Cambridge A Level Mathematics 9709 — 2025 May/June Paper 5 · Variant 2

1 Rachel has three coins. The first coin is biased so that the probability of obtaining a head when it is thrown is 1. The second coin is biased so that the probability of obtaining a head when it is thrown is 1. 3 4 The third coin is fair. Rachel throws the three coins at the same time. The random variable X is the number of tails that she obtains. Draw up the probability distribution table for X. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 B1 Table with correct values of x and at least one non- x 0 1 2 3 zero probability correct. Condone extra x values if probability stated as 0. P(X = x) 1 6 11 6 . 24 24 24 24 B1 Two more correct non-zero probabilities linked with correct outcome (3 correct probabilities 0.0416 0.25 0.458 0.25 present). 7 3 Accept probabilities not in table if clearly identified. B1 Four correct probabilities linked with the correct outcomes. Accept probabilities not in table if clearly identified. Non-exact decimals correct to at least 3SF. SCB1 for 4 non-zero probabilities (not all ¼) in table with correct x values adding to 1 if B1 max scored. 3

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2 In Millford, 70% of the residents own a bicycle. A random sample of 160 residents is selected. Use a suitable approximation to find the probability that more than 120 of these residents own a bicycle. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Mean = 160 0.7 = 112 B1 112 and 33.6 (CAO) seen, allow un-simplified. Variance = 160  0.7  0.3 = 33.6 May be in standardisation formula. ( = 33.6, 5.79655 to at least 4SF implies correct variance). Withold mark if variance clearly identified as standard deviation. Condone N 112, 33.6 if standardisation formula ( ) correct or variance/standard deviation stated correctly as well. 120.5 − M1 Substituting their 112 and their 33.6 into the P(X 120 ) = P( Z  112) ±standardising formula (any number for 120.5), 33.6 allow σ2 or √σ. M1 Use continuity correction 119.5 or 120.5 in their standardisation formula. Note: If no standardisation formula seen  8.5   8.5     or   scores M2 BOD.  33.6   5.797  [P( Z  1.466 ) = 1 − Φ (1.466 )] M1 Appropriate area Φ , from final process, must be a probability. = 1 – 0.9287 May be implied by a sketch of the required probability area. Note: correct final answer implies this M1. Expect final answer < 0.5. = 0.0713 final answer A1 Final answer AWRT. 5

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3 A bag contains 4 blue marbles and 12 red marbles. One marble is selected at random from the bag. If this marble is blue, it is replaced in the bag, but if it is red, it is not replaced. A second marble is now selected at random from the bag. (a) Find the probability that both marbles selected are the same colour. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the probability that the first marble is blue given that the second marble is red. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 4 4 12 11 M1 OE. [P(both same colour) =]  +  16 16 16 15 4 4 12 11 12 11  and either  or  seen. 16 16 16 15 16 16 11 Decimals to 4SF but condone = 0.733 . 15 No additional terms. 49 A1 ISW. = , 0.6125 49 80 CAO, (0.6125, must be seen). 80 If M mark not scored, 49 SCB1 for , 0.6125 WWW. 80 2 3(b)  P (1B  2 R )  M1 4 12  P (1B|2 R ) =   seen as a numerator of a single fraction.  P ( 2 R )  16 16 4 12  3 If 0.1875 or is seen as the numerator, the = 16 16 16 4 12 12 11  +  calculation must be seen in the working for the 16 16 16 15 denominator – or in 3(a), including by the tree diagram. The question will need to be linked if the work is in 3(a). M1   4 12   their    or correct    16 16     12 11   their    from part ( a )  59 +   16 15   or 0.7375 or   80  or correct  seen as the denominator of a single fraction. 15 A1 0.254237… to at least 3 SF. , 0.254 If one or more M not scored 59 15 SCB1 for , 0.254237… to at least 4 SF WWW. 59 3

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4 Vehicles approaching a certain road junction from Bromley must go either left, right or straight on. Over time, it is known that 30% turn left, 25% turn right and 45% go straight on. The driver of each vehicle chooses a direction independently of all other drivers. (a) Find the probability that the next three vehicles approaching this junction from Bromley all go in different directions. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the probability that, from the vehicles approaching this junction from Bromley today, the 1st vehicle to go left is before the 9th vehicle. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the probability that, from the vehicles approaching this junction from Bromley today, the 2nd vehicle to go left is the 7th vehicle. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 0.3  0.25  0.45  6 M1 OE. 0.3  0.25  0.45  k , k an integer > 1. E.g. 6 = 3! = 3P3 etc. 81 A1 CAO exact answer. 0.2025, 400 2 4(b) Method 1 1−0.78 M1 1−0.7d , d = 8, 9, 0.75, 0.3 are not misreads. = 0.942 A1 0.942 ⩽ p < 0.9425. Method 2 2 3 4 5 6 7 M1 2 3 4 0.3 + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) 0.3 + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) + 0.3 ( 0.7 ) 5 + 0.3 ( 0.7 ) 6 + 0.3 ( 0.7 ) 7  +0.3 ( 0.7 )8    = 0.942 A1 0.942 ⩽ p < 0.9425. 2 4(c) (0.3)2 (0.7)5 6 M1 (p)2 (1 – p)5 k , 0  p  1, k = 5 or 6 . = 0.0908 A1 AWRT. 2

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5 The times taken, t minutes, by 300 students to travel to Hollowton College are recorded. The results are summarised in the table below. Time (t minutes) t G 10 t G 20 t G 30 t G 40 t G 60 t G 90 Cumulative frequency 34 86 142 208 265 300 (a) On the grid, draw a cumulative frequency graph to illustrate this information. [2] (b) 120 students take more than k minutes to travel to college. Use your graph to estimate the value of k. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Calculate estimates of the mean and standard deviation of the times taken to travel to college by the 300 students. 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Mark scheme: 5(a) Cumulative frequency graph drawn B1 At least 5 points plotted accurately at class upper end points: (10, 34), (20, 86), (30, 142), (40, 208), (60, 265), (90, 300). Linear cf scale 0 ⩽ cf ⩽ 300 and linear time scale 0 ⩽ time ⩽ 90 with at least 3 values identified on each. Bar/histograms score B0. Required axes must be over 50% of grid size. Axes can be the other way round. B1 All points plotted correct, curve drawn, no line segments and joined to (0,0) passing within ½ square of points. Axes labelled cumulative frequency (cf), time (t) and minutes (min or m) – or a suitable title. Curve must be <300 for 0 ⩽ t < 90. 2 5(b) Line drawn from 180 on cf axis to meet graph M1 Use of graph must be seen. Must be an increasing graph. Annotate this mark on 5(a) grid. [k =] 35 minutes A1 FT Strict FT, reading at their curve condone use of t. 2 5(c) Frequencies: [34,] 52, 56, 66, 57, 35 B1 May be un-simplified and/or in variance calculation. Midpoints: 5, 15, 25, 35, 50, 75 B1 At least 5 correct midpoints seen, may be un- simplified, may be in calculation, may be by data table. 34 +5 52  15 + 56  25 + 66  35 + 57  50 + 35  75 M1 Correct un-simplified mean formula with their 6 Mean = midpoints (not upper bound, lower bound, upper 300 limits, lower limits, class width, frequency density,  170 + 780 + 1400 + 2310 + 2850 + 2625  frequency or cumulative frequency and must be    300  within class) and their frequencies (not cw, cf, or fd), accept un-simplified.  10135  =    300  Condone 1 value error on numerator. The ‘table’ approach may be used with multiplications and additions implied by appropriate values. = 33.8 A1 WWW. 2027 47 33.78 ⩽ mean ⩽ 33.8, , 33 . 60 60 2027 47 If M not scored, SCB1 for , 33 , 33.78 . 60 60 10135 (Note: scores A0). 300 5(c) 2 2 2 2 2 2 2 M1 Correct un-simplified Variance formula with their 34  5 + 52  15 + 56  25 + 66  35 + 57  50 + 35  75  10135  Var = − their  midpoints and their frequencies – their mean2 for 300  300  2 variance or standard deviation. 34  25 + 52  225 + 56  625 + 66  1225 + 57  2500 + 35  5625  10135  Or − their  Condone one value error on numerator. 300  300  Condone their  f from denominator of mean [= 1559.25 – 1141.31 = 417.936] calculation if not 300. sd = 20.4 A1 20.4 ⩽ σ < 20.45 WWW. At least one of the mean or the standard deviation must be linked to the value, and no incorrect identifications, for this mark to be scored. 6 6(a) Method 1 M _ _ _ _ _ _ _ _ _M (arranging the remaining letters and inserting the As) 4! × 5C1 M1 4! × m, m  , m 1 . M1 n × 5C1 or n × 5C4 or n × 5, 5 4P allow n × 5P1 or n × , n  , n .1 4! e.g. 4! × 5C1× k (oe), k  , k ⩾ 1 scores M1M1. 120 A1 Method 2 Total number of arrangements with Ms at end – arrangements with Ms at end and at least 2 As together 8! M1 8! [Total number of arrangements: = 1680 −r 5! m, m  , m 3, r = 1,2 . 4! 4! Arrangements with As together: (AAAA) ^ ^ ^ ^ 5! M1 k −5! 13  r , k  , k  1560 r = 1,2 . (AAA^) (A) ^^^ 5! × 4 4 (AA^) (AA) ^^^ 5! 2! 4  3 (A^) (A^) (2A) ^^ 5! ] 2! 8! − 5! 13 4! = 120 A1 3 6(b) Method 1 M _ _ _ M _ _ _ _ _ 8! B1 8! 4! 6 4!  b, b  , b 1 . M1 c !  6 , c = 8, 9, 10 d = 1, 2!, 4! or 2! × 4!. d 10080 A1 Method 2 M ^ ^ ^ M _ _ _ _ _ letters between Ms arranged and the treated as a single item 6! 8 B1 6!  b, b  , b 1 . 4! P 3 4! M1 c ! 8  P3 , c = 6,7,8 d = 1, 2!, 4! or 2!×4!. d 10080 A1 3 6(c) Method 1 MMAAA 1 B1 One identified outcome correct (excluding MMAA _ 4C1 4 MMAAA or MAAAA), accept un-simplified. MAA _ _ 4C2 6 MAAA _ 4C1 4 M1 Five correct scenarios added MAAAA 1 (values do not need to be correct). [Total] 16 A1 Method 2 2 Ms cannot be present in the scenarios MAA ^ ^ 5C2 10 B1 One identified outcome correct (excluding MAAA ^ 5C1 5 MAAAA), accept un-simplified. MAAAA 5C0 1 M1 Three correct scenarios added (values do not need to be correct). [Total] 16 A1 3

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6 (a) Find the number of different ways in which the 10 letters in the word AMALGAMATE can be arranged so that there is an M at the beginning, an M at the end and no As are together. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the number of different ways in which the 10 letters in the word AMALGAMATE can be arranged with exactly 3 letters between the two Ms. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Five letters are selected from the 10 letters in the word AMALGAMATE. (c) Find the number of different selections in which the five letters include at least one M and at least two As. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 6(a) Method 1 M _ _ _ _ _ _ _ _ _M (arranging the remaining letters and inserting the As) 4! × 5C1 M1 4! × m, m  , m 1 . M1 n × 5C1 or n × 5C4 or n × 5, 5 4P allow n × 5P1 or n × , n  , n .1 4! e.g. 4! × 5C1× k (oe), k  , k ⩾ 1 scores M1M1. 120 A1 Method 2 Total number of arrangements with Ms at end – arrangements with Ms at end and at least 2 As together 8! M1 8! [Total number of arrangements: = 1680 −r 5! m, m  , m 3, r = 1,2 . 4! 4! Arrangements with As together: (AAAA) ^ ^ ^ ^ 5! M1 k −5! 13  r , k  , k  1560 r = 1,2 . (AAA^) (A) ^^^ 5! × 4 4 (AA^) (AA) ^^^ 5! 2! 4  3 (A^) (A^) (2A) ^^ 5! ] 2! 8! − 5! 13 4! = 120 A1 3 6(b) Method 1 M _ _ _ M _ _ _ _ _ 8! B1 8! 4! 6 4!  b, b  , b 1 . M1 c !  6 , c = 8, 9, 10 d = 1, 2!, 4! or 2! × 4!. d 10080 A1 Method 2 M ^ ^ ^ M _ _ _ _ _ letters between Ms arranged and the treated as a single item 6! 8 B1 6!  b, b  , b 1 . 4! P 3 4! M1 c ! 8  P3 , c = 6,7,8 d = 1, 2!, 4! or 2!×4!. d 10080 A1 3 6(c) Method 1 MMAAA 1 B1 One identified outcome correct (excluding MMAA _ 4C1 4 MMAAA or MAAAA), accept un-simplified. MAA _ _ 4C2 6 MAAA _ 4C1 4 M1 Five correct scenarios added MAAAA 1 (values do not need to be correct). [Total] 16 A1 Method 2 2 Ms cannot be present in the scenarios MAA ^ ^ 5C2 10 B1 One identified outcome correct (excluding MAAA ^ 5C1 5 MAAAA), accept un-simplified. MAAAA 5C0 1 M1 Three correct scenarios added (values do not need to be correct). [Total] 16 A1 3

7 Kestrels are birds whose adult wingspans are normally distributed with mean 74.8 cm and standard deviation 3.2 cm. A random sample of 120 adult kestrels is selected. (a) How many of these 120 adult kestrels would you expect to have wingspan between 72.4 cm and 76.3 cm? [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The masses of adult kestrels are normally distributed with mean n kg and standard deviation v kg. It is known that 20% of adult kestrels have mass greater than 0.202 kg and 28% have mass less than 0.185 kg. (b) Find the value of n and the value of v. 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Mark scheme: 7(a)  72.4 − 74.8 76.3 − 74.8  M1 Use of ±standardisation formula once with 74.8, 3.2 P   Z   and either 72.4 or 76.3.  3.2 3.2  No continuity correction, not σ2, not . [ Φ ( 0.75 ) + Φ ( 0.4688 ) − 1] M1 Calculating the appropriate probability area, must be a probability = 0.7734 + 0.6804 – 1 (assume a 4SF value in the calculation it is a or ( 0.7734 − 0.5 ) + ( 0.6804 − 0.5 ) probability, assume a 3SF value in the calculation it or 0.7734 – 0.3196 is not a probability). or 0.6804 – 0.2266 (leading to their final answer, expect < 0.5). or 0.2734 + 0.1804 Unless there is a misread, the probability area calculation must be structured as shown. = 0.4538 A1 AWRT 0.454 WWW. Expected number = 0.4538  120 = 54.46 B1 FT Strict FT their 4-figure probability × 120 (check so 54 with calculator). (or 55) One integer answer. (One integer answer stated) No indication of approximation, e.g. , , about, 2 SF . 4 7(b) 0.202 −  B1 0.841 < z1 < 0.843 or −0.843 < z1 < −0.841 seen. = 0.842  B1 0.582 < z2 < 0.584 or −0.584 < z2 < −0.582 seen. 0.185 −  = − 0.583 M1 Use of the ±standardisation formula once with µ, σ  equating to a z-value (not 0.20, 0.80, 0.28, 0.72, 0.5793, 0.4207, 0.6103, 0.3897, 0.7881, 0.2119, 0.7642, 0.2358, –0.417, 0.417, -0.158 0.158, etc). No continuity correction, not σ2, not . Solve, obtaining values for µ and σ M1 Solve 2 equations in µ and σ with an attempt at the elimination method, substitution method or other = 0.192,  = 0.0119 appropriate approach to obtain values for both µ and σ . A1 AWRT = 0.192,  = 0.0119 . There must be consistency with signs in the 0.017 solution, e.g. = = 0.0119 is not −1.425 acceptable, A0. If one or both the M marks have been withheld, SCB1 for both correct WWW. 5 7(c) Method 1 [P(X < 3) = P(0, 1, 2) = ] M1 x 10 − x One term of the form 10Cx ( p ) (1 − p ) , 0 < p < 1, x  0 or 10. 0.810 + 10C1 0.89 0.2 + 10C2 0.88 0.22 [= 0.107374 + 0.268435 + 0.301989 ] A1 Correct un-simplified expression. 0.678 B1 0.677 < p ⩽ 0.678. Method 2 [P(X < 3) = 1 – P(3, 4, 5, 6, 7, 8, 9, 10) = ] M1 x 10 − x One term of the form 10Cx ( p ) (1 − p ) , 0 < p < 1, x  0 or 10. 1 – {10C3 0.87 0.23 + 10C4 0.86 0.24 + … + 0.210} A1 Correct un-simplified expression. 0.678 B1 0.677 < p ⩽ 0.678 3

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