Mathematics 9709/53 October/November 2025: question paper and mark scheme

Mark scheme

1(a) 27 B1 OE. [( 0.6 ) =] 0.216, 3 125 1 1(b) Method 1 [P(X < 9) = 1 – P(9, 10, 11) =] 1 – {11 C9(0.75)9 (0.25)2 + 11C10 (0.75)10 M1 One term of form 11 Cx ( p ) x (1 − p )11− x 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. Σ 11 11 Cx ( p ) (1 − p ) x 11− x Accept 9 with p = 0.75. = 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 + 11 C1(0.75)(0.25)10 + … + M1 One term of form 11 Cx ( p ) x (1 − p )11− x with 0 < p < 1, x ≠ 0 or 11 C (0.75) 7 (0.25) 4 + 11 C (0.75) 8 (0.25) 3 7 8 11. A1 Correct un-simplified expression, no more than 6 ‘middle’ terms omitted leading to final answer. Σ C ( p ) (1 − p ) 8 11 x 11− x Accept 0 x with p = 0.75. = 0.545 B1 0.5445 < p ≤ 0.545 . 3

Mark scheme

2 B1 Expression of the form 16 Cx × 10 Cy with x+ y = 6 linked to a Scenario S(16) D(10) correct identified scenario. SSSSDD 4 2 16 C × 10 C [81900] 4 2 SSSDDD 3 3 16 C × 10 C [67200] 3 3 SSDDDD 2 4 16 C × 10 C [25200] 2 4 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

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 6P2 b! × or b! × 6C2 or b!× . 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 10080 1 A1 WWW. [Probability =] , , 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 D1 AF1 F2 OD2 ILS =] M1 Accept 362880 as denominator in a fraction. 9! [=362880] seen as denominator in a fraction. 40320 1 A1 WWW. [ Probability =] , , 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] 5 C × 4! × 4! [=5760] 3 [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 10080 1 A1 WWW. [ Probability =] , , 0.111 90720 9 3

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 × 16 16 16 16 p < 1) that add to 1. May be calculated in variance. FT their table with 4 or 5 probabilities adding to 1. [0] + 1 + 6 + 15 + 16 = 16 38 19 = 16 = 8 1 + 3 × 22 + 5 × 32 + 4 × 42 19 2 M1FT Appropriate variance formula using their (E(X))2 value. 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 3 7 M1 3 n 1− 1 − with n = 6, 7, 8. 4 4 = 0.867 A1 = 0.8665 to 4SF. Method 2 2 3 4 1 3 1 3 1 3 1 3 1 M1 7 3 1 6 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) =] 5 6 3 1 3 1 + 4 4 4 4 = 0.867 A1 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. × × 35 5 7 6 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 × × 35 5 7 6 identification by correct (or FT) un-simplified products. P(YRB) 2 5 3 3 0.107(1) × × 28 5 8 7 P(YBR) 2 3 5 3 0.107(1) × × 28 5 8 7 [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 P(White| (b)) = 5 7 6 5 7 6 = 210 210 39 denominator. their their 39 70 70 12 or their P (WRB ) + P (WBR ) from part b Condone 35 . 39 their or unsimplified expression 70 8 A1 = , 0.615 13 2

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

Mark scheme

7(a) 15.6 − 14.8 M1 Use of ± standardisation formula with 15.6, 14.8 and 1.5 [P(X > 15.6) = P(Z >] ) 1.5 substituted, condone σ 2 , condone σ , condone continuity correction of ±0.05 . [= 1 − Φ ( 0.5333)] M1 Calculating the appropriate probability area (leading to their = 1 − 0.7029 final answer, expect < 0.5). = 0.297 A1 AWRT 0.297. If one or both M marks not awarded, SCB1 for AWRT 0.297 WWW. 3 7(b) t − 14.8 B1 −1.282 or 1.282 seen, CAO – critical value. [P(X > 15.6) = P(Z >] [P(Z > ) = 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 not σ 2 , not σ , 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 σ , no continuity 15.8 − 14.8 correction. [P(X < 15.8) = P(Z < ] 1.5 SCM1 – Condone substituting 14.8 and 1.5 and either 14.3 or Or 15.3 appropriately into one ±standardising formula, not σ2 or σ , 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 = 2 × 0.7477 −1 final answer, expect < 0.5) from their Z-values. 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