Cambridge A Level Mathematics 9709 — 2021 May/June Paper 5 · Variant 1

1 A bag contains 12 marbles, each of a different size. 8 of the marbles are red and 4 of the marbles are blue. How many different selections of 5 marbles contain at least 4 marbles of the same colour? [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 BBBBR 8C1 × 4C4 = 8 RRRRR 8C5 = 56 M1 0 ⩽ y ⩽ 4 condone 8C1 × 1 A1 Two correct outcomes evaluated M1 Add 2 or 3 identified correct scenarios only (no additional terms, not probabilities) [Total =] 344 A1 WWW, only dependent on 2nd M mark 4 SC not all (or no) scenarios identified B1 280 + 8 + 56 DB1 344

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2 A company produces a particular type of metal rod. The lengths of these rods are normally distributed with mean 25.2 cm and standard deviation 0.4 cm. A random sample of 500 of these rods is chosen. How many rods in this sample would you expect to have a length that is within 0.5 cm of the mean length? [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 ( ) ( ) 25.2 25.5 0.50 25.2 25.2 0.50 0.4 0.4      − +   − −  < <                     P z 0.5 0.5 0.4 0.4   = − < <     P z correction, σ2, √ σ ( ) 2 1.25 1 = Φ −   2 0.8944 1 = × − A1 For AWRT 0.8944 SOI M1 Appropriate area 2Φ – 1 OE, from final process, must be probability 0.7888 A1 Accept AWRT 0.789 Number of rods = 0.7888 500 × = 394 or 395 B1FT Correct or FT their 4SF (or better) probability, final answer must be positive integer, not 394.0 or 395.0, no approximation/rounding stated, only 1 answer 5

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3 (a) How many different arrangements are there of the 8 letters in the word RELEASED? 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(b) How many different arrangements are there of the 8 letters in the word RELEASED in which the letters LED appear together in that order? 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(c) An arrangement of the 8 letters in the word RELEASED is chosen at random. Find the probability that the letters A and D are not together. 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Mark scheme: 3(a) 8! 3!       = 6720 1 3(b) _ _ _ L E D _ _ : With LED together: 6! 2! M1 6! k or 5! 6 x k k ⩾ 1 and no other terms M1 2! m , m an integer, m ⩾ 5 360 A1 CAO 3 3(c) Method using _ _ _ A _ D _ _ : Arrange the 6 letters RELESE = 6! 3! [= 120] *M1 6! 3!×k seen, k an integer > 0 Multiply by number of ways of placing AD in non-adjacent places = their 120 × 7P2 [= 5040] *M1 ( ) 1 × − m n n or 2 × n m C or 2 × n m P , n = 6, 7 or 8, m an integer > 0 [Probability =] 5040 6720 their their DM1 Denominator = their (a) or correct, dependent on at least one M mark already gained. 5040 3 or or 0.75 6720 4 A1 Alternative method for Question 3(c) Method using ‘Total arrangements – Arrangements with A and D together’: Their 6720 – 7! 2 3! × [= 5040] *M1 Their 6720 – k, k a positive integer *M1 ( ) 7! , 1,2 3! × − = k m k Question Answer Marks Guidance [Probability =] 5040 6720 their their DM1 With denominator = their (a) or correct, dependent on at least one M mark already gained. 5040 3 or or 0.75 6720 4 A1 Alternative method for Question 3(c) Method using ‘1 – Probability of arrangements with A and D together’: 7! 2 3! × [= 1680] *M1 7! , 1,2 3! × = k k [Probability =] 1 680 6720 their their *M1 With denominator = their (a) or correct 1 – 1 680 6720 their their DM1 1 – m, 0 < m < 1 , dependent on at least one M mark already gained 5040 3 or or 0.75 6720 4 A1 4

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4 To gain a place at a science college, students first have to pass a written test and then a practical test. Each student is allowed a maximum of two attempts at the written test. A student is only allowed a second attempt if they fail the first attempt. No student is allowed more than one attempt at the practical test. If a student fails both attempts at the written test, then they cannot attempt the practical test. The probability that a student will pass the written test at the first attempt is 0.8. If a student fails the first attempt at the written test, the probability that they will pass at the second attempt is 0.6. The probability that a student will pass the practical test is always 0.3. (a) Draw a tree diagram to represent this information, showing the probabilities on the branches. [3] (b) Find the probability that a randomly chosen student will succeed in gaining a place at the college. 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(c) Find the probability that a randomly chosen student passes the written test at the first attempt given that the student succeeds in gaining a place at the college. 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Mark scheme: 4(a) branches clearly identifying written and practical, pass and fail for each intersection (no additional branches) B1 ‘One written test’ branch all probabilities (or %) correct B1 ‘Two written tests’ branch all probabilities (or %) correct, condone additional branches after W2F with probabilities 1 for PF and 0 for PP 3 4(b) [P(W1P) × P(PP) + P(W1F) × P(W2P) × P(PP)] 0.8 0.3 0.2 0.6 0.3 × + × × M1 Consistent with their tree diagram or correct 0.276 or 69 250 A1 2 4(c) ( ) ( ) ( ) ( ) 1 Practical 0.8 0.3 1 getting place ∩ × = = P W P W P P their b 0.24 0.276   =     M1 Correct expression or FT their (b) 20 23 or 0.87[0] A1 2

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5 The times taken by 200 players to solve a computer puzzle are summarised in the following table. Time (t seconds) 0 ≤t < 10 10 ≤t < 20 20 ≤t < 40 40 ≤t < 60 60 ≤t < 100 Number of players 16 54 78 32 20 (a) Draw a histogram to represent this information. [4] (b) Calculate an estimate of the mean time taken by these 200 players. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the greatest possible value of the interquartile range of these times. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Class width 10 10 20 20 40 Frequency Density 1.6 5.4 3.9 1.6 0.5 M1 At least 4 frequency densities calculated, accept unsimplified. May be read from graph using their scale, 3SF or correct A1 All heights correct on graph B1 Bar ends at 0, 10, 20 …, etc. with a horizontal linear scale with at least 3 values indicated, 0 ⩽ horizontal axis ⩽ 100 B1 Axes labelled: Frequency density (fd), time (t) and seconds. Linear vertical scale, with at least 3 values indicated 0 ⩽ vertical axis ⩽ 5.4 4 Question Answer Marks Guidance 5(b) Mean = 16 5 54 15 78 30 32 50 20 80 200 × + × + × + × + ×       80 810 2340 1600 1600 200 + + + + = M1 Uses at least 4 midpoint attempts (e.g. 5 ± 0.5). Accept unsimplified expression, denominator either correct or their Σfrequencies 6430 3 32 or 32.15 200 20   =     A1 Accept 32.2 2 5(c) A value in correct UQ (40–60) – a value in correct LQ (10–20) M1 Greatest possible value is 60 – 10 = 50 A1 Condone 49.9 2

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6 In Questa, 60% of the adults travel to work by car. (a) A random sample of 12 adults from Questa is taken. Find the probability that the number who travel to work by car is less than 10. 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(b) A random sample of 150 adults from Questa is taken. Use an approximation to find the probability that the number who travel to work by car is less than 81. 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(c) Justify the use of your approximation in part (b). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) [= 1 – (0.063852 + 0.017414 + 0.0021768)] A1 Correct unsimplified expression, or better. [1 – 0.083443] = 0.917 A1 AWRT Alternative method for Question 6(a) P (0,1,2,3,4,5,6,7,8,9) = 12C00.60 0.412 + 12C1 0.61 0.411+ ………….12C9 0.69 0.43 [= 0.000016777 + 0.00030199 + 0.0024914 + 0.012457 + 0.042043 + 0.10090 + 0.17658 + 0.22703 + 0.21284 + 0.14189] M1 One term: 12Cx px (1 – p)12-x for 0 < x < 12, any p allowed. A1 Correct unsimplified expression with at least the first two and last terms 0.917 A1 WWW, AWRT 3 Question Answer Marks Guidance 6(b) [Mean =] 0.6 × 150 [= 90]; [Variance =] 0.6 × 150 × 0.4 [= 36] B1 Correct mean and variance. Accept evaluated or unsimplified ( ) 80.5 90 81 6 −   < = <     P X P Z M1 Substituting their mean and variance into ±standardisation formula (with a numerical value for 80.5), allow σ2, √ σ, but not µ ± 0.5 M1 Using continuity correction 80.5 or 81.5 ( ) 1.5833 1 0.9433 Φ − = − M1 Appropriate area Φ, from final process, must be probability 0.0567 A1 AWRT 5 6(c) np = 90, nq = 60 both greater than 5 B1 At least nq evaluated and statement >5 required 1

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