Cambridge A Level Mathematics 9709 — 2021 May/June Paper 5 · Variant 2

1 An ordinary fair die is thrown repeatedly until a 5 is obtained. The number of throws taken is denoted by the random variable X. (a) Write down the mean of X. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that a 5 is first obtained after the 3rd throw but before the 8th throw. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the probability that a 5 is first obtained in fewer than 10 throws. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) 6 B1 WWW 1 1(b) 3 4 5 6 5 1 5 1 5 1 5 1 6 6 6 6 6 6 6 6         + + +                 M1 p3(1 – p) + p4(1 – p) + p5(1 – p) + p6(1 – p), 0 < p < 1 0.300 (0.2996…) A1 At least 3s.f. Award at most accurate value. Alternative method for Question 1(b) 3 7 5 5 6 6     −         M1 p3 – p7, 0 < p < 1 0.300 (0.2996…) A1 At least 3s.f. Award at most accurate value. 2 1(c) 9 5 1 6   −    M1 1 – pn, 0 < p < 1, n = 9, 10 0.806 A1 Alternative method for Question 1(c) 2 8 1 1 5 1 5 1 5 6 6 6 6 6 6 6       + + + +              M1 p + p(1 – p) + p(1 – p)2 + p(1 – p)3 + p(1 – p)4 + p(1 – p)5 + p(1 – p)6 + p(1 – p)7 + p(1 – p)8 (+ p(1 – p)9), 0 < p < 1 As per answer for minimum terms shown 0.806 A1 2

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2 The weights of bags of sugar are normally distributed with mean 1.04 kg and standard deviation 3 kg. In a random sample of 2000 bags of sugar, 72 weighed more than 1.10 kg. Find the value of 3. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 ( ) ( ) 72 1.1 0.036 2000   > = =     P X z = ±1.798 1.1 1.04 1.798 σ − = 0.06 1.798 σ   =     B1 1.1 and 1.04 substituted in ±standardisation formula, allow continuity correction, not σ2 or √ σ M1 Equate their ±standardisation formula to a z-value and to solve for the appropriate area leading to final answer (expect σ < 0.5). 0.06 Accept value σ   ± = −     z 0.0334 σ = A1 0.03335 ≤ σ ≤ 0.0334. At least 3 3s.f. 4

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3 On each day that Alexa goes to work, the probabilities that she travels by bus, by train or by car are 0.4, 0.35 and 0.25 respectively. When she travels by bus, the probability that she arrives late is 0.55. When she travels by train, the probability that she arrives late is 0.7. When she travels by car, the probability that she arrives late is x. On a randomly chosen day when Alexa goes to work, the probability that she does not arrive late is 0.48. (a) Find the value of x. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that Alexa travels to work by train given that she arrives late. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) P(not late) = 0.4 × 0.45 + 0.35 × 0.3 + 0.25 × (1 – x) or P(late) = 0.4 × 0.55 + 0.35 × 0.7 + 0.25x M1 0.4 0.35 0.25 × + × + × p q r , p = 0.45, 0.55, q = 0.3, 0.7 and r = (1 – x), x 0.18 + 0.105 + 0.25 (1 – x) = 0.48 or 0.22 + 0.245 + 0.25x = 0.52 A1 Linear equation formed using sum of 3 probabilities and 0.48 or 0.52 as appropriate. Accept unsimplified. x = 0.22 A1 Final answer 3 3(b) ( ) ( ) ( )   ∩ =       P train late P train late P late 0.35 0.7 1 0.48 × = − or 0.35 0.7 0.4 0.55 0.35 0.7 0.25 0.22 × × + × + ×their B1 0.35 0.7 × or 0.245 seen as numerator of fraction M1 P(late) seen as a denominator with their probability as numerator (Accept 0.52 0.22 0.245 0.25 0.22 + + × their p their p or their ) = 0.471 or 49 104 A1 3

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5 Every day Richard takes a flight between Astan and Bejin. On any day, the probability that the flight arrives early is 0.15, the probability that it arrives on time is 0.55 and the probability that it arrives late is 0.3. (a) Find the probability that on each of 3 randomly chosen days, Richard’s flight does not arrive late. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that for 9 randomly chosen days, Richard’s flight arrives early at least 3 times. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) 60 days are chosen at random. Use an approximation to find the probability that Richard’s flight arrives early at least 12 times. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) ( ) 3 [ 0.7 ] 0.343 = Alternative method for Question 5(a) [(0.15)3 + 3C1(0.15)2(0.55) + 3C2(0.15)(0.55)2 + (0.55)3 =] 0.343 B1 Evaluated WWW 1 5(b) 1 – (0.859 + 9C1 0.151 0.858 + 9C2 0.152 0.857) [1 – (0.231617 + 0.367862 + 0.259667)] M1 One term: 9Cx px (1 – p)9-x for 0 < x < 9, any 0 < p < 1 A1 Correct expression, accept unsimplified. 0.141 A1 0.1408 ⩽ ans ⩽ 0.141, award at most accurate value. Alternative method for Question 5(b) 9C3 0.153 0.856 + 9C4 0.154 0.855 + 9C5 0.155 0.854 + 9C6 0.156 0.853 + 9C7 0.157 0.852 + 9C8 0.158 0.85 + 0.159 M1 One term: 9Cx px (1 – p)9-x for 0 < x < 9, any 0 < p < 1 A1 Correct expression, accept unsimplified. 0.141 A1 0.1408 ⩽ ans ⩽ 0.141, award at most accurate value. 3 Question Answer Marks Guidance 5(c) Mean [ ] 60 0.15 9 = × = Variance [ ] 60 0.15 0.85 7.65 = × × = B1 Correct mean and variance, allow unsimplified. (2.765 ≤ σ ≤ 2.77 imply correct variance) ( ) 11.5 9 12 7.65 −    ≥ = >       X P Z M1 Substituting their mean and variance into ±standardisation formula (any number for 11.5), not σ2 or √ σ M1 Using continuity correction 11.5 or 12.5 in their standardisation formula. ( ) 1 0.9039 1 0.8169 −Φ = − M1 Appropriate area Φ, from final process, must be probability. 0.183 A1 Final AWRT 5

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6 (a) Find the total number of different arrangements of the 8 letters in the word TOMORROW. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the total number of different arrangements of the 8 letters in the word TOMORROW that have an R at the beginning and an R at the end, and in which the three Os are not all together. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Four letters are selected at random from the 8 letters of the word TOMORROW. (c) Find the probability that the selection contains at least one O and at least one R. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) 8! 2!3! M1 8! ! ! × k m k = 1 or 2, m = 1 or 3, not k = m = 1 no additional terms 3360 A1 2 Question Answer Marks Guidance 6(b) Method 1 Arrangements Rs at ends – Arrangements Rs at ends and Os together [Os not together = ] 6! 3! – 4! M1 6! ! k – m, 1 ⩽ k ⩽ 3, m an integer, condone 6! 2 !   × −     m k . M1 w – 4! or w – 24, w an integer Condone w – 2 × 4! 96 A1 Method 2 identified scenarios R _ _ _ R, Arrangement No Os together + 2Os and a single O 4C3 × 3! + 4C2 × 2 × 3! M1 4C3 × 3! + r or 4× 3! + r or 4P3 × 3! + r, r an integer. Condone 2 × 4C3 × 3! + r. 2 × 4× 3! + r or 2 × 4P3 × 3! + r. M1 q + 4C2 × 3! × k or q + 4P2 × 3! × k, k = 1,2, q an integer [24 + 72 =] 96 A1 3 6(c) Method 1 Identified scenarios OORR 3 2 3 2 2 0 C C C 3 1 3   × × = × =   ORR_ 3 2 3 1 2 1 C C C 3 1 3 9 × × = × × = OOR_ 3 2 3 2 1 1 C C C 3 2 3 18 × × = × × = OR_ _ 3 2 3 1 1 2 C C C 3 2 3 18 × × = × × = OOOR 3 2 3 3 1 0 C C C 1 2 2   × × = × =   B1 Outcomes for 2 identifiable scenarios correct, accept unsimplified. M1 Add 4 or 5 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. Total 50 A1 All correct and added Probability = 8 4 50 C M1 8 4 '50' their C , accept numerator unevaluated Question Answer Marks Guidance 6(c) cont’d 50 or 0.714 70 A1 Method 2 Identified outcomes ORTM 3 2 1 1 C C × = 6 ORTW 3 2 1 1 C C × = 6 ORMW 3 2 1 1 C C × = 6 ORRM 3 2 1 2 C C × = 3 ORRW 3 2 1 2 C C × = 3 ORRT 3 2 1 2 C C × = 3 OROR 3 2 2 2 C C × = 3 OROT 3 2 2 1 C C × = 6 OROM 3 2 2 1 C C × = 6 OROW 3 2 2 1 C C × = 6 OROO 3 2 3 1 C C × = 2 B1 Outcomes for 5 identifiable scenarios correct, accept unsimplified. M1 Add 9, 10 or 11 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. Total 50 A1 All correct and added Probability = 8 4 50 C M1 8 4 '50' their C , accept numerator unevaluated. 50 or 0.714 70 A1 5

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7 The heights, in cm, of the 11 basketball players in each of two clubs, the Amazons and the Giants, are shown below. Amazons 205 198 181 182 190 215 201 178 202 196 184 Giants 175 182 184 187 189 192 193 195 195 195 204 (a) State an advantage of using a stem-and-leaf diagram compared to a box-and-whisker plot to illustrate this information. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Represent the data by drawing a back-to-back stem-and-leaf diagram with Amazons on the left-hand side of the diagram. [4] (c) Find the interquartile range of the heights of the players in the Amazons. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Four new players join the Amazons. The mean height of the 15 players in the Amazons is now 191.2 cm. The heights of three of the new players are 180 cm, 185 cm and 190 cm. (d) Find the height of the fourth new player. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Includes all data B1 Reference to either including all/raw data or further statistical processes are possible that cannot be found using data from box-and-whisker, eg frequency, mean, mode or standard deviation not only median, IQR, range or spread which can be found from both. 1 7(b) Amazons Giants 8 17 5 4 2 1 18 2 4 7 9 8 6 0 19 2 3 5 5 5 5 2 1 20 4 5 21 Key: 1|18|2 means 181 cm for Amazons and 182 cm for Giants B1 Correct stem can be upside down, ignore extra values B1 Correct Amazons labelled on left, leaves in order from right to left and lined up vertically (less than halfway to next column), no commas or other punctuation. B1 Correct Giants labelled on same diagram, leaves in order and lined up vertically (less than halfway to next column), no commas or other punctuation. B1 Correct single key for their diagram, need both teams identified and ‘cm’ stated at least once here or in leaf headings or title. SC for if 2 separate diagrams drawn, award SCB1 if both keys meet these criteria (Max B1, B0, B0, B1) 4 7(c) [UQ = 202 (cm), LQ = 182 (cm)] [IQR =] 202 – 182 = 20 (cm) M1 201 ⩽ UQ ⩽ 205 – 181 ⩽ LQ ⩽ 184 A1 WWW 2 Question Answer Marks Guidance 7(d) 11 [Σ 2132 = 15 Σ 191.2 15 2868 = × = ] B1 Both Σ11 and Σ15 found. Accept unevaluated. their 2868 = their 2132 + (180 + 185 + 190) + h M1 Forming an equation for the height using their Σ11 and Σ15. 181 (cm) A1 Alternative method for Question 7(d) 15 [Σ 191.2 15 2868 = × = 15 Σ 2687 = + h ] B1 Σ15 found using the mean and raw data methods. Accept unevaluated. their 2868 = their 2687 + h M1 Forming an equation for the height using their Σ15 expressions. 181 (cm) A1 Alternative method for Question 7(d) 15 [Σ 2687 = + h 15 Σ 191.2 15 = ] B1 Σ15 found using raw data method and statement on calculating new mean. Accept unevaluated. 2687 191.2 15 + = their h M1 Forming an equation for the height using their Σ15 expressions 181 (cm) A1 3 N.B. All methods can be presented as a logical numerical argument which can be condoned if clear.

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