Cambridge A Level Mathematics 9709 — 2021 Feb/March Paper 5 · Variant 2

1 A fair spinner with 5 sides numbered 1, 2, 3, 4, 5 is spun repeatedly. The score on each spin is the number on the side on which the spinner lands. (a) Find the probability that a score of 3 is obtained for the first time on the 8th spin. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that fewer than 6 spins are required to obtain a score of 3 for the first time. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) 7 4 1 5 5     =           16384 390625 or 0∙0419[43…] 1 1(b) 5 4 1 5  −  or 2 3 4 1 4 1 4 1 4 1 4 1 5 5 5 5 5 5 5 5 5       + × × + × + ×             M1 1 – pn n = 5,6 or p + pq + pq2+pq3+ pq4 (+ pq5) 0 < p < 1, p + q = 1, Sum of a geometric series may be used. 2101 3125 or 0∙672[32] A1 Final answer. Alternative method for question 1(b) [P(at least 1 three scored in 5 throws) =] 5 4 3 2 2 3 4 5 5 5 5 4 3 2 4 1 1 4 1 4 1 4 1 4 C C C C 5 5 5 5 5 5 5 5 5              + + + +                           M1 5 5 4 5 3 2 5 2 3 5 4 4 3 2 1 ( ) C ( ) ( ) C ( ) ( ) C ( ) ( ) C ( )( ) p p q p q p q p q + + + + or 6 6 5 6 4 2 6 3 3 5 4 3 6 2 4 6 5 2 1 , 0 1, 1 ( ) C ( ) ( ) C ( ) ( ) C ( ) ( ) C ( ) ( ) C ( )( ) p p p q p q p q p q p q p q + + < < + + + = + At least first, last and one intermediate term is required to show pattern of terms if not all terms stated. 2101 3125 or 0∙672[32] A1 Final answer. 2

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2 Georgie has a red scarf, a blue scarf and a yellow scarf. Each day she wears exactly one of these scarves. The probabilities for the three colours are 0.2, 0.45 and 0.35 respectively. When she wears a red scarf, she always wears a hat. When she wears a blue scarf, she wears a hat with probability 0.4. When she wears a yellow scarf, she wears a hat with probability 0.3. (a) Find the probability that on a randomly chosen day Georgie wears a hat. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that on a randomly chosen day Georgie wears a yellow scarf given that she does not wear a hat. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) 0.2 1 0.45 0.4 0.35 0.3     × + × + × M1 0∙2 [× 1] + 0∙45 × b + 0∙35 × c, b = 0∙4, 0∙6 c = 0∙3, 0∙7 0∙485 or 97 200 A1 2 2(b) ( ) ( ) ( ) 0.35 0.7 0.245 | 1 0.515 ∩ × = = = − P Y H P Y H their P H (a) B1 0∙35 × 0∙7 or 0∙245 seen as numerator or denominator of fraction. M1 0∙515 or 1 – their (a) or [0∙3 × 0 +] 0∙45 × d + 0∙35 × e, where d = their b′, e = their c′ seen as denominator of fraction. 0∙476 or 49 103 A1 0∙4757 ⩽ p ⩽ 0∙476 3

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3 The time spent by shoppers in a large shopping centre has a normal distribution with mean 96 minutes and standard deviation 18 minutes. (a) Find the probability that a shopper chosen at random spends between 85 and 100 minutes in the shopping centre. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 88% of shoppers spend more than t minutes in the shopping centre. (b) Find the value of t. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) 85 96 100 96 P 18 18 z − − < <                   M1 Use of ±standardisation formula once with appropriate values substituted, no continuity correction, not σ2 or √σ. ( ) ( ) ( ) P 0.6111 0.2222 0.2222 0.6111 1 0.5879 0.7294 1 z − < < = Φ + Φ − = + − M1 Appropriate area Φ, from final process, must be probability. Use of (1 – z) implies M0. 0∙317 A1 Final answer which rounds to 0∙317. 3 Question Answer Marks Guidance 3(b) z = ±1∙175 B1 1∙17 ⩽ z ⩽ 1∙18 or –1∙18 ⩽ z ⩽ –1∙17 96 1.175 18 − − = t M1 An equation using ±standardisation formula with a z-value, condone σ2, √σ or continuity correction. E.g. equating to 0∙88, 0∙12, 0∙8106, 0∙1894, 0∙5478, 0∙4522, ±0∙175 or ±2∙175 implies M0. 74∙85 or 74∙9 A1 74∙85 ⩽ t ⩽ 74∙9 3

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5 A driver records the distance travelled in each of 150 journeys. These distances, correct to the nearest km, are summarised in the following table. Distance (km) 0 −4 5 −10 11 −20 21 −30 31 −40 41 −60 Frequency 12 16 32 66 20 4 (a) Draw a cumulative frequency graph to illustrate the data. [4] (b) For 30% of these journeys the distance travelled is d km or more. Use your graph to estimate the value of d. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Calculate an estimate of the mean distance travelled for the 150 journeys. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Distance 0-4 5-10 11-20 21-30 31-40 41-60 Upper boundary 4∙5 10∙5 20∙5 30∙5 40∙5 60∙5 Cumulative frequency 12 28 60 126 146 150 B1 Correct cumulative frequencies seen (may be by table or plotted accurately on graph), condone 12 not stated. B1 Axes labelled ‘distance (or d) [in] km’ from 0 to 60 and ‘cumulative frequency’ (or cf) from 0 to 150. M1 At least 5 points plotted at upper end points for d (allow upper boundary ±0∙5) with a linear scale for distance, condone 0 – 4 interval inaccurate, no scale break on axis. Not bar graph/histogram unless clear indication of upper end point only of each bar. A1 All plotted correctly at correct upper end points (4.5 etc.) with both scales linear (0 ⩽ d ⩽ 60, 0 ⩽ cf ⩽ 150), curve drawn accurately joined to (0,0), cf line>150, no daylight if >150. 4 5(b) 70% of 150 = 105 M1 105 seen or implied by indication on grid. Approx. 27 A1 FT Strict FT their increasing cumulative frequency graph, use of graph must be seen. If no clear evidence of use of graph: SC B1 FT correct value from their increasing cumulative frequency graph. 2 Question Answer Marks Guidance 5(c) Midpoints: 2.25, 7.5, 15.5, 25.5, 35.5, 50.5 B1 At least 5 correct midpoints seen. Mean 2.25 12 7.5 16 15.5 32 25.5 66 35.5 20 50.5 4 150 × + × + × + × + × + × = = 27 120 496 1683 710 202 150 + + + + + M1 Using 6 midpoint attempts (e.g. 2∙25 ±0∙5), condone one error not omission, multiplied by frequency, accept unevaluated, denominator either correct or their Σ frequencies. 3238 44 21.6, 21 150 75       = = A1 Evaluated, WWW, accept 21∙5[866…]. 3

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6 (a) Find the total number of different arrangements of the 11 letters in the word CATERPILLAR. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the total number of different arrangements of the 11 letters in the word CATERPILLAR in which there is an R at the beginning and an R at the end, and the two As are not together. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the total number of different selections of 6 letters from the 11 letters of the word CATERPILLAR that contain both Rs and at least one A and at least one L. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) 11! 2!2!2! 2! × m! × n! on denominator, m = 1, 2, n = 1, 2. no additional terms, no additional operations. 4989600 A1 Exact answer only. 2 Question Answer Marks Guidance 6(b) Method 1 R ^ ^ ^ ^ ^ ^ ^ R Arrange the 7 letters CTEPILL = 7! 2! Number of ways of placing As in non-adjacent places = 8 2 C 8 2 7! 2!× C B1 7! 2!× k seen, k an integer > 1. M1 ( )1 × − m n n or 2 × n m C or 2 × n m P , n = 7, 8 or 9, m an integer > 1. M1 8 2 7! !× C p or 8 2 7! !× P p , p integer ⩾ 1, condone 2520×28. = 70560 A1 Exact answer only. SC B1 70560 from M0, M1 only. Method 2 [Arrangements Rs at ends – Arrangements Rs at ends and As together] Total arrangements with R at beg. and end = 9! 2!2! Arrangements with R at ends and As together = 8! 2! With As not together = 9! 8! 2!2! 2! − M1 9! 2! ! m – k, 90720 > k integer > 1, m = 1, 2. B1 s – 8! 2! , s an integer >1 M1 9! 8! − p q , p, q integers ⩾ 1, condone 90720 – 20160. [90720 – 20160] = 70560 A1 Exact answer only. SC B1 70560 from M0, M1 only. 4 Question Answer Marks Guidance 6(c) Method 1 R R A L _ _ 5C2 = 10 R R A L L _ 5C1 = 5 R R A A L _ 5C1 = 5 R R A A L L = 1 M1 5Cx seen alone or 5Cx × k, 2⩾ k ⩾ 1, k an integer, 0 < x < 5 linked to an appropriate scenario. A1 5C2 × k, k = 1 oe or 5C1 × m, m = 1,2 oe alone. SC if 5Cx not seen. B2 for 5 or 10 linked to the appropriate scenario WWW. M1 Add outcomes from 3 or 4 identified correct scenarios only, accept unsimplified. 2Cw × 2Cx × 2Cy × 5Cz, w+x+y+z=6 identifies w Rs, × As and y Ls. [Total =] 21 A1 WWW, only dependent on 2nd M mark. Note: 5C2 + 5C1 + 5C1 + 1 = 21 is sufficient for 4/4. SC not all (or no) scenarios identified. B1 10 + 5 + 5 + 1 DB1 = 21 Method 2 – Fixing RRAL first. N.B. No other scenarios can be present anywhere in solution. R R A L ^ ^ = 7C2 M1 7Cx seen alone or 7Cx × k, 2⩾ k⩾ 1, k an integer, 0<x<7. Condone 7Px or 7Px × k, 2⩾ k⩾ 1, k an integer, 0<x<7. M1 7C2 × k, 2⩾ k⩾ 1oe A1 7C2 × k, k = 1oe no other terms. [Total =] 21 A1 Value stated. 4

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7 There are 400 students at a school in a certain country. Each student was asked whether they preferred swimming, cycling or running and the results are given in the following table. Swimming Cycling Running Female 104 50 66 Male 31 57 92 A student is chosen at random. (a) (i) Find the probability that the student prefers swimming. [1] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) Determine whether the events ‘the student is male’ and ‘the student prefers swimming’ are independent, justifying your answer. [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ On average at all the schools in this country 30% of the students do not like any sports. (b) (i) 10 of the students from this country are chosen at random. Find the probability that at least 3 of these students do not like any sports. [3] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) 90 students from this country are now chosen at random. Use an approximation to find the probability that fewer than 32 of them do not like any sports. [5] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................

Mark scheme: 7(a)(i) 104 31 135 27 , 400 400 80 +   =     , 0∙3375 B1 Evaluated, exact value. 1 7(a)(ii) Method 1 ( ) 180 P 400 = M , 0∙45 ( ) 135 P 400 = S , 0∙3375 ( ) 31 P 400 ∩ = M S , 0∙0775 180 135 243 31 ,0.151875 400 400 1600 400 × = ≠ so NOT independent M1 Their P(M) × their P(S) seen, accept unsimplified. A1 P(M), P(S) and P(M ∩ S) notation seen, numerical comparison and correct conclusion, WWW. Method 2 ( ) 31 P 400 ∩ = M S ( ) 135 P 400 = S ( ) 180 P 400 = M ( ) 31 31 400 P | , 0.2296 135 135 400 = = … M S 180 400 ≠ so NOT independent M1 ( ) ( ) ( ) P [P | ] P ∩ = their M S M S their S (oe) seen, accept unsimplified. A1 P(M), P(S) and P(M ∩ S) notation seen, numerical comparison and correct conclusion, WWW. 2 Question Answer Marks Guidance 7(b)(i) Method 1 [1 – P(0,1,2)] = 1 – (10C0 0∙30 0∙710 + 10C1 0∙31 0∙79 + 10C2 0∙32 0∙78) M1 10Cx px (1 – p)10-x for 0 < × < 10, 0 < p < 1, any p. = 1 – (0∙028248 + 0∙121061 + 0∙233474) A1 Correct expression, accept unsimplified, condone omission of final bracket, condone recovery from poor notation. = 0∙617 A1 Accept 0∙61715 ⩽ p ⩽ 0∙61722, WWW. Method 2 [P(3,4,5,6,7,8,9,10) =] 10C3 0∙33 0∙77 + 10C4 0∙34 0∙76 + 10C5 0∙35 0∙75 + 10C6 0∙36 0∙74 + 10C7 0∙37 0∙73 + 10C8 0∙38 0∙72 + 10C9 0∙39 0∙71 + 10C10 0∙310 0∙70 M1 10Cx px (1 – p)10–x for 0 < × < 10, 0 < p < 1, any p. A1 Correct unsimplified expression. = 0∙617 A1 Accept 0∙61715 ⩽ p ⩽ 0∙61722, WWW. 3 Question Answer Marks Guidance 7(b)(ii) [p = 0∙3] Mean = 0∙3 × 90 = 27; variance = 0∙3 × 90 × 0∙7 = 18∙9 B1 Correct mean and variance, allow unsimplified. Condone σ = 4∙347 evaluated. P(X < 32) = 31.5 27 18.9 −   <     P z M1 Substituting their μ and σ (not σ2, √σ) into the ±standardising formula with a numerical value for ‘31∙5’. M1 Using either 31∙5 or 32∙5 within a ±standardising formula with numerical values for their μ and σ (condone σ2, √σ). ( ) 1.035 =Φ M1 Appropriate area Φ, from standardisation formula P(z<…) in final solution, must be probability. = 0∙850 A1 Allow 0∙8495 < p ⩽ 0∙85(0), final answer WWW. 5

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