Mathematics 9709/52 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
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.
(a) Find the probability that Rajesh is successful for the first time on his 7th attempt. [1]
(b) Find the probability that Rajesh is successful for the first time before his 6th attempt. [2]
(c) Find the probability that Rajesh is successful for the second time on his 10th attempt. [2]
Mark scheme
1(a) [( 0.7 ) × 0.3] = 0.0353 B1 352947
6
or 0.03529… to at least 3sf.
10000000
1
1(b) Method 1
[P(X < 6) =] 1 − 0.75 M1 1 – 0.7d, d = 5, 6.
= 0.832 A1 Accept 0.83193 to at least 3sf.
If M0 scored, SC B1 for 0.8319[3].
Method 2
[P(X < 6) =] (M1) 0.3 + ( 0.3)( 0.7 ) + ( 0.3)( 0.7 ) + ( 0.3)( 0.7 ) + ( 0.3 )( 0.7 ) + ( 0.3 )( 0.7
2 3 4
0.3 + ( 0.3)( 0.7 ) + ( 0.3)( 0.7 ) + ( 0.3)( 0.7 ) + ( 0.3)( 0.7 )
2 3 4
= 0.832 (A1) Accept 0.83193 to at least 3sf.
If M0 scored, SC B1 for 0.8319[3].
2
1(c) ( 0.7 ) × ( 0.3) × 9 C1 M1 ( 0.7 ) × ( 0.3) × k , k a positive integer, 1 may be implied.
8 2 8 2
or ( 0.7 ) × ( 0.3) × 9 C1 × ( 0.3) No addition/subtraction/additional terms.
8
= 0.0467 A1
2
Seva has a coin which is biased so that when it is thrown the probability of obtaining a head is 13 . 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.
(a) Find the probability that Seva selects at least one red marble. [3]
(b) Find the probability that Seva obtains a head given that he selects no red marbles. [2]
Mark scheme
2(a) Method 1
P(HR) + P(TR) + P(TBR) B1 Two of the calculations for P(HR), P(TBR), either P(TR) or
1 4 2 4 2 5 4 P(TRR) + P(TRB) unsimplified, ignore any identification.
× + × + × ×
3 9 3 9 3 9 8 4 1
Condone = in the unsimplified calculation.
8 2
or Condone use of tree diagram to show calculation if values correct
P(HR) + (P(TRR) + P(TRB)) + P(TBR) at end.
1 4 2 4 3 2 4 5 2 5 4
× + × × + × × + × × M1 Values of all correct identified scenarios added.
3 9 3 9 8 3 9 8 3 9 8
Correct branches may be identified on the tree diagram.
4 8 5 17 A1 0.6296…, 0.630
= 27 + 27 + 27 = 27 If M0 scored SC B1 for acceptable answers, WWW.
Method 2
1 5 2 5 4 5 5 (B1) One calculation of P(HB), P(TBB), unsimplified, ignore any
1 – P(HB) – P(TBB) = 1 − × + × × = 1 − − identification.
3 9 3 9 8 27 27
1 – probability must be seen.
Condone use of tree diagram to show calculation if values correct
at end.
(M1) 1 – values of two correct identified scenarios subtracted.
Correct branches may be identified on the tree diagram.
17 (A1) 0.6296…, 0.630
= If M0 scored SC B1 for acceptable answers, WWW.
27
2(a) Method 3
P(HR) + P(T, (1 – no R)) = (B1) Calculation for P(T, (1 – no R)) seen unsimplified.
Condone use of tree diagram to show calculation if values correct
1 4 2 5 4 at end.
× + 1 − × =
3 9 3 9 8
(M1) Values of two correct identified scenarios added.
Correct branches may be identified on the tree diagram.
4 2 20
27 + 3 1 − 27
17 (A1) 0.6296…, 0.630
= If M0 scored SC B1 for acceptable answers, WWW.
27
3
2(b) Method 1
P ( head no reds ) M1 d
or
d d
or 10 , 0 < d < 1.
P(head | no reds) = 1 − their ( a ) 1 − 27
P ( no reds )
17
27
1 5 5 10
× Condone = 0.3704 or more accurate.
3 9 27 27
= = =
17 10
1−
27 27
1 A1 OE
=
2 Condone 0.499 [9].
Method 2
P ( head blue ) (M1) d
or
d
P(head | no reds) = , 0 < d < 1.
P ( HB ) + P ( TBB ) 1 5 2 5 4 10
× + × ×
3 9 3 9 8 27
1 5 5
× 10
3 9 Condone = 0.3704 or more accurate.
= = 27 = 27
1 5 2 5 4 10
× + × ×
3 9 3 9 8 27
1 (A1) OE
=
2 Condone 0.499 [9] .
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.
(a) Find the percentage of oranges that are classified as large. [3]
(b) Find the greatest possible weight of a small orange. [3]
Mark scheme
3(a) 184 − 131 M1 Use of ± standardisation formula with 184, 131 and 54, no
P(X >184) = P Z > [= P( Z > 0.9815) ] continuity correction.
54
Condone use of σ 2 , σ .
1 – 0.837 [= 0.163] M1 Calculating the appropriate probability area (leading to their final
answer).
Percentage [= 0.163 ×100] = 16.3 A1 AWRT
3
3(b) w − 131 B1 −0.842 ⩽ z < −0.8415 or 0.8415 < z ⩽ 0.842 seen.
[P(X < w) = P(Z < ) = 0.2]
54
w − 131 M1 Use of the ± standardisation formula with 131, 54, w and a z-value
= −0.842 (not 0.2, 0.8, 0.158, 0.508[0], 0.492[0], 0.7881, 0.2119, 0.5593,
54
0.4407).
w = 85.5 A1 85.5 ⩽ p ⩽ 85.6
Signs must be consistent to create a positive answer.
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.
(a) Find the median and the interquartile range of the salaries of the Petral employees. [3]
The median salary of the Ravon employees is $33 800, the lower quartile is $32 000 and the upper quartile is $34 400.
(b) 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]
(c) Comment on whether the mean or the median would be a better representation of the data for the employees at Petral. [1]
Mark scheme
4(a) Median = 32000 B1 Clearly identified, e.g. Q2, med.
Accept 32 k.
[UQ = 33500, LQ = 31200] M1 33300 ⩽ UQ ⩽ 33700 – 31100 ⩽ LQ ⩽ 31200
[IQR =] 33500 – 31200 Implied if both quartile values are stated and an appropriate IQR is
calculated accurately.
= 2300 A1 WWW
Ignore $ signs.
If M0 scored, SC B1 for 2300 WWW.
If key ignored consistently:
B0 Median = 320
SC M1 325 ⩽ UQ ⩽ 335 – 311 ⩽ LQ ⩽ 312
SC A1 23.
3
4(b) Box-and-whisker plot on provided grid B1 All five key values for R plotted accurately in standard format
using a linear scale with at least three linear values. Labelled R.
Condone whiskers through box or at corners of boxes or extending
1
2
square beyond limit.
Scale no less than 1 cm = $1000.
Daylight rule applied to vertical lines of box.
B1FT 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.
Condone whiskers through box or at corners of boxes or extending
1
2
square beyond limit.
R 30 200 32 000 33 800 34 400 36 900 Scale no less than 1 cm = $1000.
Daylight rule applied to vertical lines of box.
P 30 000 31 200 32 000 33 500 36 800
B1 Whiskers not through box (condone 1 square in box) for either, not
2
drawn at corners of boxes. single linear scale for the diagram and
labelled ‘salaries’ (OE) and $.
If only one plot attempted, SC B1 for meeting all the requirements
above.
3
4(c) Median because there is an extreme value ($36 800) B1 Do not accept ‘values’.
Must identify median and reference either the extreme value
(anomaly, outlier, 36 800) or the skew in context (e.g. concentrated
in lower values, positive skew).
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.
(a) Show that P (X = 7) = 0.2 . [1]
(b) Draw up the probability distribution table for X . [3]
(c) Find the value of Var (X ) . [3]
Mark scheme
5(a) [$7 =] [$]5 + [$]2 B1 AG
1 2 1 Must include [$7], 5, 2 and link the probabilities to the appropriate
[Probability =] × × 2 = = 0.2 value
5 4 5
Or 1 C1 × 2 C1
= 0.2.
[Probability =] 0.2 × 0.5 × 2 = 0.2 5
C2
1 2 2 1 1 2 1 2
× + × , not × + × unless 5 and 2 and 2 and 5 seen in
5 4 5 4 5 4 5 4
solution.
If all possibilities identified (e.g. outcome table), must be clearly
labelled and terms fulfilling the condition identified.
1
5(b) B1 Table with correct x values and at least one further non-zero
x 2 3 4 6 7
probability correct.
P(X = x) 0.1 0.4 0.1 0.2 0.2 Condone extra x values if probability stated as 0.
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.
SC B1 for four further non-zero probabilities adding to 0.8 if B1
max scored.
3
5(c) [E(X) = 0.1 × 2 + 0.4 × 3 + 0.1 × 4 + 0.2 × 6 + 0.2 × 7] M1 Accept unsimplified expression. May be calculated in the variance,
0.2 + 1.2 + 0.4 + 1.2 + 1.4 [ = 4.4] FT their table with at least 5 probabilities, 0 < p < 1, that sum to 1.
FT acceptable at the bold partially evaluated stage.
[Var ( X ) = 0.1 × 22 + 0.4 × 32 + 0.1 × 42 + 0.2 × 62 M1 Appropriate variance formula using their (E(X))2 value. FT their
table with at least 4 probabilities, 0 < p < 1, that may not sum to 1.
+0.2 × 7 2 − 4.42 FT acceptable at the bold partially evaluated stage.
0.1× 4 + 0.4 × 9 + 0.1×16 + 0.2 × 36 + 0.2 × 49 − 4.42 Note: if table is correct, 22.6 – (4.42 or 19.36) implies this M1.
= 3.24 A1 CAO
81 6
,3 scores A0.
25 25
Only dependent upon previous M1 (M0 M1 A1 possible).
If M0 M0 scored, SC B1 for 3.24 WWW.
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) 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]
(b) 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]
(c) 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]
Mark scheme
6(a) Mean [=110 ×0.25] = 27.5 B1 27.5 and 20.625 (CAO) seen, allow unsimplified.
165 May be in standardisation formula
Variance [=110 ×0.25 × 0.75] = 20.625,
8 165 330
(4.541475… to at least 4sf or or implies correct
8 4
variance).
Penalise incorrect identification, condone no identification.
21.5 − 27.5 M1 Substituting their 27.5 and their 20.625 into the ± standardising
P(X < 22) = P Z <
20.625 formula (any number for 21.5), not σ 2 , not σ .
M1 Using continuity correction 21.5 or 22.5 in their standardisation
formula.
[P( Z < −1.3212) = 1 − Φ (1.3212 )] M1 Appropriate probability area, from final process, must be a
1 – 0.9068 = probability.
May be implied by a sketch of the required probability area.
Expect final answer < 0.5.
0.0932 A1 0.0932 ⩽ p < 0.09325
If either M1 M1 not awarded for standardisation and/or M1 not
awarded for finding probability area, SC B1 0.0932 ⩽ p < 0.09325
WWW.
5
6(b) Method 1
[1 – P(8, 9, 10) = ] M1 One term 10Cx ( p ) x (1 − p )10− x , 0 < p < 1, x ≠ 0 or 10.
1 – (10C8 0.858 0.152 + 10C9 0.8590.151 + 0.8510)
A1 Correct unsimplified expression.
Condone omission of last bracket only.
[ = 1 – (0.275897 + 0.347425 + 0.196874)]
= 0.180 B1 0.1795 < p ⩽ 0.180
Method 2
[P(0, 1, 2, 3, 4, 5, 6, 7) = ] (M1) One term 10Cx ( p ) x (1 − p )10− x , 0 < p < 1, x ≠ 0 or 10.
0.1510 + 10C1 0.85×0.159 + … + 10C7 0.8570.153
(A1) Correct unsimplified expression.
= 0.180 (B1) 0.1795 < p ⩽ 0.180
3
6(c) 0.25 × 0.6 × 0.15 × 6 M1 0.25 × 0.6 × 0.15 × k , k an integer > 1.
27 A1
0.135,
200
2
(a) How many different arrangements are there of the 10 letters in the word REGENERATE? [1]
(b) 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]
(c) 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]
Additional page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) 10! B1
2!4! = 75600
1
7(b) 4! × 3! 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.
144 B1 WWW
4
7(c) Method 1: If denominator is from 7(a), no denominator or incorrect denominator
[Numerator = Number of required arrangements =] B1 5!
seen (arrangements of consonants).
2!
5! 5!
× × 2 [ = 600]
2! 4! B1 5!
seen (arrangements of vowels).
4!
M1 5! 5!
× × 2 , r = 1 or 2, s = 1, 4, 4! or 24.
r s
their 600 M1 their 600 their 600
[Probability =] or .
their 75600 their ( a ) 75600
1 A1 600
= , 0.00794 Accept OE.
126 75600
Method 2: If denominator 10!
[Numerator = Number of required arrangements =] (B1) 5! seen (arrangements of consonants).
5! × 5! × 2 [ = 28800] (B1) A second 5! seen (arrangements of vowels).
(M1) 5! × 5! × k , k = 1 or 2.
their 28800 (M1)
[Probability = ]
10!
1 (A1) 600
= , 0.00794 Accept OE.
126 75600
7(c) Method 3: Using probabilities
5 5 4 4 3 3 2 2 1 1 (B1) 5 4 3 2 1
× × × × × × × × × ×2 × × × × seen.
10 9 8 7 6 5 4 3 2 1 a b c d e
10 ≥ a > b > c > d > e ≥ 1 (arrangements of consonants).
(B1) 5 4 3 2 1
A second × × × × seen.
f g h i j
10 ≥ f > g > h > i > j ≥ 1 (arrangements of vowels).
(M1) 5 4 3 2 1 5 4 3 2 1
× × × × × × × × × ×k
a b c d e f g h i j
k = 1 or 2.
(M1) their 5 × 4 × 3 × 2 × 1 × 5 × 4 × 3 × 2 × 1
.
10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
1 (A1) 28800
= , 0.00794 Accept OE.
126 362800
5
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