Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 5 · Variant 1

1 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) 6 3 1 4 4       ) 6 1 , 0 1 − < < p p p 0.0445, 729 16384 A1 2 1(b) 9 3 4       M1 3 , 0 1, 8, 9,1 0 4   < < =     n n or p p n 0.0751, 19683 262144 A1 2

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2 A summary of 40 values of x gives the following information: Σ x −k = 520, Σ x −k 2 = 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) ( ) 40 40   −   − =     x k x k 40 34 520 40 40 × − = k Accept at a numeric stage with k. [ ] 34 13 21 = − = k A1 Evaluated. 2 Question Answer Marks Guidance 2(b) Var = ( ) ( ) 2 2 40 40    −  −    −        x k x k 2 9640 520 40 40   = − =     [241 – 132 =] M1 Values substituted into an appropriate variance formula, accept unsimplified. 72 A1 2

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3 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 P | P T B T B B   ∩ =      ′  ′ ′ ( ) P 0.45 0.7 0.35 0.4 0.2 1 B × ′ = × + + × 131 0.655, 200   =     M1 [ ] 0.45 0.35 0.2 1 , 0.7,0.3 0.4,0.6 × + × + × = = a b a b , seen anywhere. A1 Correct, accept unsimplified. ( ) P ' 0.35 0.4 T B ∩ = × [= 0.14, 7 50 ] M1 Seen as numerator or denominator of a fraction. ( ) 0.14 P | ' 0.655 their T B their = M1 Values substituted into conditional probability formula correctly. Accept unsimplified. Denominator sum of 3 two-factor probabilities (condone omission of 1 from final factor). If clearly identified, condone from incomplete denominator. 0.214, 28 131 A1 If 0 marks awarded, SC B1 0.214 WWW. 5

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5 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 ( ) 1 × − m n n or 2 × n m C or 2 × n m P , 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 5(c) 9C5 [× 4C4] M1 9Cx [× 9–xC9–x,] x = 4, 5. Condone × 1 for 9–xC9–x. Condone use of P. 126 A1 WWW 2 Question Answer Marks Guidance 5(d) [Number of ways with Raman and Sanjay together on back row =] 7C3 [Number of ways with Raman and Sanjay together on front row =] 7C2 M1 7Cx seen, x = 3 or 2. [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. Probability = ( ) 56 56 4 , 126 9 = their their c , 0.444 B1 FT FT their 56 from adding 2 or more scenarios in numerator and their (c) or correct as denominator. 4

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6 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 6 6 8 9 8 5 7 1 2 4 5 5 6 8 9 6 5 4 3 2 2 0 8 3 3 4 5 6 9 5 3 9 2 2 10 Key: 8 | 7 | 2 means 78 kg for Rebels and 72 kg for Sharks B1 Correct stem, ignore extra values (not in reverse). B1 Correct Rebels labelled on left, leaves in order from right to left and lined up vertically, no commas. B1 Correct Sharks labelled on same diagram, leaves in order and lined up vertically, no commas. 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 Question Answer Marks Guidance 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

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7 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] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a)(i) P(X > 142) = P 142 125 24 Z −   >     M1 Substitution of correct values into the ±Standardisation formula, allow continuity correction, not σ2, √σ. [= P( 0.7083) ]1 0.7604 > = − Z 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 [= 0.064628 + 0.20364] M1 One term: 10Cx px (1 – p)10–x for 0 < x < 10, any p. 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) 1.282 = ± z B1 Correct value only, critical value. 125 1.282 24 − = − t M1 Use of ± Standardisation formula with correct values substituted, allow continuity correction, σ2, √ σ, to form an equation with a z-value and not probability. 94.2 = t A1 AWRT, condone AWRT 94.3. Not dependent on B mark. 3

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