 + 0}{4} = \frac{46}{4} = 11.5$

(b) [1 mark]

Answer: The sample must be a random sample from the population (or the plants are independently selected from a normally distributed population).

Marking:

• (a) M1: Correct calculation of sample mean; A1: Both mean and variance correct
• (b) B1: Valid assumption stated

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Question 4 [4 marks]

$X \sim \mathrm{N}(800, 50^2)$

(a) [2 marks]

$\mathrm{P}(X > 860) = \mathrm{P}\left(Z > \frac{860 - 800}{50}\right) = \mathrm{P}(Z > 1.2)$

$= 1 - \Phi(1.2) = 1 - 0.8849 = 0.1151$

Answer: 0.115 (to 3 s.f.)

(b) [2 marks]

We need $k$ such that $\mathrm{P}(X < k) = 0.90$.

$\mathrm{P}\left(Z < \frac{k - 800}{50}\right) = 0.90$

From tables, $\Phi(1.282) = 0.90$, so:

$\frac{k - 800}{50} = 1.282$

$k = 800 + 50 \times 1.282 = 864.1$

Answer: $k = 864$ hours (to 3 s.f.)

Marking:

• (a) M1: Standardising correctly; A1: Correct probability
• (b) M1: Using inverse normal correctly; A1: Correct value of $k$

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Question 5 [3 marks]

(a) [1 mark]

Answer: $X \sim \mathrm{B}(4, \frac{1}{6})$

(b) [2 marks]

$\mathrm{P}(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2 = 6 \times \frac{1}{36} \times \frac{25}{36} = \frac{150}{1296} = \frac{25}{216}$

Answer: $\frac{25}{216} \approx 0.1157$ (or 0.116 to 3 s.f.)

Marking:

• (a) B1: Correct distribution stated with both parameters
• (b) M1: Correct binomial probability formula applied; A1: Correct answer

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Question 6 [4 marks]

Let $X$ = number of emails per hour, $X \sim \mathrm{Po}(3.5)$

(a) [2 marks]

$\mathrm{P}(X = 4) = \frac{e^{-3.5} \times 3.5^4}{4!} = \frac{e^{-3.5} \times 150.0625}{24}$

$= \frac{150.0625}{24} \times e^{-3.5} = 6.2526 \times 0.030197 = 0.1888$

Answer: 0.189 (to 3 s.f.)

(b) [2 marks]

For a 2-hour period, the mean is $\lambda = 3.5 \times 2 = 7$.

Let $Y \sim \mathrm{Po}(7)$.

$\mathrm{P}(Y \geq 5) = 1 - \mathrm{P}(Y \leq 4)$

$= 1 - \left[\frac{e^{-7}7^0}{0!} + \frac{e^{-7}7^1}{1!} + \frac{e^{-7}7^2}{2!} + \frac{e^{-7}7^3}{3!} + \frac{e^{-7}7^4}{4!}\right]$

$= 1 - e^{-7}\left[1 + 7 + \frac{49}{2} + \frac{343}{6} + \frac{2401}{24}\right]$

$= 1 - e^{-7}\left[1 + 7 + 24.5 + 57.167 + 100.042\right]$

$= 1 - e^{-7} \times 189.708 = 1 - 0.0009119 \times 189.708$

$= 1 - 0.1730 = 0.8270$

Answer: 0.827 (to 3 s.f.)

Marking:

• (a) M1: Correct Poisson formula with $\lambda = 3.5$; A1: Correct answer
• (b) M1: Correct adjustment of $\lambda$ to 7 for 2 hours and use of complement; A1: Correct answer

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Question 7 [3 marks]

Answer: $\frac{21}{44}$ or approximately 0.477

Working:

Total balls = 12. Selecting 3 balls without replacement.

$\mathrm{P}(2 \text{ red, } 1 \text{ blue}) = \frac{\binom{7}{2} \times \binom{5}{1}}{\binom{12}{3}} = \frac{21 \times 5}{220} = \frac{105}{220} = \frac{21}{44}$

Marking:

• M1: Correct numerator (combinations of red and blue)
• M1: Correct denominator (total combinations)
• A1: Correct simplified answer

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Question 8 [4 marks]

$X \sim \mathrm{B}(20, 0.3)$

(a) [2 marks]

$\mathrm{E}(X) = np = 20 \times 0.3 = 6$

$\mathrm{Var}(X) = np(1-p) = 20 \times 0.3 \times 0.7 = 4.2$

(b) [2 marks]

Since $n = 20$ is moderately large and $np = 6 > 5$, $n(1-p) = 14 > 5$, we can use the normal approximation:

$X \stackrel{\text{approx}}{\sim} \mathrm{N}(6, 4.2)$

Using continuity correction:

$\mathrm{P}(X \geq 10) \approx \mathrm{P}\left(Z \geq \frac{9.5 - 6}{\sqrt{4.2}}\right) = \mathrm{P}\left(Z \geq \frac{3.5}{2.049}\right) = \mathrm{P}(Z \geq 1.708)$

$= 1 - \Phi(1.708) = 1 - 0.9562 = 0.0438$

Answer: 0.0438 (to 3 s.f.)

Marking:

• (a) B1: Each correct (E(X) and Var(X))
• (b) M1: Correct normal approximation with continuity correction; A1: Correct probability

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Question 9 [4 marks]

(a) [2 marks]

For a valid PDF, $\int_0^4 f(x)\,dx = 1$:

$\int_0^4 kx(4-x)\,dx = k\int_0^4 (4x - x^2)\,dx = k\left[2x^2 - \frac{x^3}{3}\right]_0^4$

$= k\left[\left(2(16) - \frac{64}{3}\right) - 0\right] = k\left[32 - \frac{64}{3}\right] = k\left[\frac{96 - 64}{3}\right] = k \times \frac{32}{3}$

Setting equal to 1:

$k \times \frac{32}{3} = 1 \implies k = \frac{3}{32} \quad \text{(shown)}$

(b) [2 marks]

$\mathrm{E}(X) = \int_0^4 x \cdot \frac{3}{32}x(4-x)\,dx = \frac{3}{32}\int_0^4 (4x^2 - x^3)\,dx$

$= \frac{3}{32}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_0^4 = \frac{3}{32}\left[\frac{256}{3} - \frac{256}{4}\right]$

$= \frac{3}{32}\left[\frac{256}{3} - 64\right] = \frac{3}{32}\left[\frac{256 - 192}{3}\right] = \frac{3}{32} \times \frac{64}{3} = \frac{64}{32} = 2$

Answer: $\mathrm{E}(X) = 2$

Marking:

• (a) M1: Correct integration; A1: Correct derivation of $k = \frac{3}{32}$
• (b) M1: Correct setup of $\mathrm{E}(X)$ integral; A1: Correct answer

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Section B: Applied Statistics & Data Interpretation (30 marks)

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Question 10 [5 marks]

(a) [1 mark]

$\bar{x} = \frac{320 + 280 + 350 + 410 + 390 + 460}{6} = \frac{2210}{6} = 368.33$

Answer: $368 (to 3 s.f.)

(b) [2 marks]

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}$

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
320	−48.33	2336.11
280	−88.33	7802.78
350	−18.33	336.11
410	41.67	1736.11
390	21.67	469.44
460	91.67	8402.78

$\sum(x_i - \bar{x})^2 = 21083.33$

$s^2 = \frac{21083.33}{5} = 4216.67$

$s = \sqrt{4216.67} = 64.94$

Answer: s = \64.9$ (to 3 s.f.)

(c) [2 marks]

Adding Sunday's sales of $520:

New mean: $\bar{x}_{\text{new}} = \frac{2210 + 520}{7} = \frac{2730}{7} = 390$

The mean increases from $368 to $390 because $520 is above the original mean, pulling the average up.

The standard deviation will also increase because $520 is far from the original mean, increasing the spread of the data. The new data point is an outlier relative to the original dataset, so both measures of central tendency and dispersion are affected.

Marking:

• (a) B1: Correct mean
• (b) M1: Correct method for standard deviation; A1: Correct answer
• (c) B1: Mean increases (with reasoning); B1: Standard deviation increases (with reasoning)

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Question 11 [6 marks]

(a) [3 marks]

Calculating summary statistics:

$n = 8$

$\sum x = 2.0 + 3.5 + 4.0 + 5.5 + 6.0 + 7.5 + 8.0 + 9.0 = 45.5$

$\sum y = 15 + 22 + 25 + 30 + 33 + 38 + 40 + 45 = 248$

$\bar{x} = \frac{45.5}{8} = 5.6875$

$\bar{y} = \frac{248}{8} = 31$

$\sum xy = (2.0 \times 15) + (3.5 \times 22) + (4.0 \times 25) + (5.5 \times 30) + (6.0 \times 33) + (7.5 \times 38) + (8.0 \times 40) + (9.0 \times 45)$

$= 30 + 77 + 100 + 165 + 198 + 285 + 320 + 405 = 1580$

$\sum x^2 = 4 + 12.25 + 16 + 30.25 + 36 + 56.25 + 64 + 81 = 299.75$

$S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 1580 - \frac{45.5 \times 248}{8} = 1580 - 1410.5 = 169.5$

$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 299.75 - \frac{45.5^2}{8} = 299.75 - 258.781 = 40.969$

$b = \frac{S_{xy}}{S_{xx}} = \frac{169.5}{40.969} = 4.137$

$a = \bar{y} - b\bar{x} = 31 - 4.137 \times 5.6875 = 31 - 23.530 = 7.470$

Answer: $y = 7.47 + 4.14x$ (to 3 s.f.)

(b) [1 mark]

Answer: For every additional $1,000 spent on advertising, the monthly sales revenue increases by approximately $4,140.

(c) [2 marks]

When $x = 6.5$:

$y = 7.47 + 4.14 \times 6.5 = 7.47 + 26.91 = 34.38$

Answer: Estimated sales revenue is approximately $34,400.

Comment: Since $x = 6.5$ lies within the range of the data (2.0 to 9.0), this is an interpolation and the estimate is reasonably reliable.

Marking:

• (a) M1: Correct calculation of $S_{xy}$ and $S_{xx}$; M1: Correct $b$ and $a$; A1: Correct equation to 3 s.f.
• (b) B1: Correct interpretation in context
• (c) M1: Correct substitution; A1: Correct estimate with valid reliability comment

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Question 12 [5 marks]

(a) [1 mark]

$H_0: \mu = 5.0$ mmol/L
$H_1: \mu > 5.0$ mmol/L

(b) [1 mark]

Since $n = 36$ is large, by the Central Limit Theorem, we use the $z$-test:

$z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{5.3 - 5.0}{1.2 / \sqrt{36}} = \frac{0.3}{1.2 / 6} = \frac{0.3}{0.2} = 1.5$

(c) [3 marks]

This is a one-tailed test at the 5% significance level.

Critical value: $z_{0.05} = 1.645$

Since $z = 1.5 < 1.645$, we do not reject $H_0$.

Conclusion: There is insufficient evidence at the 5% significance level to support the claim that the mean cholesterol level is greater than 5.0 mmol/L.

Marking:

• (a) B1: Both hypotheses correct (one-tailed)
• (b) B1: Correct test statistic
• (c) M1: Correct critical value; M1: Correct comparison and decision; A1: Correct conclusion in context

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Question 13 [5 marks]

(a) [3 marks]

Total outcomes = $6 \times 6 = 36$

$s$	2	3	4	5	6	7	8	9	10	11	12
$\mathrm{P}(S = s)$	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

(b) [2 marks]

$\mathrm{E}(S) = \sum s \cdot \mathrm{P}(S = s)$

$= \frac{1}{36}[2(1) + 3(2) + 4(3) + 5(4) + 6(5) + 7(6) + 8(5) + 9(4) + 10(3) + 11(2) + 12(1)]$

$= \frac{1}{36}[2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12] = \frac{252}{36} = 7$

$\mathrm{E}(S^2) = \frac{1}{36}[4(1) + 9(2) + 16(3) + 25(4) + 36(5) + 49(6) + 64(5) + 81(4) + 100(3) + 121(2) + 144(1)]$

$= \frac{1}{36}[4 + 18 + 48 + 100 + 180 + 294 + 320 + 324 + 300 + 242 + 144] = \frac{1974}{36} = 54.833$

$\mathrm{Var}(S) = \mathrm{E}(S^2) - [\mathrm{E}(S)]^2 = 54.833 - 49 = 5.833$

Answer: $\mathrm{E}(S) = 7$, $\mathrm{Var}(S) = 5.83$ (to 3 s.f.) or $\frac{35}{6}$

Marking:

• (a) B1: Correct numerator pattern (1,2,3,4,5,6,5,4,3,2,1); B1: Correct denominator 36; B1: All probabilities correct
• (b) M1: Correct method for E(S) and Var(S); A1: Both correct

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Question 14 [4 marks]

$X \sim \mathrm{N}(180, 15^2)$

(a) [2 marks]

$\mathrm{P}(165 < X < 195) = \mathrm{P}\left(\frac{165 - 180}{15} < Z < \frac{195 - 180}{15}\right) = \mathrm{P}(-1 < Z < 1)$

$= \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6826$

Answer: 0.683 (to 3 s.f.)

(b) [2 marks]

For the sample mean $\bar{X}$ of $n = 9$ apples:

$\bar{X} \sim \mathrm{N}\left(180, \frac{15^2}{9}\right) = \mathrm{N}(180, 25)$

$\mathrm{P}(\bar{X} > 185) = \mathrm{P}\left(Z > \frac{185 - 180}{5}\right) = \mathrm{P}(Z > 1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587$

Answer: 0.159 (to 3 s.f.)

Marking:

• (a) M1: Correct standardisation; A1: Correct probability
• (b) M1: Correct distribution of sample mean with $\sigma/\sqrt{n} = 5$; A1: Correct probability

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Question 15 [5 marks]

(a) [1 mark]

$\mathrm{P}(\text{MRT}) = \frac{72}{200} = 0.36$

(b) [2 marks]

$\mathrm{P}(\text{both bus}) = \frac{55}{200} \times \frac{54}{199} = \frac{2970}{39800} = 0.07462$

Answer: 0.0746 (to 3 s.f.)

(c) [2 marks]

$\mathrm{P}(\text{at least one walks}) = 1 - \mathrm{P}(\text{none walk})$

$\mathrm{P}(\text{none walk}) = \frac{165}{200} \times \frac{164}{199} \times \frac{163}{198} = \frac{4465980}{7880400} = 0.5667$

$\mathrm{P}(\text{at least one walks}) = 1 - 0.5667 = 0.4333$

Answer: 0.433 (to 3 s.f.)

Marking:

• (a) B1: Correct probability
• (b) M1: Correct multiplication of conditional probabilities (without replacement); A1: Correct answer
• (c) M1: Correct use of complement and multiplication; A1: Correct answer

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Mark Summary

Section	Marks
Section A: Questions 1–9	30
Section B: Questions 10–15	30
Total	60