Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 5 · Variant 3

1 There are a large number of students at Greenfield college. Each student travels to college by car, by bus or on foot, independently of any other student. The probability that any student travels by car is 0.4. The probability that any student travels by bus is 0.35. (a) 3 students from Greenfield college are selected at random. Find the probability that none of these 3 students travel to college by car. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) 11 students from Greenfield college are selected at random. Find the probability that fewer than 9 of these 11 students travel to college by car or by bus. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 3 27 B1 OE. [( 0.6 ) =] 0.216,125 1 1(b) Method 1 [P(X < 9) = 1 – P(9, 10, 11) =] 1 – {11C9(0.75)9 (0.25)2 + 11C10 (0.75)10 M1 x 11− x One term of form 11Cx ( p ) (1 − p ) with 0  p  1, x  0 or (0.25) + (0.75)11} 11. [= 1 – (0.258104 + 0.154862 + 0.042235)] A1 Correct un-simplified expression, no terms omitted leading to final answer. Condone omission of final bracket ‘}’ If other brackets omitted, allow recovery if 1 – 0.455 (or better) seen. 11 11 x 11− x Accept C x ( p ) (1 − p ) with p = 0.75.  9 = 0.545 B1 0.5445 p 0.545 . Method 2 [P(X < 9) = P(0,1,2,3,4,5,6,7,8) =] (0.25)11 + 11C1(0.75)(0.25)10 + … + M1 One term of form 11Cx ( p ) x (1 − p )11− x with 0  p  1, x  0 or 11C7(0.75)7(0.25)4 + 11C8(0.75)8(0.25)3 11. A1 Correct un-simplified expression, no more than 6 ‘middle’ terms omitted leading to final answer. 8 11 x 11− x Accept C x ( p ) (1 − p ) with p = 0.75.  0 = 0.545 B1 0.5445  p  0.545 . 3

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2 The Splash Club has 26 members, of whom 16 are swimmers and 10 are divers. No member is both a swimmer and a diver. The club committee consists of 6 of these 26 members. In how many ways can the club committee be selected if it must include at least 2 swimmers and at least 2 divers? [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 B1 Expression of the form 16Cx  10Cy with x+ y = 6 linked to a Scenario S(16) D(10) correct identified scenario. SSSSDD 4 2 16C4  10C2 [81900] SSSDDD 3 3 16C3  10C3 [67200] SSDDDD 2 4 16C2  10C4 [25200] M1 Two identified outcomes evaluated accurately, accept un- simplified. Identification can be implied by un-simplified expression. Condone consistent use of permutations. M1 Sum of their values of 3 correct identified scenarios, no incorrect/repeated scenarios. Identification can be implied by un-simplified expression. Total = 174 300 A1 If either or both Ms not awarded, SCB1 for 174 300 WWW. 4

More questions on Permutations and combinations

3 (a) Find the number of different arrangements of the 9 letters in the word DAFFODILS in which there is a D at each end and the two Fs are not next to each other. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the probability that a randomly chosen arrangement of the 9 letters in the word DAFFODILS has exactly 4 letters between the two Ds. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) Method 1 Total arrangements with Ds at ends – arrangements with Ds at ends and Fs together 7! M1 7! − 6! seen alone, not multiplied or divided in a calculation nor 2! 2! subtracted from another value. [= 2520 – 720] M1 l ! − 6!, l = 7, 8 or 9, m = 1 or 2 m ! = 1800 A1 Method 2 D ^ ^ ^ ^ ^ D, Fs inserted not next to each other 6  5 M1 5!  a, 1 a 30 1 can be implied, a is an integer. No other 5!  2 terms added or subtracted. M1 6  5 6 P 2 b !  or b ! 6C2 orb ! . 2 2! Where b = 5, 6 or 7 . = 1800 A1 3 3(b) Method 1 [Number of outcomes with 4 letters between Ds (D ^ ^ ^ ^ D ^ ^ ^)] M1 Accept 7! 7!  4 [= 10080]  4 or 10080 alone or as the numerator or denominator of a 2! 2 fraction. 9! M1 Accept 90720 as denominator in a fraction. [Total number of arrangements =] [= 90720] seen as 2!2! denominator in a fraction 1 A1 WWW.  Probability = 10080 , , 0.111 90720 9 Method 2 [Total number of outcomes with 4 letters between Ds (D1 ^ ^ ^ ^ D2 ^ M1 Accept 7!  4  2! or 40320 alone or as the numerator or ^ ^)] denominator of a fraction. 7!  4  2! [= 40320] [Total number of arrangements of D1AF1F2OD2ILS =] M1 Accept 362880 as denominator in a fraction. 9! [=362880] seen as denominator in a fraction. 1 A1 WWW.  Probability = 40320 , , 0.111 362880 9 3(b) Method 3 [When 2 F’s are not part of the 4 letters between the D’s] M1 Accept 10080 alone or as the numerator or denominator of a fraction. 5P4  4! 5!  4! or [= 1440] 2! 2! [When 1F is part of the 4 letters between the D’s] 5C3  4!  4! [=5760] [When 2F’s are part of the 4 letters between the D’s] 5C2  4!  4! [=2880] 2! [1440 + 5760 + 2880] = 10080 9! M1 Accept 90720 as denominator in a fraction. [Total number of arrangements =] [= 90720] seen as 2!2! denominator in a fraction 1 A1 WWW.  Probability = 10080 , , 0.111 90720 9 3

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4 A fair red spinner has 4 sides, numbered 1, 2, 3, 4. A fair blue spinner has 4 sides, numbered 0, 1, 2, 3. When a spinner is spun, the score is the number on the side on which it lands. The two spinners are spun at the same time. The random variable X denotes the higher of the two scores obtained. If the two scores are equal, then the value of X is 0. (a) Draw up the probability distribution table for X. 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(b) Find Var(X ). 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The red spinner, with sides numbered 1, 2, 3, 4, is spun repeatedly. (c) Find the probability that it lands on 4 for the first time before the 8th spin. 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Mark scheme: 4(a) B1 Table with correct x values and one correct probability. x 0 1 2 3 4 B1 2 more probabilities correct linked to the correct x value, need P(X = x) 3 1 3 5 4 not be in table, accept un-simplified. 16 16 16 16 16 B1 5 correct probabilities linked with correct outcomes. 0.1875 0.0625 0.1875 0.3125 0.25 SCB1 5 non-zero probabilities (not all 1/5) summing to 1 Accept 0.188 0.0625 0.188 0.313 0.25 placed in a probability distribution table with correct x values if only one of the first two B marks are scored. 3 4(b) 1 3 5 4 M1FT Accept un-simplified or bold expression with probabilities (0 < [E(X) = ] 0 + 1  + 2  + 3  + 4  p < 1) that add to 1. May be calculated in variance. 16 16 16 16 FT their table with 4 or 5 probabilities adding to 1.  0 + 1 + 6 + 15 + 16   =   16   38 19  = =    16 8  2 2 2 2 M1FT Appropriate variance formula using their (E(X))2 value. 1 + 3  2 + 5  3 + 4  4  19  Var(X) = −  FT their table with 4 or more probabilities (0 < p < 1) which 16  8  need not sum to 1 or with an expression no more evaluated than  1 + 3  4 + 5  9 + 4  16 361  shown. = −   16 64      122  19  2   = −     16  8   127 63 A1 Accept 1.984375 to 3 or more SF. = , 1 , 1.98 SCB1 for 1.98 if M0 M1 scored. 64 64 3 4(c) Method 1 7 M1 n  3   3  1 −  1 −  with n = 6, 7, 8.  4   4  = 0.867 A1 = 0.8665 to 4SF. Method 2 2 3 4 M1 7 6 1  3  1   3  1   3  1   3  1   3  1   3  1  + + + + + Condone extra term or the term missing.                   4  4  4   4  4   4  4   4  4   4  4   4  4  [P(X < 8) =]  3 5 1   3 6 1  +        4  4   4  4  = 0.867 A1 2

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5 Chen has three boxes. Box A contains 5 counters, of which 3 are white and 2 are yellow. Box W contains 4 red marbles and 3 blue marbles. Box Y contains 5 red marbles and 3 blue marbles. Chen chooses one counter at random from box A. If the counter is white, he chooses two marbles from box W, at random and without replacement. If the counter is yellow, he chooses two marbles from box Y, at random and without replacement. (a) Draw a fully labelled tree diagram to illustrate this information, including all the probabilities. [3] (b) Find the probability that one red marble and one blue marble are obtained. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the probability that a white counter is chosen from box A, given that one red marble and one blue marble are obtained. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) B1 Correct tree diagram structure. (3 tiers). B1 Counter and first marble probabilities correct with outcomes labelled. B1 Second marble probabilities correct with outcomes labelled. 3 5(b) M1 At least 2 correct 3-term products linked to correct scenarios. P(WRB) 3 4 3  6  0.171(4) May be clearly identified on tree diagram.     5 7 6  35  FT from a 3-tier tree. M1 4 correct identified scenarios added. P(WBR) 3 3 4  6  0.171(4)   May be clearly identified on the tree diagram, condone   5 7 6  35  identification by correct (or FT) un-simplified products. P(YRB) 2 5 3  3  0.107(1)     5 8 7  28  P(YBR) 2 3 5  3  0.107(1)     5 8 7  28  [Total 39] 0.557 70 39 A1 39 = , 0.557 SCB1 for if one or both M marks not scored WWW. 70 70 3 5(c) 3 4 3 3 3 4  36 36  M1 Correct formula for conditional probability using their   +   +   probabilities from 3-tier tree with their (b) < 1 or correct in 5 7 6 5 7 6 210 210 P(White| (b)) =  =  denominator. 39 39 their  their  70  70  12 ortheir P (WRB ) + P (WBR ) from partb 35 Condone . 39 their orunsimplified expression 70 8 A1 = , 0.615 13 2

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6 Last Saturday, a cycling competition for teams of 11 cyclists took place. For each cyclist, the time taken to complete the course was recorded to the nearest minute. The times taken by the cyclists from two teams, the Linnets and the Puffins, are shown in the following table. Linnets 48 51 54 57 59 60 64 64 65 68 70 Puffins 45 49 51 55 55 58 59 62 64 64 74 (a) Draw a back-to-back stem-and-leaf diagram to represent this information, with Linnets on the left-hand side. [4] (b) Find the interquartile range of the times taken by the Linnets. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) On the grid below, draw a box-and-whisker plot to represent the information for the Linnets and the Puffins. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) Make one comparison between the times taken by the Linnets and the times taken by the Puffins. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) Linnets Puffins B1 Correct stem, ignore extra values (not in reverse, not split) 8 4 5 9 If a split stem-and-leaf plot is used (i.e. stem values are 9 7 4 1 5 1 5 5 8 9 repeated) the remaining B marks are available. 8 5 4 4 0 6 2 4 4 0 7 4 B1 Correct Linnets labelled on left, leaves in order from right to left and lined up vertically, no commas or other punctuation. Key: 4|5|5 means 54 minutes for Linnets and 55 minutes for Puffins B1 Correct Puffins labelled on same diagram, leaves in order and lined up vertically, no commas or other punctuation. Penalise each error only once in question (apart from omission of data) e.g. commas in both sets of data. B1 Correct non-symmetrical key, for their diagram, need both team names and ‘mins’ at least once here or in leaf headings (or title). If 2 separate diagrams drawn, max marks B1 if both stems correct, B1 if Linnets correct to the left of the stem, B1 if both keys correct including ‘mins’ and team name. 4 6(b) [IQR =] 65 – 54 M1 64 ⩽ UQ ⩽ 68−51 ⩽ LQ ⩽ 57 Implied if both quartile values are stated and an appropriate IQR calculated accurately. Condone 54 and 65 clearly identified in the table or their stem and leaf diagram. = 11 A1 2 6(c) B1 Single linear scale from 45 to 75, at least 3 equally spaces values marked, using a scale of 2cm = 10 or 20 minutes only. Label to include time and minutes/min/m. SCB1 for 2 labelled box plots correct relative to each other and no scale is present. No further marks can be gained. B1FT Correct plot labelled ‘Linnets’ with 5 key values using their linear scale. Daylight rule applied. Whiskers not at top/bottom of box or through box (penalise only once). Condone missing vertical lines on max/min values. Linnets 48 54 60 65 70 LQ and UQ correct or FT their values from 6(b). Puffins 45 51 58 64 74 B1 Correct plot labelled ‘Puffins’ with 5 key values using the same linear scale as Linnets. Whiskers not at top/bottom of box or through box (penalise only once). Condone missing vertical lines on max/min values. SCB1 for neither plot identified assuming Linnets at top of grid and both correct. SCB1 for only one unidentified plot – mark as Linnets. 3 6(d) Acceptable comment about spread/central tendency B1 Not just a comment comparing median or range. e.g. the times taken by the Puffins are more spread out/less consistent Ignore extra comment(s) if not contradicting a correct than those of the Linnets, statement. the times taken by the Linnets are less varied/more consistent than for the Puffins, Linnets are slower [on average], Linnets have slower times Puffins are faster [on average], Puffins have faster times 1

More questions on Representation of data

7 The times taken by Obi to walk to work each morning are normally distributed with mean 14.8 minutes and standard deviation 1.5 minutes. (a) Find the probability that, on a randomly chosen day, Obi takes more than 15.6 minutes to walk to work. 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(b) On 90% of days, Obi takes more than t minutes to walk to work. Find the value of t. 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Obi walks to work 5 days a week for 45 weeks in a year. (c) On how many days in a year would you expect Obi to take within one minute of the mean time to walk to work? 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) 15.6 − M1 Use of ± standardisation formula with 15.6, 14.8 and 1.5 [P(X > 15.6) = P(Z >] 14.8) 2 1.5 substituted, condone , condone , condone continuity correction of 0.05 . [= 1 − Φ ( 0.5333 )] M1 Calculating the appropriate probability area (leading to their final answer, expect < 0.5). = 1 − 0.7029 = 0.297 A1 AWRT 0.297. If one or both M marks not awarded, SCB1 for AWRT 0.297 WWW. 3 7(b) t − B1 −1.282 or 1.282 seen, CAO – critical value. [P(X > 15.6) = P(Z >] [P(Z > 14.8) = 0.9] 1.5 M1 Using ± standardisation formula with 14.8, 1.5 and t equating t − 14.8 = −1.282 to a z-value (not 0.8159, 0.1841, 0.9, 0.1, 0.5398, 0.4602 …) 1.5 2 not  , not , no continuity correction. t = 12.9 A1 AWRT 12.9. 3 7(c)  13.8 − 14.8  M1 Substituting 14.8 and 1.5 and either 13.8 or 15.8 appropriately [P(X > 13.8) = P(Z >]   and/or  1.5  into one ±standardising formula, not σ2 or , no continuity correction.  15.8 − 14.8  [P(X < 15.8) = P(Z < ]    1.5  SCM1 – Condone substituting 14.8 and 1.5 and either 14.3 or 15.3 appropriately into one ±standardising formula, not σ2 or Or , no continuity correction.  − 1  [P(Z) ]  and/or  1.5   1  [P(Z) ]   1.5  [= 2 Φ ( 0.6667 ) −]1 M1 Calculating the appropriate probability area (leading to their final answer, expect < 0.5) from their Z-values. = 2  0.7477 − 1 Or 2  ( 0.7477 − 0.5 ) Or 0.7477 − (1 − 0.7477 ) Or 0.7477 − 0.2523 = 0.495(4) A1 AWRT 0.495. If one or both M marks not awarded, SCB1 for AWRT 0.495 SOI. [Expected number of days =] 0.4954  5  45 = 111.465, B1FT Strict FT their at least 4 figure probability × 225 (Check with so 111 days calculator). One integer answer, expect 111 or 112. No indication of ‘approximation’, e.g. , , about , 3SF. 4

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