A Level Mathematics 9709 Paper 52, May/June 2024

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Question paper

9709/52 · May/June 2024

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Rajesh applies once every year for a ticket to a music festival. The probability that he is successful in any particular year is 0.3, independently of other years.

Find the probability that Rajesh is successful for the first time on his 7th attempt. [1]

[1]

Find the probability that Rajesh is successful for the first time before his 6th attempt. [2]

[2]

Find the probability that Rajesh is successful for the second time on his 10th attempt. [2]

[2]

Seva has a coin which is biased so that when it is thrown the probability of obtaining a head is $\frac{1}{3}$ . He also has a bag containing 4 red marbles and 5 blue marbles. Seva throws the coin. If he obtains a head, he selects one marble from the bag at random. If he obtains a tail, he selects two marbles from the bag at random and without replacement.

Find the probability that Seva selects at least one red marble. [3]

[3]

Find the probability that Seva obtains a head given that he selects no red marbles. [2]

[2]

The weights of oranges can be modelled by a normal distribution with mean 131 grams and standard deviation 54 grams. Oranges are classified as small, medium or large. A large orange weighs at least 184 grams and 20% of oranges are classified as small.

Find the percentage of oranges that are classified as large. [3]

[3]

Find the greatest possible weight of a small orange. [3]

[3]

The back-to-back stem-and-leaf diagram shows the annual salaries of 19 employees at each of two companies, Petral and Ravon.

Petral							Ravon
			$3$	$0$	$0$	$30$	$2$	$6$
$9$	$9$	$8$	$2$	$2$	$1$	$31$	$1$	$5$
		$5$	$5$	$4$	$0$	$32$	$0$	$0$	$2$
			$7$	$5$	$3$	$33$	$0$	$4$	$8$	$9$
				$1$	$0$	$34$	$1$	$1$	$3$	$4$	$6$
						$35$	$3$
					$8$	$36$	$7$	$9$

Key: $2 ∣ 31 ∣ 5$ means $$ 31 200$ for a Petral employee and $$ 31 500$ for a Ravon employee.

Find the median and the interquartile range of the salaries of the Petral employees. [3]

[3]

The median salary of the Ravon employees is $$ 33 800$ , the lower quartile is $$ 32 000$ and the upper quartile is $$ 34 400$ . Represent the data shown in the back-to-back stem-and-leaf diagram by a pair of box-and-whisker plots in a single diagram. [3]

[3]

Comment on whether the mean or the median would be a better representation of the data for the employees at Petral. [1]

[1]

Jasmine has one $$ 5$ coin, two $$ 2$ coins and two $$ 1$ coins. She selects two of these coins at random. The random variable $X$ is the total value, in dollars, of these two coins.

Show that $P (X = 7) = 0.2$ . [1]

[1]

Draw up the probability distribution table for $X$ . [3]

[3]

Find the value of $Var (X)$ . [3]

[3]

The residents of Mahjing were asked to classify their local bus service: • 25% of residents classified their service as good. • 60% of residents classified their service as satisfactory. • 15% of residents classified their service as poor.

A random sample of 110 residents of Mahjing is chosen. Use a suitable approximation to find the probability that fewer than 22 residents classified their bus service as good. [5]

[5]

For a random sample of 10 residents of Mahjing, find the probability that fewer than 8 classified their bus service as good or satisfactory. [3]

[3]

Three residents of Mahjing are selected at random. Find the probability that one resident classified the bus service as good, one as satisfactory and one as poor. [2]

[2]

How many different arrangements are there of the 10 letters in the word REGENERATE? [1]

[1]

How many different arrangements are there of the 10 letters in the word REGENERATE in which the 4 Es are together and the 2 Rs have exactly 3 letters in between them? [4]

[4]

Find the probability that a randomly chosen arrangement of the 10 letters in the word REGENERATE is one in which the consonants (G, N, R, R, T) and vowels (A, E, E, E, E) alternate, so that no two consonants are next to each other and no two vowels are next to each other. [5]

[5]

Mark scheme

9709/52 · May/June 2024

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1a · 1 mark

B1
[(0.7)^6 × 0.3] = 0.0353

1b · 2 marks

M1
[P(X < 6) =] 1 - 0.7^5
A1
= 0.832

1c · 2 marks

M1
(0.7)^8 × (0.3)^2 × ^9C_1 or (0.7)^8 × (0.3) × ^9C_1 × (0.3)
A1
= 0.0467

2a · 3 marks

B1
Two of the calculations for P(HR), P(TBR), either P(TR) or P(TRR) + P(TRB) unsimplified, ignore any identification.
M1
Values of all correct identified scenarios added.
A1
= 17/27

2b · 2 marks

M1
d / (1 - their(a)) or d / (1 - 17/27) or d / (10/27), 0 < d < 1.
A1
= 1/2

3a · 3 marks

M1
Use of ±standardisation formula with 184, 131 and 54, no continuity correction. Condone use of σ^2, √σ.
M1
Calculating the appropriate probability area (leading to their final answer).
A1
Percentage [= 0.163 × 100] = 16.3

3b · 3 marks

B1
-0.842 ⩽ z < -0.8415 or 0.8415 < z ⩽ 0.842 seen.
M1
Use of the ±standardisation formula with 131, 54, w and a z-value (not 0.2, 0.8, 0.158, 0.508[0], 0.492[0], 0.7881, 0.2119, 0.5593, 0.4407).
A1
w = 85.5

4a · 3 marks

B1
Median = 32000
M1
[UQ = 33500, LQ = 31200] [IQR =] 33500 – 31200
A1
= 2300

4b · 3 marks

B1
All five key values for R plotted accurately in standard format using a linear scale with at least three linear values. Labelled R.
B1
All five key values for P, FT from (a), plotted accurately in standard format using a linear scale with at least three linear values. Labelled P.
B1
Whiskers not through box (condone 1/2 square in box) for either, not drawn at corners of boxes. single linear scale for the diagram and labelled ‘salaries’ (OE) and $.

4c · 1 mark

B1
Median because there is an extreme value ($36800)

5a · 1 mark

B1
[ $7 =] [$ ]5 + [$]2 [Probability =] 1/5 × 2/4 × 2 = 1/5 = 0.2 Or [Probability =] 0.2 × 0.5 × 2 = 0.2

5b · 3 marks

B1
Table with correct x values and at least one further non-zero probability correct.
B1
Two more correct non-zero probabilities linked with correct outcomes. Accept probabilities not in table if clearly identified.
B1
All five probabilities correct. Accept probabilities not in table if clearly identified.

5c · 3 marks

M1
[E(X) = 0.1 × 2 + 0.4 × 3 + 0.1 × 4 + 0.2 × 6 + 0.2 × 7] 0.2 + 1.2 + 0.4 + 1.2 + 1.4 [ = 4.4]
M1
[Var(X) = 0.1 × 2^2 + 0.4 × 3^2 + 0.1 × 4^2 + 0.2 × 6^2 + 0.2 × 7^2 - 4.4^2 0.1 × 4 + 0.4 × 9 + 0.1 × 16 + 0.2 × 36 + 0.2 × 49 - 4.4^2
A1
= 3.24

6a · 5 marks

B1
Mean [=110 × 0.25] = 27.5 Variance [=110 × 0.25 × 0.75] = 20.625, 165/8
M1
P(X < 22) = P(Z < (21.5 - 27.5) / √20.625)
M1
Using continuity correction 21.5 or 22.5 in their standardisation formula.
M1
[P(Z < -1.3212) = 1 - Φ(1.3212)] 1 – 0.9068 =
A1
0.0932

6b · 3 marks

M1
One term ^10C_x (p)^x (1 - p)^(10-x), 0 < p < 1, x ≠ 0 or 10.
A1
Correct unsimplified expression.
B1
= 0.180

6c · 2 marks

M1
0.25 × 0.6 × 0.15 × 6
A1
0.135, 27/200

7a · 1 mark

B1
[ 10! / (2!4!) = ] 75600

7b · 4 marks

M1
4! SOI in all terms leading to final answer. Allow 24 if 4! = 24 is seen.
M1
Ignoring any values used to justify 4!. Either 3! SOI in expression leading to final answer, or at least 6 distinct scenarios identified and added in expression leading to final answer. Condone 3 distinct scenarios × 2. Ignore repeated scenarios.
A1
4! × 3! Fully correct unsimplified expression leading to final answer.
B1
144

7c · 5 marks

B1
5! / 2! seen (arrangements of consonants).
B1
5! / 4! seen (arrangements of vowels).
M1
(5! / r) × (5! / s) × 2, r = 1 or 2, s = 1, 4, 4! or 24.
M1
[Probability =] their 600 / their 75600 or their 600 / their (a).
A1
= 1/126 , 0.00794

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