Cambridge A Level Mathematics 9709 — 2022 Feb/March Paper 5 · Variant 2

2 In a certain country, the probability of more than 10cm of rain on any particular day is 0.18, independently of the weather on any other day. (a) Find the probability that in any randomly chosen 7-day period, more than 2 days have more than 10cm of rain. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) For 3 randomly chosen 7-day periods, find the probability that exactly two of these periods have at least one day with more than 10cm of rain. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) [P(>2) = 1 – P(0,1,2) =] 1 – (7C0 0 7 0.18 0.82 + 7C1 1 6 0.18 0.82 + 7C2 2 5 0.18 0.82 ) M1 One term 7Cx ( ) 7 1 , 0 1, 0 7 − − < < < < x x p p p x = 1 – (0.249285 + 0.383048 + 0.252251) = 1 – 0.88458 A1 Correct unsimplified expression or better Condone omission of brackets if recovered 0.115 B1 WWW. 0∙115 ⩽ p < 0∙1155 not from wrong working 3 2(b) [P(at least 1 day of rain) = 1 – P(0) = ( ) 7 1 0.82 ] 0.7507 − = B1 AWRT 0.751 seen [P(exactly 2 periods) =] ( ) 2 0.7507 1 0.7507 3 × − × M1 FT their 7 1−p or their 0.7507 if identified, not 0.18, 0.82 Accept ×3Cr, r=1,2 or ×3P1 for ×3 Condone ×2 0.421 A1 Accept 0.421 ⩽ p ⩽ 0.4215 SC B1 if 0/3 scored for final answer only 0.421 ⩽ p ⩽ 0.4215 3

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3 At a summer camp an arithmetic test is taken by 250 children. The times taken, to the nearest minute, to complete the test were recorded. The results are summarised in the table. Time taken, in minutes 1 −30 31 −45 46 −65 66 −75 76 −100 Frequency 21 30 68 86 45 (a) Draw a histogram to represent this information. [4] (b) State which class interval contains the median. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Given that an estimate of the mean time is 61.05 minutes, state what feature of the distribution accounts for the median and the mean being different. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Class Width 30 15 20 10 25 Frequency Density 0.7 2 3.4 8.6 1.8 M1 At least 4 frequency densities calculated A1 All heights correct on graph B1 Bar ends at 0∙5, 30∙5, 45∙5, 65.5, 75.5, 100.5 (at axis), 5 bars drawn, condone 0 in first bar 0.5 ⩽ time axis ⩽ 100.5, linear scale with at least 3 values indicated. B1 Axes labelled: Frequency density (fd), time (t) and mins (or appropriate title). Linear fd scale, with at least 3 values indicated 0 ⩽ fd axis ⩽ 8∙6 4 3(b) 66 – 75 B1 Condone 65.5 – 75.5 1 3(c) Distribution is not symmetrical B1 Or skewed, ignore nature of skew 1

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4 The weights of male leopards in a particular region are normally distributed with mean 55kg and standard deviation 6kg. (a) Find the probability that a randomly chosen male leopard from this region weighs between 46 and 62kg. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The weights of female leopards in this region are normally distributed with mean 42kg and standard deviation 3 kg. It is known that 25% of female leopards in the region weigh less than 36kg. (b) Find the value of 3. 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The distributions of the weights of male and female leopards are independent of each other. A male leopard and a female leopard are each chosen at random. (c) Find the probability that both the weights of these leopards are less than 46kg. 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Mark scheme: 4(a) P( 46 55 62 55 46 62) P 6 6 X Z − −   < < = < <     M1 46 or 62, 55 and 6 substituted into ±standardisation formula once. Condone 62 and continuity correction ±0.5 7 P 1.5 6   = − < <     Z B1 Both standardisation values correct, accept unsimplified ( ) ( ) 7 =Φ 1 Φ 1.5 6    − −         = ( ) 0.8784 0.9332 1 + − M1 Calculating the appropriate area from stated Φs of z-values, must be probabilities. 0.812 A1 0∙8115 < p ⩽ 0∙812 4 4(b) z = ±0.674 B1 CAO, critical z-value 36 42 0.674 σ − = − M1 36 and 42 substituted in ±standardisation formula, no continuity correction, not σ2, √ σ, equated to a z-value [ ] 8.9 0 σ = A1 WWW. Only dependent on M. 3 Question Answer Marks Guidance 4(c) P(male < 46) = 1−their 0.9332 = 0.0668 M1 FT value from part (a) or Correct: 46 55 1 Φ 6 −   −     ,condone continuity correction, σ2, √ σ, and probability found. Condone unsupported correct value stated. P(female < 46) = P( ( ) 46 42 ) Φ 0.449 8.90 − < =    Z their 0.6732 = M1 46, 42 and their 4(b) σ (or correct σ) substituted in ±standardisation formula, condone continuity correction, σ2, √ σ, and probability found Condone 4 8.90 their . P(both) = 0.0668 ×0.6732 M1 Product of their 2 probabilities (0 < both < 1) Not 0.25 or their final answer to 4(a) used. 0.0450 or 0.0449 A1 0∙0449 ⩽ p ⩽ 0∙0450 4

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5 A group of 12 people consists of 3 boys, 4 girls and 5 adults. (a) In how many ways can a team of 5 people be chosen from the group if exactly one adult is included? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) In how many ways can a team of 5 people be chosen from the group if the team includes at least 2 boys and at least 1 girl? [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The same group of 12 people stand in a line. (c) How many different arrangements are there in which the 3 boys stand together and an adult is at each end of the line? 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Mark scheme: 5(a) M1 Condone 5P1 for M1 only 175 A1 2 Question Answer Marks Guidance 5(b) 2B 1G 2A 3C2 × 4C1 × 5C2 = 120 2B 2G 1A 3C2 × 4C2 × 5C1 = 90 2B 3G 3C2 × 4C3 = 12 3B 1G 1A 3C3 × 4C1 × 5C1 = 20 3B 2G 3C3 × 4C2 = 6 M1 3Cx × 4Cy × 5Cz , x + y + z = 5, x,y,z integers ⩾1 Condone use of permutations for this mark B1 2 appropriate identified outcomes correct, allow unsimplified M1 Summing their values for 4 or 5 correct identified scenarios only (no repeats or additional scenarios), condone identification by unsimplified expressions [Total =] 248 A1 Note: Only dependent upon M marks 4 5(c) 8! × 3! × 5P2 M1 8! × m, m an integer ⩾ 1 Accept 8 × 7! for 8! M1 3! × n, n an integer > 1 M1 p × 5P2, p × 5C2 × 2, p × 20, p an integer > 1 If extra terms present, maximum 2/3 M marks available 4 838 400 A1 Exact value required 4

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6 A factory produces chocolates in three flavours: lemon, orange and strawberry in the ratio 3 : 5 : 7 respectively. Nell checks the chocolates on the production line by choosing chocolates randomly one at a time. (a) Find the probability that the first chocolate with lemon flavour that Nell chooses is the 7th chocolate that she checks. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that the first chocolate with lemon flavour that Nell chooses is after she has checked at least 6 chocolates. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ‘Surprise’ boxes of chocolates each contain 15 chocolates: 3 are lemon, 5 are orange and 7 are strawberry. Petra has a box of Surprise chocolates. She chooses 3 chocolates at random from the box. She eats each chocolate before choosing the next one. (c) Find the probability that none of Petra’s 3 chocolates has orange flavour. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Find the probability that each of Petra’s 3 chocolates has a different flavour. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (e) Find the probability that at least 2 of Petra’s 3 chocolates have strawberry flavour given that none of them has orange flavour. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) [Probability of lemon = 3 1] 15 5 = 6 4 1 4096 , 0.0524 5 5 78125    × =           B1 0.0524288 rounded to more than 3SF if final answer 1 6(b) 6 1 1 5   −     M1 or 6 4 5       . FT their 1 5 or correct. From final answer Condone 5 5 6 4 1 4 4 or 5 5 5 5         × +                 4096 , 15625 0.262 A1 0.262144 rounded to more than 3SF Alternative method for question 6(b) [1 – P(1,2,3,4,5,[6]) =] 1 – 2 3 4 5 1 4 1 4 1 4 1 4 1 4 1 5 5 5 5 5 5 5 5 5 5 5           + × + × + × + × + ×                       M1 From final answer Condone omission of 5 4 1 5 5  ×     4096 , 15625 0.262 A1 0.262144 rounded to more than 3SF 2 Question Answer Marks Guidance 6(c) 10 9 8 15 14 13 × × M1 1 2 15 14 13 − − × × a a a , no additional terms 24 91 , 0∙264 A1 0.263736 rounded to more than 3SF Alternative method for question 6(c) 3 2 1 3 2 7 3 15 14 13 15 14 13 × × + × × × 3 7 6 7 6 5 3 15 14 13 15 14 13 + × × × + × × M1 [3Ls + 2Ls1S + 1L2Ss + 3Ss] Condone one numerator error. Condone no multiplications seen if tree diagram complete with probabilities on each branch, scenarios listed and attempt at evaluation 24 91 , 0∙264 A1 0.263736 rounded to more than 3SF Alternative method for question 6(c) 5 4 3 5 4 10 5 10 9 1 3 3 15 14 13 15 14 13 15 14 13   − × × + × × × + × × ×     M1 1 ‒ P(3,2,1 oranges) Condone one numerator error. 24 91 , 0∙264 A1 0.263736 rounded to more than 3SF Alternative method for question 6(c) 10 3 15 3 C c M1 24 91 , 0∙264 A1 0.263736 rounded to more than 3SF 2 Question Answer Marks Guidance 6(d) 7 5 3 3! 15 14 13 × × × M1 All probabilities of the form: 7 5 3 × × a b c , 13 ⩽ a,b,c ⩽ 15 M1 3! × × × e g i f h j e,f,g,h,i.j positive integers forming probabilities or 6 identical probability calculations or values added, no additional terms 3 13 , 0∙231 A1 0∙230769 rounded (not truncated) to more than 3SF Alternative method for question 6(d) 3 5 7 1 1 1 15 3 C C C C × × M1 3 5 7 1 1 1 C C C × × k , k integer > 1 Condone use of permutations M1 3 5 7 15 3 C C C C × × a b c , 0<a<3, 0<b<5, 0<c<7, Condone use of permutations 3 13 , 0∙231 A1 0∙230769 rounded (not truncated) to more than 3SF 3 Question Answer Marks Guidance 6(e) ( ) 7 6 5 3 7 6 3 15 14 13 15 14 13 × × + × × × their c 14 24 65 91   = ÷     B1 3 7 6 3 15 14 13 × × × seen (SSL, SLS, LSS) SC B1 3 126 3, 3 65 2730 × × seen B1 7 6 5 15 14 13 × × seen in numerator (SSS) SCB1 210 1 , 2730 13 seen in numerator M1 Fraction with their (c) or correct in denominator 720 24 , , 0.263736 2730 91       = 49 60 , 0∙817 A1 Accept 0.816 Alternative method for question 6(e) 7 3 7 2 1 3 10 3 C C C C × + B1 7 3 2 1 C C × seen (SSL, SLS, LSS) SCB1 21 × 3 seen or use of permutations B1 7 3 C seen in numerator (SSS) SCB1 35 seen in numerator or use of permutations M1 Fraction with 10 3 C or consistent with their numerator of 6(c) in denominator = 49 60 , 0∙817 A1 Accept 0.816 4

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