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Secondary 4 Additional Mathematics Statistics Probability Quiz

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Questions

Secondary 4 Additional Mathematics Quiz - Statistics Probability

Name: _________________________________ Class: __________ Date: __________

Duration: 40 minutes
Total Marks: 40 marks
Instructions: Answer all questions. Show all working clearly. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified.

Section A: Standard Questions [1 – 10] (20 marks)

Answer all questions. Each question carries 2 marks.

1.
A bag contains 5 red marbles, 7 blue marbles, and 3 green marbles. One marble is drawn at random from the bag. Find the probability that the marble drawn is
(a) red,
(b) not blue.

Working space:

[2 marks]

2.
The probability that a student passes Mathematics is $\frac{3}{4}$ , and the probability that the student passes Science is $\frac{2}{3}$ . Assuming the two events are independent, find the probability that the student
(a) passes both subjects,
(b) fails at least one subject.

Working space:

[2 marks]

3.
A fair six-sided die is rolled twice. Find the probability that
(a) the sum of the two scores is 7,
(b) the first score is greater than the second score.

Working space:

[2 marks]

4.
In a group of 80 students, 45 study Physics, 30 study Chemistry, and 15 study both Physics and Chemistry. A student is selected at random from the group. Find the probability that the student studies
(a) Physics or Chemistry or both,
(b) neither Physics nor Chemistry.

Working space:

[2 marks]

5.
The events $A$ and $B$ are such that $P(A) = 0.4$ , $P(B) = 0.5$ , and $P(A \cup B) = 0.7$ .
(a) Find $P(A \cap B)$ .
(b) State, with a reason, whether events $A$ and $B$ are independent.

Working space:

[2 marks]

6.
A random variable $X$ has probability distribution given by $P(X = k) = \frac{k}{15}$ for $k = 1, 2, 3, 4, 5$ .
Find
(a) $P(X \leq 3)$ ,
(b) the expected value $E(X)$ .

Working space:

[2 marks]

7.
The heights of 200 plants are measured and grouped into the following frequency distribution:

Height ( $h$ cm)	$140 \leq h < 150$	$150 \leq h < 160$	$160 \leq h < 170$	$170 \leq h < 180$	$180 \leq h < 190$
Frequency	24	56	64	44	12

(a) State the modal class.
(b) Estimate the mean height of the plants.

Working space:

[2 marks]

8.
For a set of 10 data values, the sum of the values is 250 and the sum of the squares of the values is 6,750.
Find
(a) the mean,
(b) the standard deviation.

Working space:

[2 marks]

9.
The discrete random variable $X$ takes values $-2, 0, 2, 4$ with probabilities $\frac{1}{4}, \frac{1}{8}, \frac{3}{8}, \frac{1}{4}$ respectively.
(a) Find $E(X)$ .
(b) Find $E(X^2)$ .

Working space:

[2 marks]

10.
A frequency distribution has mean $\bar{x} = 12$ and standard deviation $\sigma = 3$ . Each value in the distribution is transformed using the formula $y = 2x + 5$ .
Find
(a) the new mean,
(b) the new standard deviation.

Working space:

[2 marks]

Section B: Intermediate Questions [11 – 15] (10 marks)

Answer all questions. Each question carries 2 marks.

11.
Two cards are drawn without replacement from a standard pack of 52 playing cards. Find the probability that
(a) both cards are aces,
(b) the two cards are of different suits.

Working space:

[2 marks]

12.
In a factory, machines $A$ , $B$ , and $C$ produce 30%, 50%, and 20% of the total output respectively. The percentages of defective items produced by machines $A$ , $B$ , and $C$ are 2%, 3%, and 5% respectively. An item is selected at random from the total output.

(a) Find the probability that the item is defective.

(b) Given that the item is defective, find the probability that it was produced by machine $B$ .

Working space:

[2 marks]

13.
The cumulative frequency curve below shows the distribution of marks scored by 120 students in a test.

<image_placeholder> id: Q13-fig1 type: graph linked_question: Q13 description: Cumulative frequency curve (ogive) showing marks distribution labels: x-axis "Mark", y-axis "Cumulative frequency", points at (20, 8), (40, 32), (60, 78), (80, 108), (100, 120) values: x-axis 0 to 100 in intervals of 20, y-axis 0 to 120 in intervals of 20 must_show: Smooth S-shaped curve passing through given points, labeled axes with scale, grid lines </image_placeholder>

Use the curve to estimate
(a) the median mark,
(b) the interquartile range,
(c) the percentage of students who scored more than 70 marks.

Working space:

[2 marks]

14.
A discrete random variable $X$ has the following probability distribution:

$x$	0	1	2	3	4
$P(X=x)$	0.1	$p$	0.3	$q$	0.2

Given that $E(X) = 2.2$ , find
(a) the values of $p$ and $q$ ,
(b) $Var(X)$ .

Working space:

[2 marks]

15.
The masses of 500 apples are normally distributed with mean 150 g and standard deviation 20 g.
Find
(a) the number of apples with mass less than 130 g,
(b) the value of $k$ such that 5% of apples have mass greater than $k$ g.

Working space:

[2 marks]

Section C: Advanced Questions [16 – 20] (10 marks)

Answer all questions. Each question carries 2 marks.

16.
A committee of 5 people is to be chosen from 6 men and 4 women. Find the number of ways the committee can be formed if
(a) there are no restrictions,
(b) there must be at least 2 women,
(c) there must be more men than women.

Working space:

[2 marks]

17.
The probability distribution of a discrete random variable $X$ is given by

$P(X = r) = kr(5-r)$

for $r = 1, 2, 3, 4$ .

(a) Show that $k = \frac{1}{10}$ .
(b) Find $P(1 \leq X < 3)$ .
(c) Find $E(3X - 2)$ .

Working space:

[2 marks]

18.
In a class of 40 students, 18 study History, 15 study Geography, and 8 study neither History nor Geography.
(a) Find the probability that a randomly chosen student studies both History and Geography.
(b) Given that a student studies History, find the probability that the student also studies Geography.

Working space:

[2 marks]

19.
The random variable $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$ . Given that $P(X < 25) = 0.0668$ and $P(X > 35) = 0.0228$ , find
(a) the values of $\mu$ and $\sigma$ ,
(b) $P(28 < X < 40)$ .

Working space:

[2 marks]

20.
A box contains 4 red balls and 6 blue balls. Two balls are drawn at random from the box, one after the other, without replacement.
(a) Show that the probability that the two balls are of the same colour is $\frac{7}{15}$ .
(b) If three balls are drawn without replacement, find the probability that exactly two of them are red.

Working space:

[2 marks]

END OF QUIZ

Answers

Secondary 4 Additional Mathematics Quiz - Statistics Probability

Answer Key and Marking Scheme

Total Marks: 40 marks
Duration: 40 minutes

Section A: Standard Questions [1 – 10]

Question 1 [2 marks]

(a) $P(\text{red}) = \frac{5}{5+7+3} = \frac{5}{15} = \frac{1}{3}$ [1 mark]

Teaching note: The probability of an event equals the number of favorable outcomes divided by the total number of equally likely outcomes. Here there are 15 marbles in total and 5 are red.

(b) $P(\text{not blue}) = 1 - P(\text{blue}) = 1 - \frac{7}{15} = \frac{8}{15}$ [1 mark]

Alternative: Count non-blue marbles directly: $\frac{5+3}{15} = \frac{8}{15}$

Common mistake: Forgetting that "not blue" includes both red and green marbles.

Question 2 [2 marks]

(a) $P(\text{both}) = P(M) \times P(S) = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$ [1 mark]

Teaching note: For independent events, $P(A \cap B) = P(A) \times P(B)$ . The occurrence of one does not affect the other.

(b) $P(\text{fails at least one}) = 1 - P(\text{both pass}) = 1 - \frac{1}{2} = \frac{1}{2}$ [1 mark]

Alternative using complement: "At least one fail" is the complement of "both pass."

Common mistake: Trying to enumerate all cases (fail M pass S, pass M fail S, fail both) without using the complement rule, which is longer but also valid: $P(\text{fail M}) \times P(\text{pass S}) + P(\text{pass M}) \times P(\text{fail S}) + P(\text{fail M}) \times P(\text{fail S})$ $= \frac{1}{4} \times \frac{2}{3} + \frac{3}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{3} = \frac{2+3+1}{12} = \frac{6}{12} = \frac{1}{2}$

Question 3 [2 marks]

Sample space: 36 equally likely outcomes $(1,1)$ to $(6,6)$ when two dice are rolled.

(a) Sum equals 7: Favorable outcomes are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — six outcomes. [0.5 mark for identifying correct pairs]

$P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$ [0.5 mark]

(b) First score greater than second: Count outcomes where first > second:

First = 2: second = 1 (1 way)
First = 3: second = 1, 2 (2 ways)
First = 4: second = 1, 2, 3 (3 ways)
First = 5: second = 1, 2, 3, 4 (4 ways)
First = 6: second = 1, 2, 3, 4, 5 (5 ways)

Total: $1+2+3+4+5 = 15$ outcomes [0.5 mark]

$P(\text{first} > \text{second}) = \frac{15}{36} = \frac{5}{12}$ [0.5 mark]

Teaching note: By symmetry, $P(\text{first} > \text{second}) = P(\text{second} > \text{first})$ , and $P(\text{equal}) = \frac{6}{36} = \frac{1}{6}$ . So each of the first two probabilities equals $\frac{1}{2}\left(1-\frac{1}{6}\right) = \frac{5}{12}$ .

Question 4 [2 marks]

Using the principle of inclusion-exclusion: $|P \cup C| = |P| + |C| - |P \cap C| = 45 + 30 - 15 = 60$ students study Physics or Chemistry or both. [0.5 mark for method]

(a) $P(\text{Physics or Chemistry}) = \frac{60}{80} = \frac{3}{4}$ [0.5 mark]

(b) $P(\text{neither}) = 1 - \frac{60}{80} = \frac{20}{80} = \frac{1}{4}$ [1 mark]

Alternative for (b): Or count directly: $80 - 60 = 20$ students study neither.

Teaching note: Venn diagrams help visualize this. The region "neither" is outside both circles.

Question 5 [2 marks]

(a) Using $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ : [0.5 mark for formula]

$0.7 = 0.4 + 0.5 - P(A \cap B)$

$P(A \cap B) = 0.9 - 0.7 = 0.2$ [0.5 mark]

(b) For independence, check if $P(A \cap B) = P(A) \times P(B)$ [0.5 mark for test]

$P(A) \times P(B) = 0.4 \times 0.5 = 0.2$ [0.25 mark]

Since $P(A \cap B) = 0.2 = P(A) \times P(B)$ , events $A$ and $B$ are independent. [0.25 mark]

Question 6 [2 marks]

(a) $P(X \leq 3) = P(X=1) + P(X=2) + P(X=3) = \frac{1}{15} + \frac{2}{15} + \frac{3}{15} = \frac{6}{15} = \frac{2}{5}$ [1 mark]

(b) $E(X) = \sum x \cdot P(X=x) = 1 \times \frac{1}{15} + 2 \times \frac{2}{15} + 3 \times \frac{3}{15} + 4 \times \frac{4}{15} + 5 \times \frac{5}{15}$ [0.5 mark for setting up]

$= \frac{1 + 4 + 9 + 16 + 25}{15} = \frac{55}{15} = \frac{11}{3}$ or $3\frac{2}{3}$ or $3.67$ (3 s.f.) [0.5 mark]

Teaching note: The expected value is the weighted average of all possible values, where weights are probabilities. It represents the long-run average if the experiment were repeated many times.

Question 7 [2 marks]

(a) Modal class is $160 \leq h < 170$ (highest frequency of 64) [1 mark]

(b) Using mid-values: 145, 155, 165, 175, 185 [0.5 mark for mid-values]

Estimated mean $= \frac{24 \times 145 + 56 \times 155 + 64 \times 165 + 44 \times 175 + 12 \times 185}{200}$ [0.25 mark for formula]

$= \frac{3480 + 8680 + 10560 + 7700 + 2220}{200} = \frac{32640}{200} = 163.2$ cm [0.25 mark]

Teaching note: For grouped data, we use mid-values as estimates since we don't know exact values. This gives an estimate, not the exact mean.

Question 8 [2 marks]

(a) Mean $\bar{x} = \frac{\sum x}{n} = \frac{250}{10} = 25$ [1 mark]

(b) Variance $= \frac{\sum x^2}{n} - \bar{x}^2 = \frac{6750}{10} - 25^2 = 675 - 625 = 50$ [0.5 mark]

Standard deviation $= \sqrt{50} = 5\sqrt{2} \approx 7.07$ (3 s.f.) [0.5 mark]

Teaching note: Standard deviation measures spread. The formula $\sigma = \sqrt{\frac{\sum(x-\bar{x})^2}{n}}$ is equivalent to $\sqrt{\frac{\sum x^2}{n} - \bar{x}^2}$ but the second is computationally easier when you have $\sum x^2$ and $\sum x$ .

Question 9 [2 marks]

(a) $E(X) = (-2) \times \frac{1}{4} + 0 \times \frac{1}{8} + 2 \times \frac{3}{8} + 4 \times \frac{1}{4}$ [0.5 mark]

$= -\frac{1}{2} + 0 + \frac{3}{4} + 1 = \frac{-2+0+3+4}{4} = \frac{5}{4} = 1.25$ or $1\frac{1}{4}$ [0.5 mark]

(b) $E(X^2) = (-2)^2 \times \frac{1}{4} + 0^2 \times \frac{1}{8} + 2^2 \times \frac{3}{8} + 4^2 \times \frac{1}{4}$ [0.5 mark]

$= 4 \times \frac{1}{4} + 0 + 4 \times \frac{3}{8} + 16 \times \frac{1}{4} = 1 + 0 + 1.5 + 4 = 6.5$ or $\frac{13}{2}$ [0.5 mark]

Teaching note: $E(X^2)$ is needed to find variance: $Var(X) = E(X^2) - [E(X)]^2$ . Be careful: $E(X^2) \neq [E(X)]^2$ .

Question 10 [2 marks]

Given transformation $y = 2x + 5$ :

(a) New mean $\bar{y} = 2\bar{x} + 5 = 2(12) + 5 = 29$ [1 mark]

(b) New standard deviation: Adding a constant (5) doesn't change spread, but multiplying by 2 scales spread by 2.

New standard deviation $= 2 \times 3 = 6$ [1 mark]

Teaching note: For $y = ax + b$ : $E(Y) = aE(X) + b$ and $SD(Y) = |a| \times SD(X)$ . The shift $b$ affects location (mean) but not spread (standard deviation). The scale factor $a$ affects both.

Section B: Intermediate Questions [11 – 15]

Question 11 [2 marks]

(a) $P(\text{both aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$ [1 mark]

Teaching note: This uses the multiplication rule for dependent events. After drawing one ace, only 3 aces remain from 51 cards.

(b) $P(\text{different suits}) = 1 - P(\text{same suit})$ [0.25 mark for strategy]

$P(\text{same suit}) = \frac{52}{52} \times \frac{12}{51} = \frac{12}{51} = \frac{4}{17}$ (first card any suit, second must match) [0.5 mark]

$P(\text{different suits}) = 1 - \frac{4}{17} = \frac{13}{17}$ [0.25 mark]

Alternative for (b): Direct count: $\frac{52 \times 39}{52 \times 51} = \frac{39}{51} = \frac{13}{17}$

Question 12 [2 marks]

Let $D$ = "item is defective".

(a) Using law of total probability: [0.25 mark for method]

$P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$

$= 0.02 \times 0.30 + 0.03 \times 0.50 + 0.05 \times 0.20$ [0.25 mark]

$= 0.006 + 0.015 + 0.010 = 0.031$ [0.5 mark]

(b) Using Bayes' theorem: $P(B|D) = \frac{P(D|B)P(B)}{P(D)}$ [0.25 mark for formula]

$= \frac{0.03 \times 0.50}{0.031} = \frac{0.015}{0.031} = \frac{15}{31} \approx 0.484$ (3 s.f.) [0.5 mark]

Teaching note: This is a classic Bayesian probability problem. Machine $B$ produces the most output, so even with moderate defect rate, it contributes significantly to defective items.

Question 13 [2 marks]

Expected visual features for answer verification: The ogive passes through points (20, 8), (40, 32), (60, 78), (80, 108), (100, 120). Cumulative frequency ranges from 0 to 120.

(a) Median: $\frac{120}{2} = 60$ th value. From curve, estimate mark ≈ 58±2 (reading across from 60 on y-axis to curve, down to x-axis) [0.67 mark, accept 56-60]

(b) Lower quartile $Q_1$ : 30th value ≈ 42±2. Upper quartile $Q_3$ : 90th value ≈ 72±2. [0.33 mark for each, accept range]

IQR = $Q_3 - Q_1 \approx 72 - 42 =$ 30±4 [0.33 mark]

(c) Students scoring > 70: Read cumulative frequency at mark = 70, approximately 95 students. [0.33 mark for method]

Students scoring more than 70 = $120 - 95 = 25$ students [0.17 mark]

Percentage = $\frac{25}{120} \times 100\% \approx 20.8\%$ or about 21% [0.17 mark, accept 15%-25%]

Teaching note: The interquartile range measures middle 50% spread. Reading from cumulative frequency curves requires careful estimation; small variations accepted due to graph reading.

Question 14 [2 marks]

(a) Using $\sum P(X=x) = 1$ : $0.1 + p + 0.3 + q + 0.2 = 1$ , so $p + q = 0.4$ [0.25 mark]

Using $E(X) = 2.2$ : $0 \times 0.1 + 1 \times p + 2 \times 0.3 + 3 \times q + 4 \times 0.2 = 2.2$ [0.25 mark] $p + 0.6 + 3q + 0.8 = 2.2$ $p + 3q = 0.8$ [0.25 mark]

Solving: $(p + 3q) - (p + q) = 0.8 - 0.4 = 0.4$ , so $2q = 0.4$ , thus $q = 0.2$ and $p = 0.2$ [0.25 mark]

(b) $E(X^2) = 0^2 \times 0.1 + 1^2 \times 0.2 + 2^2 \times 0.3 + 3^2 \times 0.2 + 4^2 \times 0.2$ [0.25 mark]

$= 0 + 0.2 + 1.2 + 1.8 + 3.2 = 6.4$ [0.25 mark]

$Var(X) = E(X^2) - [E(X)]^2 = 6.4 - 2.2^2 = 6.4 - 4.84 =$ $1.56$ [0.5 mark]

Question 15 [2 marks]

Given: $X \sim N(150, 20^2)$

(a) $Z = \frac{130-150}{20} = -1$ [0.25 mark]

$P(X < 130) = P(Z < -1) = 1 - 0.8413 = 0.1587$ [0.25 mark]

Number of apples = $500 \times 0.1587 =$ 79 or 80 apples [0.5 mark, accept 79-80]

(b) $P(X > k) = 0.05$ , so $P(X \leq k) = 0.95$ [0.25 mark]

$P(Z \leq z) = 0.95$ gives $z \approx 1.645$ [0.25 mark]

$\frac{k - 150}{20} = 1.645$

$k = 150 + 32.9 =$ $182.9$ g or 183 g (3 s.f.) [0.25 mark for equation, 0 mark already counted]

Section C: Advanced Questions [16 – 20]

Question 16 [2 marks]

(a) No restrictions: $\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$ ways [0.67 mark]

(b) At least 2 women: Cases are (2 women, 3 men), (3 women, 2 men), (4 women, 1 man)

$= \binom{4}{2}\binom{6}{3} + \binom{4}{3}\binom{6}{2} + \binom{4}{4}\binom{6}{1}$ [0.33 mark for correct approach]

$= 6 \times 20 + 4 \times 15 + 1 \times 6 = 120 + 60 + 6 = 186$ ways [0.33 mark]

(c) More men than women: Cases are (3 men, 2 women), (4 men, 1 woman), (5 men, 0 women)

$= \binom{6}{3}\binom{4}{2} + \binom{6}{4}\binom{4}{1} + \binom{6}{5}\binom{4}{0}$ [0.33 mark]

$= 20 \times 6 + 15 \times 4 + 6 \times 1 = 120 + 60 + 6 = 186$ ways [0.33 mark]

Teaching note: Combinations $\binom{n}{r}$ count unordered selections. The key is identifying which cases satisfy the condition and summing them.

Question 17 [2 marks]

(a) For a valid probability distribution, $\sum P(X=r) = 1$ : [0.33 mark for condition]

$k \times 1 \times 4 + k \times 2 \times 3 + k \times 3 \times 2 + k \times 4 \times 1 = 1$ [0.33 mark]

$4k + 6k + 6k + 4k = 20k = 1$ [0.17 mark]

Therefore $k = \frac{1}{20} = 0.05$ [0.17 mark]

Note: Original question stated $\frac{1}{10}$ which is incorrect based on working; correct value is $\frac{1}{20}$ .

(b) $P(1 \leq X < 3) = P(X=1) + P(X=2) = \frac{4}{20} + \frac{6}{20} = \frac{10}{20} = \frac{1}{2}$ [0.33 mark]

(c) First find $E(X) = 1 \times \frac{4}{20} + 2 \times \frac{6}{20} + 3 \times \frac{6}{20} + 4 \times \frac{4}{20} = \frac{4+12+18+16}{20} = \frac{50}{20} = 2.5$ [0.33 mark for method]

$E(3X-2) = 3E(X) - 2 = 3(2.5) - 2 = 7.5 - 2 = 5.5$ or $\frac{11}{2}$ [0.33 mark]

Teaching note: Linearity of expectation: $E(aX+b) = aE(X) + b$ . We don't need to find the full distribution of $3X-2$ .

Question 18 [2 marks]

Using principle of inclusion-exclusion: $|H \cup G| = 40 - 8 = 32$ students study History or Geography. [0.25 mark]

$|H \cap G| = |H| + |G| - |H \cup G| = 18 + 15 - 32 = 1$ student studies both. [0.25 mark]

(a) $P(\text{both}) = \frac{1}{40}$ [0.5 mark]

(b) $P(G|H) = \frac{P(G \cap H)}{P(H)} = \frac{|H \cap G|}{|H|} = \frac{1}{18}$ [1 mark]

Teaching note: Conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$ . "Given History" restricts our sample space to just the 18 History students.

Question 19 [2 marks]

(a) $P(X < 25) = 0.0668 \Rightarrow P\left(Z < \frac{25-\mu}{\sigma}\right) = 0.0668$

Since $P(Z < -1.5) \approx 0.0668$ from tables, $\frac{25-\mu}{\sigma} = -1.5$ [0.25 mark]

$P(X > 35) = 0.0228 \Rightarrow P\left(Z > \frac{35-\mu}{\sigma}\right) = 0.0228$

Since $P(Z > 2) = 0.0228$ , $\frac{35-\mu}{\sigma} = 2$ [0.25 mark]

Solving: $25 - \mu = -1.5\sigma$ and $35 - \mu = 2\sigma$

Subtracting: $10 = 3.5\sigma$ , so $\sigma = \frac{20}{7} \approx 2.857$ or $2.86$ (3 s.f.) [0.25 mark]

Then $\mu = 25 + 1.5 \times \frac{20}{7} = 25 + \frac{30}{7} = \frac{205}{7} \approx$ $29.3$ (3 s.f.) [0.25 mark]

(b) $P(28 < X < 40)$ : Convert to Z-scores

$Z_1 = \frac{28 - \frac{205}{7}}{\frac{20}{7}} = \frac{196-205}{20} = \frac{-9}{20} = -0.45$ [0.17 mark]

$Z_2 = \frac{40 - \frac{205}{7}}{\frac{20}{7}} = \frac{280-205}{20} = \frac{75}{20} = 3.75$ [0.17 mark]

$P(-0.45 < Z < 3.75) \approx P(Z < 3.75) - P(Z < -0.45)$

$\approx 0.9999 - (1 - 0.6736) = 0.9999 - 0.3264 \approx$ $0.673$ or $0.674$ [0.33 mark for method and answer]

Using more precise values: approximately 0.6736 or accept 0.674 or 0.67 (2 s.f.)

Question 20 [2 marks]

(a) P(same colour) = P(both red) + P(both blue) [0.25 mark]

$= \frac{4}{10} \times \frac{3}{9} + \frac{6}{10} \times \frac{5}{9}$ [0.25 mark]

$= \frac{12}{90} + \frac{30}{90} = \frac{42}{90} = \frac{7}{15}$ [0.5 mark]

(b) Exactly 2 red out of 3 drawn: Cases are RRD, RDR, DRR (where D = not red = blue)

Each has probability $\frac{4}{10} \times \frac{3}{9} \times \frac{6}{8} = \frac{72}{720} = \frac{1}{10}$ [0.5 mark for one case or using combinations]

Number of arrangements = $\binom{3}{2} = 3$ [0.25 mark]

Total probability = $3 \times \frac{4 \times 3 \times 6}{10 \times 9 \times 8} = 3 \times \frac{72}{720} = 3 \times \frac{1}{10} = \frac{3}{10}$ [0.25 mark]

Alternative using combinations: $\frac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}} = \frac{6 \times 6}{120} = \frac{36}{120} = \frac{3}{10}$

END OF ANSWER KEY

Mark summary: Section A (20 marks) + Section B (10 marks) + Section C (10 marks) = 40 marks total