Cambridge A Level Mathematics 9709 — 2019 May/June Paper 6 · Variant 3

1 The time taken, in minutes, by a ferry to cross a lake has a normal distribution with mean 85 and standard deviation 6.8. (i) Find the probability that, on a randomly chosen occasion, the time taken by the ferry to cross the lake is between 79 and 91 minutes. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Over a long period it is found that 96% of ferry crossings take longer than a certain time t minutes. Find the value of t. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) P(79 < X < 91) = P 79 85 91 85 6.8 6.8 − −   < <     Z = P(–0.8824 < Z < 0.8824) correction = ( ) ( ) Φ 0.8824 Φ 0.8824 − − = 0.8111 – (1 – 0.8111) M1 Correct area ( Φ Φ − ) with one +ve and one –ve z-value or 2 Φ – 1 or 2(Φ 0.5) − = 0.622 A1 Correct answer 3 1(ii) z = –1.751 B1 ± 1.751 seen –1.751 = 85 6.8 − t M1 An equation using ± standardisation formula with a z-value, condone σ2 or √σ t = 73.1 A1 Correct answer 3

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2 Megan sends messages to her friends in one of 3 different ways: text, email or social media. For each message, the probability that she uses text is 0.3 and the probability that she uses email is 0.2. She receives an immediate reply from a text message with probability 0.4, from an email with probability 0.15 and from social media with probability 0.6. (i) Draw a fully labelled tree diagram to represent this information. [2] (ii) Given that Megan does not receive an immediate reply to a message, find the probability that the message was an email. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) B1 Fully correct labelled tree with correct probabilities for ‘Send’ B1 Fully correct labelled branches with correct probabilities for the ‘reply’ 2 Question Answer Marks Guidance 2(ii) P ( ) email NR = ( ) ( ) P email NR 0.2 0.85 P NR 0.3 0.6 0.2 0.85 0.5 0.4 ∩ × = × + × + × M1 P(email) × P(NR) seen as numerator of a fraction, consistent with their tree diagram = 0.17 0.18 0.17 0.2 + + = 0.17 0.55 M1 Summing three appropriate 2-factor probabilities, consistent with their tree diagram, seen anywhere 0.55 oe (can be unsimplified) seen as denom of a fraction = 0.309, 17 55 A1 A1 Correct answer 4

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3 Mr and Mrs Keene and their 5 children all go to watch a football match, together with their friends Mr and Mrs Uzuma and their 2 children. Find the number of ways in which all 11 people can line up at the entrance in each of the following cases. (i) Mr Keene stands at one end of the line and Mr Uzuma stands at the other end. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The 5 Keene children all stand together and the Uzuma children both stand together. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) 9! × 2 = 725760 B1 Exact value 2 3(ii) Eg (K1K2K3K4K5) A A A (U1U2) A = 5! × 2! × 6! B1 2! or 5! seen mult by k > 1, no addition (arranging Us or Ks) B1 6! Seen mult by k > 1, no addition (arranging AAAAKU) = 172800 B1 Exact value 3

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4 (i) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if there must be at least twice as many men as there are women on the committee. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if 2 particular men refuse to be on the committee together. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) M(8) W(4) 4 2 in 8C4 × 4C2 = 420 ways 5 1 in 8C5 × 4C1 = 224 ways 6 0 in 8C6 × 4C0 = 28 ways M1 Summing the number of ways for 2 or 3 correct scenarios (can be unsimplified), no incorrect scenarios Total 672 ways A1 Correct answer 3 Question Answer Marks Guidance 4(ii) Total number of selections = 12C6 = 924 (A) M1 12Cx – (subtraction seen), accept unsimplified Selections with males together = 10C4 = 210 (B) A1 Correct unsimplified expression Total = (A) – (B) = 714 A1 Correct answer Alternative method for question 4(ii) No males + Only male 1 + Only male 2 = 10C6 + 10C5 + 10C5 M1 10Cx + 2 x 10Cy , x ≠ y seen, accept unsimplified = 210 + 252 + 252 A1 Correct unsimplified expression = 714 A1 Correct answer Alternative method for question 4(ii) Pool without male 1 + Pool without male 2 – Pool without either male M1 2 x 11Cx – 10Cx = 11C6 + 11C6 – 10C6 = 462 + 462 – 210 A1 Correct unsimplified expression = 714 A1 Correct answer 3

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5 On average, 34% of the people who go to a particular theatre are men. (i) A random sample of 14 people who go to the theatre is chosen. Find the probability that at most 2 people are men. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Use an approximation to find the probability that, in a random sample of 600 people who go to the theatre, fewer than 190 are men. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) = 0.0029758 + 0.02146239 + 0.071866 A1 Correct unsimplified answer = 0.0963 A1 Correct answer 3 5(ii) Mean =600 × 0.34 = 204, Var = 600 × 0.34 × 0.66 = 134.64 B1 Correct unsimplified np and npq (or sd = 11.603 or Variance = 3366/25) P(< 190) = P 189.5 204 134.64 −   <     z = P(z < –1.2496) M1 Substituting their µ and σ, (no σ2 or √σ) into the Standardisation Formula with a numerical value for ‘189.5’. Condone ± standardisation formula M1 Using continuity correction 189.5 or 190.5 within a Standardisation formula = 1 – Φ (1.2496) M1 Appropriate area Φ from standardisation formula P(z<….) in final solution, (<0.5 if z is –ve, >0.5 if z is +ve) = 1 – 0.8944 = 0.106 A1 Correct final answer 5

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6 A fair five-sided spinner has sides numbered 1, 1, 1, 2, 3. A fair three-sided spinner has sides numbered 1, 2, 3. Both spinners are spun once and the score is the product of the numbers on the sides the spinners land on. (i) Draw up the probability distribution table for the score. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the mean and the variance of the score. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the probability that the score is greater than the mean score. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) score 1 2 3 4 6 9 prob 3 15 4 15 4 15 1 15 2 15 1 15 values if probability of zero stated B1 2 probabilities (with correct score) correct B1 3 or more correct probabilities with correct scores B1 FT Σp = 1, at least 4 probabilities 4 6(ii) mean = (3 8 12 4 12 9) 15 + + + + + = 48 15 (3.2) B1 Var = ( ) 2 (3 16 36 16 72 81) 3.2 15 their + + + + + − M1 FT Substitute their attempts at scores in correct var formula, must have “– mean2 ” (condone probabilities not summing to 1) = 224 15 – 3.22 = 4.69 352 75       A1 3 6(iii) Score of 4, 6, 9 M1 Identifying relevant scores from their mean and their table Prob 4 15 (0.267) A1 Correct answer SC B1 for 4/15 with no working 2

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