Cambridge A Level Mathematics 9709 — 2019 May/June Paper 6 · Variant 1

2 Jameel has 5 plums and 3 apricots in a box. Rosa has x plums and 6 apricots in a box. One fruit is chosen at random from Jameel’s box and one fruit is chosen at random from Rosa’s box. The probability that both fruits chosen are plums is 4.1 Write down an equation in x and hence find x. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Jameel: P(plum) = 5 8 , Rosa: P(plum) = 6 + x x 5 1 8 6 4 × = + x x A1 Correct equation oe (x =) 4 A1 SC correct answer with no appropriate equations i.e. common sense B1 3

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3 A fair six-sided die is thrown twice and the scores are noted. Event X is defined as ‘The total of the two scores is 4’. Event Y is defined as ‘The first score is 2 or 5’. Are events X and Y independent? Justify your answer. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 P(X) = 3 36 1 12 oe       P(Y) = 12 36 1 3 oe       B1 P(X∩Y) = 1 36 M1 Independent method to find P(X∩Y) without multiplication, either stated or by listing or circling numbers on a probability space diagram. OR condititional prob with a single fraction numerator P(X) × P(Y) = P(X∩Y), independent A1 Numerical comparison and conclusion, www 4

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4 The Mathematics and English A-level marks of 1400 pupils all taking the same examinations are shown in the cumulative frequency graphs below. Both examinations are marked out of 100. 1500 1400 1300 1200 1100 English Mathematics 1000 900 frequency 800 700 Cumulative 600 500 400 300 200 100 0 0 10 20 30 40 50 60 70 80 90 100 Marks Use suitable data from these graphs to compare the central tendency and spread of the marks in Mathematics and English. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Median Maths = 40 M1 Indication of finding medians, such as mark on graph or reference marks to 700 pupils, condone poor terminology such as ‘mean’ Median English = 55 A1 Both values correct, condone 54<English<56 but 54, 56 get A0 Median of English is larger than median of Maths B1 Correct statement, median must be referenced within answer. No credit if statement references ‘means’ Range Maths is 100 or IQ range Maths = 80 – 12 = 68 M1 Evidence of finding either both ranges or both IQ ranges i.e. see a minus Range English is 60 or IQ range English = 62 – 42 = 20 A1 Both ranges or IQR correct Maths marks have more spread then English marks B1 Correct conclusion. Accept standard deviation but must see some figures 6

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5 In a certain country the probability that a child owns a bicycle is 0.65. (i) A random sample of 15 children from this country is chosen. Find the probability that more than 12 own a bicycle. 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(ii) A random sample of 250 children from this country is chosen. Use a suitable approximation to find the probability that fewer than 179 own a bicycle. 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Mark scheme: 5(i) (P > 12) = P(13, 14, 15) = 15C13(0.65)13(0.35)2 + 15C14(0.65)14(0.35)1 + (0.65)15 A1 Correct unsimplified answer = 0.0617 A1 SC if use np and npq with justification give (12.5 – 9.75)/√3.41 M1 1–F(1.489) A1 0.0681 A0 3 5(ii) mean = 250 × 0.65 = 162.5 variance = 250 × 0.65 × 0.35 = 56.875 B1 Correct unsimplified np and npq P(< 179) = P(z <178.5 162.5 56.875 − ) = P(z < 2.122) M1 Substituting their µ and σ (condone σ2) into the Standardisation Formula with a numerical value for ‘178.5’. Continuity correct not required for this M1. Condone ± standardisation formula Using continuity correction 178.5 or 179.5 M1 = 0.983 A1 Correct final answer 4

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6 At a funfair, Amy pays $1 for two attempts to make a bell ring by shooting at it with a water pistol. ³ If she makes the bell ring on her first attempt, she receives $3 and stops playing. This means that overall she has gained $2. ³ If she makes the bell ring on her second attempt, she receives $1.50 and stops playing. This means that overall she has gained $0.50. ³ If she does not make the bell ring in the two attempts, she has lost her original $1. The probability that Amy makes the bell ring on any attempt is 0.2, independently of other attempts. (i) Show that the probability that Amy loses her original $1 is 0.64. 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(ii) Complete the probability distribution table for the amount that Amy gains. [4] Amy’s gain ($) Probability 0.64 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Calculate Amy’s expected gain. 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Mark scheme: 6(i) P(loses $1) = P( F and F) = 0.8 × 0.8 M1 0.8 x 0.8 or (1 – 0.2)(1-0.2) or P(F) × P(F) or P(F)+P(F) seen or implied = 0.64 AG A1 Must see probabilities multiplied together with final answer and a clear probability statement or implied by labelled tree diagram 2 Question Answer Marks Guidance 6(ii) Amount gained ($) –1 0.50 2 Prob 0.16 0.2 B1 –1 linked with 0.64 in table B1 0.5 seen in table B1 0.16 seen in table linked to their 0.5 B1 FT P(2.00 gained) = 0.36 – P(0.50 gained) or correct, and all amount gained linked correctly in table 4 6(iii) E(winnings) = –1 × 0.64 + 0.5 × 0.16 + 2 × 0.2 = –($)0.16, –16 cents B1 FT Accept ($)0.16 or 16 cents loss. FT unsimplified E(winnings) from their table provided Σp = 1 1

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8 Freddie has 6 toy cars and 3 toy buses, all different. He chooses 4 toys to take on holiday with him. (i) In how many different ways can Freddie choose 4 toys? [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) How many of these choices will include both his favourite car and his favourite bus? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Freddie arranges these 9 toys in a line. (iii) Find the number of possible arrangements if the buses are all next to each other. 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(iv) Find the number of possible arrangements if there is a car at each end of the line and no buses are next to each other. 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Mark scheme: 8(i) B1 1 8(ii) 7C2 B1 7Cx or yC2 (implied by correct answer) or 7Px or 7Py, seen alone = 21 B1 correct answer 2 Question Answer Marks Guidance 8(iii) _ C1 (B1 B2 B3 ) C2 _ C3 _ C4 _ C5 _ C6 B1 3! or 6! seen alone or multiplied by k > 1 need not be an integer 3! × 6! × 7 B1 3! and 6! seen multiplied by k > 1, integer, no division = 30240 B1 Exact value Alternative method for question 8(iii) C1 (B1 B2 B3 ) C2 C3 C4 C5 C6 B1 3! or 7! seen alone or multiplied by k > 1 need not be an integer 3! × 7! B1 3! and 7! seen multiplied by k > or = 1, no division = 30240 B1 Exact value 3 8(iv) C1 _ C2 _ C3 _ C4 _ C5 _C6 B1 6! or 4! X 6P2 seen alone or multiplied by k > 1, no division (arrangements of cars) 6! × 5P3 or 6! × 5 × 4 × 3 or 6! x 3! x10 B1 Multiply by 5P3 oe i.e. putting Bs in between 4 of the Cs OR multiply by 3! x n where n = 7, 8, 9, 10 (number of options) = 43200 B1 Correct answer 3

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