Mathematics 9709/51 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Two fair coins are thrown at the same time. The random variable X is the number of throws of the two coins required to obtain two tails at the same time.
(a) Find the probability that two tails are obtained for the first time on the 7th throw. [2]
(b) Find the probability that it takes more than 9 throws to obtain two tails for the first time. [2]
Mark scheme
1(a) 3 1
6 M1 (1 − p ) p, 0 < p < 1
6
4 4
729 A1
0.0445,
16384
2
1(b) 3
9 M1 3
n
n
or p , 0 < p < 1, n = 8, 9,10
4 4
19683 A1
0.0751,
262144
2
A summary of 40 values of x gives the following information:
Σ x − k = 520, Σ x − k2 = 9640, where k is a constant.
(a) Given that the mean of these 40 values of x is 34, find the value of k. [2]
(b) Find the variance of these 40 values of x. [2]
Mark scheme
2(a) Σ x Σ(x − k ) M1 Forms an equation involving Σx, Σ(x-k) and k.
−k = Accept at a numeric stage with k.
40 40
40 × 34 520
−k =
40 40
k [ = 34 − 13] = 21 A1 Evaluated.
2
2(b) Σ ( x − k )2 Σ ( x − k ) 2 9640 520 2 M1 Values substituted into an appropriate variance formula,
Var = − = − 2
= [241 – 13 =] accept unsimplified.
40 40 40 40
72 A1
2
For her bedtime drink, Suki has either chocolate, tea or milk with probabilities 0.45, 0.35 and 0.2 respectively. When she has chocolate, the probability that she has a biscuit is 0.3. When she has tea, the probability that she has a biscuit is 0.6. When she has milk, she never has a biscuit.
Find the probability that Suki has tea given that she does not have a biscuit. [5]
Mark scheme
3 P (T ∩ B′ ) M1 0.45 × a + 0.35 × b + 0.2 [×1] , a = 0.7,0.3 b = 0.4,0.6 , seen
P ( T |B ′ ) =
P ( B ′ ) anywhere.
P ( B′ ) = 0.45 × 0.7 + 0.35 × 0.4 + 0.2 × 1 A1 Correct, accept unsimplified.
131
= 0.655, 200
7 M1 Seen as numerator or denominator of a fraction.
P (T ∩ B ') = 0.35 × 0.4 [= 0.14, ]
50
their 0.14 M1 Values substituted into conditional probability formula
P (T | B ') = correctly. Accept unsimplified.
their 0.655
Denominator sum of 3 two-factor probabilities (condone
omission of 1 from final factor).
If clearly identified, condone from incomplete
denominator.
28 A1 If 0 marks awarded, SC B1 0.214 WWW.
0.214,
131
5
A fair spinner has edges numbered 0, 1, 2, 2. Another fair spinner has edges numbered −1, 0, 1. Each spinner is spun. The number on the edge on which a spinner comes to rest is noted. The random variable X is the sum of the numbers for the two spinners.
(a) Draw up the probability distribution table for X . [3]
(b) Find Var X . [3]
Mark scheme
4(a) x -1 0 1 2 3 B1 0 1 2 2
p 1 2 4 3 2
= 0.0833 = 0.167 = 0.333 = 0.25 = 0.167
12 12 12 12 12 –1 –1 0 1 1
0 0 1 2 2
1 1 2 3 3
Table with x values and at least one probability
substituted, 0 < p < 1.
Condone any additional x values if probability stated as 0.
B1 2 correct identified probabilities.
B1 All probabilities correct (accept to 3sf).
SC if less than 2 correct probabilities:
SC B1 4 or 5 probabilities summing to one.
3
4(b) 1 4 6 6 15 M1 May be implied by use in Variance, accept unsimplified
E(X) = − + + + =
12 12 12 12 12 expression.
Probabilities must sum to 1 ± 0.001.
1 4 12 18 15
2 M1 Appropriate variance formula using their (E(X))2 . FT
Var (X) = +0+ + + − accept probabilities not summing to 1.
12 12 12 12 12 2
35 15 35 25
Condone − or − from correct table.
12 12 12 9
35 25 65 A1 WWW
12 − 16 = 48 , 1.35
3
Raman and Sanjay are members of a quiz team which has 9 members in total. Two photographs of the quiz team are to be taken.
For the first photograph, the 9 members will stand in a line.
(a) How many different arrangements of the 9 members are possible in which Raman will be at the centre of the line? [1]
(b) How many different arrangements of the 9 members are possible in which Raman and Sanjay are not next to each other? [3]
For the second photograph, the members will stand in two rows, with 5 in the back row and 4 in the front row.
(c) In how many different ways can the 9 members be divided into a group of 5 and a group of 4?
[2]
(d) For a random division into a group of 5 and a group of 4, find the probability that Raman and
Sanjay are in the same group as each other. [4]
Mark scheme
5(a) [8! =] 40 320 B1 Evaluated, exact value only.
1
5(b) Method 1 [^ ^ ^ R ^ ^ S ^ ^]
7! × 8C2 × 2 M1 7! × k seen, k an integer > 1.
M1 m × n ( n − 1) or m × nC2 or m × n P2 , n = 7, 8 or 9, m an
integer > 1.
282 240 A1 Exact value only.
SC B1 for final answer 282 240 WWW.
Method 2 [Total number of arrangements – Arrangements with R & S together]
9! – 8! × 2 M1 9! – k, k an integer < 362 880 .
M1 m – 8! × n, m an integer > 40 320, n = 1,2.
282 240 A1 Exact value only.
SC B1 for final answer 282 240 WWW.
3
9
5(c) C5 [× 4C4] M1 9
Cx [× 9–xC9–x,] x = 4, 5. Condone × 1 for 9–xC9–x.
Condone use of P.
126 A1 WWW
2
5(d) [Number of ways with Raman and Sanjay together on back row =] 7C3 M1 7
Cx seen, x = 3 or 2.
[Number of ways with Raman and Sanjay together on front row =] 7C2
[Total =] 35 + 21 M1 Summing two correct scenarios.
56 A1 Evaluated – may be seen used in probability.
If M0 scored, SC B1 for 56 WWW.
their 56 56 4 B1 FT FT their 56 from adding 2 or more scenarios in numerator
Probability = = , , 0.444 and their (c) or correct as denominator.
their ( c ) 126 9
4
The weights, in kg, of 15 rugby players in the Rebels club and 15 soccer players in the Sharks club are shown below.
Rebels 75 78 79 80 82 82 83 84 85 86 89 93 95 99 102
Sharks 66 68 71 72 74 75 75 76 78 83 83 84 85 86 92
(a) Represent the data by drawing a back-to-back stem-and-leaf diagram with Rebels on the left-hand side of the diagram. [4]
(b) Find the median and the interquartile range for the Rebels. [3]
A box-and-whisker plot for the Sharks is shown below.
Sharks
60 70 80 90 100 110
Weight (kg)
(c) On the same diagram, draw a box-and-whisker plot for the Rebels. [2]
(d) Make one comparison between the weights of the players in the Rebels club and the weights of the players in the Sharks club. [1]
Mark scheme
6(a) Rebels Sharks B1 Correct stem, ignore extra values (not in reverse).
6 6 8
9 8 5 7 1 2 4 5 5 6 8 B1 Correct Rebels labelled on left, leaves in order from right
9 6 5 4 3 2 2 0 8 3 3 4 5 6 to left and lined up vertically, no commas.
9 5 3 9 2
2 10 B1 Correct Sharks labelled on same diagram, leaves in order
and lined up vertically, no commas.
Key: 8 | 7 | 2 means 78 kg for Rebels and 72 kg for Sharks
B1 Correct key for their diagram, need both teams identified
and ‘kg’ stated at least once here or in leaf headings or
title.
SC If 2 separate diagrams drawn, SC B1 if both keys
meet these criteria.
4
6(b) Median = 84 (kg) B1
[UQ = 93, LQ = 80] 93 – 80 M1 95 ⩽ UQ ⩽ 89 – 79 ⩽ LQ ⩽ 82
[IQR =] 13 (kg) A1 WWW
3
6(c) Box and whisker with end points 75 and 102 B1 Whiskers drawn to correct end points not through box,
not joining at top or bottom of box.
Median and quartiles plotted as found in (b) B1 FT Quartiles and median plotted as box graph.
2
6(d) e.g. Average weight of Rebels is higher than average weight of Sharks B1 Acceptable answers refer to: Range, skew, central
tendency within context.
E.g. range of Rebels is greater B0.
Range of weights of the rebels is greater B1.
Simple value comparison insufficient.
1
The times, in minutes, that Karli spends each day on social media are normally distributed with mean 125 and standard deviation 24.
(a) (i) On how many days of the year (365 days) would you expect Karli to spend more than
142 minutes on social media? [5]
(ii) Find the probability that Karli spends more than 142 minutes on social media on fewer than
2 of 10 randomly chosen days. [3]
(b) On 90% of days, Karli spends more than t minutes on social media.
Find the value of t. [3]
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)(i) 142 − 125 M1 Substitution of correct values into the ±Standardisation
P(X > 142) = P Z > formula, allow continuity correction, not σ2, √σ.
24
[= P( Z > 0.7083) =]1 − 0.7604 M1 Appropriate numerical area Φ, from final process, must
be probability, expect p < 0.5.
0.2396 A1 0.239 ⩽ p ⩽ 0.240 to at least 3sf.
Their 0.2396 × 365 [= 87.454] M1 FT their 4sf (or better) probability.
87 or 88 A1 FT Final answer must be positive integer, no indication of
approximation/rounding, only dependent on previous M
mark.
SC B1 FT for their 3sf probability × 365 = integer value,
condone 0.24 used.
5
7(a)(ii) P(0, 1) = 0.760410 + 10C1 × 0.23961 × 0.76049 M1 One term: 10Cx px (1 – p)10–x for 0 < x < 10, any p.
[= 0.064628 + 0.20364]
A1 FT Correct unsimplified expression using their probability to
at least 3sf from (a)(i) or correct.
0.268 A1 AWRT, WWW.
3
7(b) z = ±1.282 B1 Correct value only, critical value.
t − 125 M1 Use of ± Standardisation formula with correct values
= −1.282 substituted, allow continuity correction, σ2, √ σ, to form
24
an equation with a z-value and not probability.
t = 94.2 A1 AWRT, condone AWRT 94.3. Not dependent on B mark.
3
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/51, October/November 2021 (9709_w21_ms_51), and the question paper (9709_w21_qp_51), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/51 for October/November 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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