Cambridge A Level Mathematics 9709 — 2015 Oct/Nov Paper 6 · Variant 3

2 In country X, 25% of people have fair hair. In country Y, 60% of people have fair hair. There are 20 million people in country X and 8 million people in country Y. A person is chosen at random from these 28 million people. (i) Find the probability that the person chosen is from country X. [1] (ii) Find the probability that the person chosen has fair hair. [2] (iii) Find the probability that the person chosen is from country X, given that the person has fair hair. [2]

Mark scheme: 20  5  2 (i) P(X) =   (0.714),71.4% B1 1 oe 28  7  20 1 8 6 7 (ii) P(F) = × × × = M1 Summing two 2-factor probs created by 28 4 28 10 20 One of ¼ or ¾ multiplied by 20/28 or 8/28 Added to 4/10 or 6/10 × altn population prob A1 2 Correct answer 5 / 28 25 (iii) P ( X | F ) = = ( .0 510) M1 Their unsimplified country X probability 7 / 20 49 (5/28) as num or denom of a fraction Or (their fair hair population) ÷ (total fair hair pop) A1 2 Correct answer 3

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3 Ellie throws two fair tetrahedral dice, each with faces numbered 1, 2, 3 and 4. She notes the numbers on the faces that the dice land on. Event S is ‘the sum of the two numbers is 4’. Event T is ‘the product of the two numbers is an odd number’. (i) Determine whether events S and T are independent, showing your working. [5] (ii) Are events S and T exclusive? Justify your answer. [1]

Mark scheme: 3 3 (i) P(S) = M1 Sensible attempt at P(S) 16 4 P(T) = M1 Sensible attempt at P(T) 16 2 P(S∩T) = B1 Correct P(S∩T) 16 3 2 P(S) × P(T) = ≠ M1 comp P(S) × P(T) with P(S∩T) (their 64 16 values), evaluated Not independent A1 5 Correct conclusion following all correct working (ii) not exclusive since P(S∩T) ≠ 0 FT their P(S∩T), not obtained from P(S) × Or counter example e.g. 1 and 3 P(T), with value and statement. Or P(SUT) ≠ P(S)+P(T) with values B1 1

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4 The time taken for cucumber seeds to germinate under certain conditions has a normal distribution with mean 125 hours and standard deviation 3 hours. (i) It is found that 13% of seeds take longer than 136 hours to germinate. Find the value of 3. [3] (ii) 170 seeds are sown. Find the expected number of seeds which take between 131 and 141 hours to germinate. [4]

Mark scheme: 4 (i) z = 1.127 B1 ± 1.127 seen accept rounding to ±1.13 136 − 125 .1127 = M1 Standardising no cc no sq rt, with attempt σ at z σ = 9.76 A1 3 (not ±0.8078, ±0.5517, ±0.13, ±0.87) Correct ans 131− 125 141− 125 (ii) P(131<x<141)= P ( <z< ) M1 Standardising once with their sd, no √,2, .976 .9 76 allow cc = Φ( 1.639) – Φ(0.6147) M1 Correct area Φ2 – Φ1 = 0.9493 – 0.7307 = 0.2186 M1 Mult by 170, P<1 Number = 0.2186 × 170 = 37 or 38 or awrt 37.2 A1 4 Correct answer, nfww

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5 (a) Find the number of different ways that the 13 letters of the word ACCOMMODATION can be arranged in a line if all the vowels (A, I, O) are next to each other. [3] (b) There are 7 Chinese, 6 European and 4 American students at an international conference. Four of the students are to be chosen to take part in a television broadcast. Find the number of different ways the students can be chosen if at least one Chinese and at least one European student are included. [5]

Mark scheme: 5 (a) e.g. **(AAOOOI)***** B1 8! (8 × 7!) or 6! seen anywhere, either alone or in numerator) !8 !6 × = 604800 M1 Dividing by at least 3 of 2!2!2!3! (may be !2!2 !3!2 fractions added) A1 3 Correct answer (b) C(7) E(6) A(4) M1 Mult 3 appropriate combinations together 1 1 2 = 7 × 6 × 4C2 = 252 assume 6=6C1, 1=4C0 etc., ∑r=4, C&E both 1 2 1 = 7 × 6C2 × 4 = 420 present 1 3 0 = 7 × 6C3 × 1 = 140 2 1 1 = 7C2 × 6 × 4 = 504 A1 At least 3 correct unsimplified products 2 2 0 = 7C2 × 6C2 × 1 = 315 3 1 0 = 7C3 × 6 × 1 = 210 M1* Listing at least 4 different correct options DM1 Summing at least 4 outcomes, involving 3 combs or perms, ∑r=4 Total = 1841 A1 5 Correct answer SC if CE removed, M1 available for listing at least 4 different correct options for remaining 2. DM1 for 7C1×6C1×(sum of at least 4 outcomes)

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6 The heights to the nearest metre of 134 office buildings in a certain city are summarised in the table below. Height (m) 21 −40 41 −45 46 −50 51 −60 61 −80 Frequency 18 15 21 52 28 (i) Draw a histogram on graph paper to illustrate the data. [4] (ii) Calculate estimates of the mean and standard deviation of these heights. [5]

Mark scheme: 6 (i) fd 0.9, 3, 4.2, 5.2, 1.4 M1 Attempt at scaled freq [f/(attempt at cw)] fd 5 4 A1 Correct heights seen on diagram 3 Scale no less than 1cm to 1 unit 2 B1 Correct bar widths visually no gaps 1 B1 4 Labels (ht/metres and fd or freq per 20 m 20.5 30.5 40.5 50.5 60.5 70.5 80.5 etc.) and end points at 20.5 etc. condone 2 ht metres end point errors, scale no less than 1cm to 5m for 20,30… unless clearly accurate, linear scale between 20.5 and 80 (ii) (30.5 × 18 + 43 × 15 + 48 × 21 + 55.5 × 52 + M1 Attempt at unsimplified, mid points (at least 70.5 × 28)/134 4 within 0.5) 7062 = = 52.701 M1 Attempt at Σfx their mid points ÷ 134 134 A1 Correct mean rounding to 53 Var = (30.52 × 18 + 432 × 15 + 482 × 21 + 55.52 M1 Attempts at Σfx2 their mid points ÷ their Σf – × 52 + 70.52 × 28)/134 – 52.7012 mean2 = 392203.5/134 – 52.7012 = 149.496 A1 5 Correct answer, nfww sd = 12.2 19 19

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7 A factory makes water pistols, 8% of which do not work properly. (i) A random sample of 19 water pistols is taken. Find the probability that at most 2 do not work properly. [3] (ii) In a random sample of n water pistols, the probability that at least one does not work properly is greater than 0.9. Find the smallest possible value of n. [3] (iii) A random sample of 1800 water pistols is taken. Use an approximation to find the probability that there are at least 152 that do not work properly. [5] (iv) Justify the use of your approximation in part (iii). [1]

Mark scheme: 7 (i) P(0, 1, 2) = M1 Binomial term 19Cxpx(1 – p)19-x seen 0<p<1 (0.92)19+19C1(0.08)(0.92)18+ 19C2(0.08)2(0.92)17 M1 Correct unsimplified expression = 0.809 A1 3 Correct answer (no working SC B2) (ii) P(at least 1) = 1 – P(0) = 1 – P(0.92)n > 0.90 M1 Eqn with their 0.92n, 0.9 or 0.1, 1 not nec 0.1 > (0.92)n M1 Solving attempt by logs or trial and error, n > 27.6 power eqn with one unknown power Ans 28 A1 3 Correct answer, not approx., ≈, ⩾, >, ⩽, < (iii) np = 1800 × 0.08 = 144 B1 correct unsimplified np and npq seen npq = 132.48 accept 132.5, 132, 11.5, awrt 11.51   1515. − 144   M1 standardising, with √ P( at least 152) = P  z >      132.48   M1 cont correction 151.5 or 152.5 seen = P(z > 0.6516) M1 correct area 1 – Φ (probability) = 1 – 0.7429 = 0.257 A1 5 correct answer (iv) Use because 1800 ×0.08 (and 1800 × 0.92 are B1 1 1800 ×0.08 > 5 is sufficient both) > 5 np>5 is sufficient if clearly evaluated in (iii) If npq>5 stated then award B0

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