Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 6 · Variant 3

1 A group consists of 5 men and 2 women. Find the number of different ways that the group can stand in a line if the women are not next to each other. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Method 1 ... M … M … M … M … M … M1 k × 5! (120) or k × 6P2 (30), k is an integer ⩾ 1, No. ways men placed × No. ways women placed in gaps = 5! × 6P2 M1 Correct unsimplified expression = 3600 A1 Correct answer Method 2 Number with women together = 6! × 2 (1440) Total number of arrangements = 7! (5040) M1 6! × 2 or 7! – k seen, k is an integer ⩾ 1 Number with women not together = 7! – 6! × 2 M1 Correct unsimplified expression = 3600 A1 Correct answer 3

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3 A box contains 3 red balls and 5 blue balls. One ball is taken at random from the box and not replaced. A yellow ball is then put into the box. A second ball is now taken at random from the box. (i) Complete the tree diagram to show all the outcomes and the probability for each branch. [2] First ball Second ball (ii) Find the probability that the two balls taken are the same colour. 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(iii) Find the probability that the first ball taken is red, given that the second ball taken is blue. 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Mark scheme: 3(i) B1 Fully correct labelled tree and correct probabilities for ‘First Ball’ B1 Correct probabilities (with corresponding labels) for ‘Second Ball’ 2 3(ii) P(RR) + P(BB) = 3/8 × 2/8 + 5/8 × 4/8 = 3/32 + 5/16 M1 Correct unsimplified expression from their tree diagram, Σp = 1 on each branch = 13/32 (0.406) A1 Correct answer 2 Question Answer Marks Guidance 3(iii) P(RB) = 3 / 8 5 / 8 × = 15/64 M1 ( ) ( ) 1st ball red 2nd ball blue P P × from their tree diagram seen unsimplified as numerator or denominator of a fraction Allow Σp ≠ 1 on each branch P(B) = 3/8 × 5/8 + 5/8 × 4/8 = 35/64 M1 Correct unsimplified expression for P(B) from their tree diagram seen as denominator of a fraction. Allow Σp ≠ 1 on each branch P(R|B) = P(RB) / P(B) = (15/64) ÷ (35/64) = 3/7 (0.429) A1 Correct answer 3

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4 Out of a class of 8 boys and 4 girls, a group of 7 people is chosen at random. (i) Find the probability that the group of 7 includes one particular boy. 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(ii) Find the probability that the group of 7 includes at least 2 girls. 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Mark scheme: 4(i) B1 Seen as denominator of fraction Selections with boy included = 11C6 or 12C7 – 11C7 = 462 M1 Correct unsimplified expression for selections with boy included seen as numerator of fraction Probability = 462/792 = 7/12 (0.583) A1 Correct answer OR prob of boy not included = 11/12 × 10/11 × …. × 5/6= 5/12 B1 Correct unsimplified prob 1 – 5/12 M1 Subtracting prob from 1 = 7/12 A1 Correct answer 3 Question Answer Marks Guidance 4(ii) Method 1 Scenarios are: 2G + 5B: 4C2 × 8C5 = 336 B1 One unsimplified product correct 3G + 4B: 4C3 × 8C4 = 280 4G + 3B: 4C4 × 8C3 = 56 M1 No of selections (products of n C r and n P r ) added for 2, 3 and 4 girls with no of girls and no of boys summing to 7 Total = 672 A1 Correct total Probability = 672/792 (28/33) (0.848) A1ft Correct answer – ‘total’/( ‘total no of selections’ from i) Method 2 0G + 7B 4C0 × 8C7 = 8 B1 One unsimplified no of selections correct 1G + 6B 4C1 × 8C6 = 112 Total = 8 + 112 = 120 M1 No of selections (products of n C r and n P r ) added for 0 and 1 girls with no of girls and no of boys summing to 7 (12C7 – 120)/792 or 1 – 120/792 A1 792 – 120 = 672 or 1 – 120/792 Probability = 672/792 (28/33) (0.848) A1ft ‘672’ over ‘792’ from i Method 3 (probability) 1 – P(0) – P(1) = 1 – (8/12 × 7/11 × …… × 2/6) – (8/12 × … × 3/7 × 4/6 × 7) B1 One correct unsimplified prob for 0 or 1 = 1 – 1/99 –14/99 M1 Subtracting ‘P(0)’ and ‘P(1)’ (using products of 7 fractions with denominators from 12 to 6) from 1 A1 Both probs correct unsimplified = 84/99 = 28/33 A1ft 1 – ‘P(0)’ – ‘P(1)’ Question Answer Marks Guidance 4(ii) Method 4 (probability) P(2) + P(3) + P(4) = B1 One correct unsimplified prob for 2, 3 or 4 42/99 + 35/99 + 7/99 M1 Adding ‘P(2)’, ‘P(3)’ and P(4)’ (using products of 7 fractions with denominators from 12 to 6) A1 Three probs correct unsimplified = 84/99 = 28/33 A1ft ‘P(2)’+ ‘P(3)’ + ‘P(4)’ 4

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5 The weights of apples sold by a store can be modelled by a normal distribution with mean 120 grams and standard deviation 24 grams. Apples weighing less than 90 grams are graded as ‘small’; apples weighing more than 140 grams are graded as ‘large’; the remainder are graded as ‘medium’. (i) Show that the probability that an apple chosen at random is graded as medium is 0.692, correct to 3 significant figures. 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(ii) Four apples are chosen at random. Find the probability that at least two are graded as medium. 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Mark scheme: 5(i) 1 90 120 24 z = ± = – 5 4 , 2 140 120 24 z = ± = 5 6 and either 90 or 140 = 20 30 Φ Φ 24 24     − −         A1 –5/4 and 5/6 unsimplified = ( ) ( ) Φ 0.8333 (1 Φ 1.25 ) − − = 0.7975 – (1 – 0.8944) or 0.8944 – 0.2025 = 0.6919 M1 Correct area Φ – Φ legitimately obtained and evaluated from phi(their z2) – phi (their z1) = 0.692 AG A1 Correct answer obtained from 0.7975 and 0.1056 oe to 4sf or 0.6919 seen www 4 Question Answer Marks Guidance 5(ii) Method 1 Probability = P(2, 3, 4) = 0.6922(1 – 0.692)2 × 4C2 + 0.6923(1 – 0.692) × 4C3 + 0.6924 M1 Any binomial term of form ( ) 4 4 1 x x x C p p − − , x ≠ 0 or 4 B1 One correct bin term with 4 n = and 0.692 p = , = 0.27256 + 0.40825 + 0.22931 M1 Correct unsimplified expression using 0.692 or better = 0.910 A1 Correct answer Method 2: 1 – P(0, 1) = M1 Any binomial term of form ( ) 4 4 1 x x x C p p − − , x≠0 or 4 1 − 0.6920(1 – 0.692)4 × 4C0 − 0.6921(1 – 0.692)3 × 4C1 B1 One correct bin term with 4 n = and 0.692 p = = 1 – 0.00899 – 0.0808757 M1 Correct unsimplified expression using 0.692 or better = 0.910 A1 Correct answer 4

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6 The lifetimes, in hours, of a particular type of light bulb are normally distributed with mean 2000 hours and standard deviation 3 hours. The probability that a randomly chosen light bulb of this type has a lifetime of more than 1800 hours is 0.96. (i) Find the value of 3. 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New technology has resulted in a new type of light bulb. It is found that on average one in five of these new light bulbs has a lifetime of more than 2500 hours. (ii) For a random selection of 300 of these new light bulbs, use a suitable approximate distribution to find the probability that fewer than 70 have a lifetime of more than 2500 hours. 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(iii) Justify the use of your approximate distribution in part (ii). [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) P(X >1800) = 0.96, so P( 1800 2000 Z σ > ) = 0.96 B1 ± 1.75 seen 200 Φ( ) σ = 0.96 200 1.751 σ = M1 1800 2000 z σ − = ± , allow cc, allow sq rt, allow sq equated to a z-value 114 σ = A1 Correct final answer www 3 6(ii) Mean = 300 × 0.2 = 60 and variance = 300 × 0.2 × 0.8 = 48 B1 Correct unsimplified mean and variance P(X < 70) = P( 69.5 60 48 Z − > ) M1 Z = ± 60 48 x their their − = ( ) Φ 1.371 M1 69.5 or 70.5 seen in an attempted standardisation expression as cc =0.915 A1 Correct final answer 4 6(iii) np = 60, nq = 240: both > 5, (so normal approximation holds) B1 Both parts evaluated are required 1

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7 The heights, in cm, of the 11 members of the Anvils athletics team and the 11 members of the Brecons swimming team are shown below. Anvils 173 158 180 196 175 165 170 169 181 184 172 Brecons 166 170 171 172 172 178 181 182 183 183 192 (i) Draw a back-to-back stem-and-leaf diagram to represent this information, with Anvils on the left-hand side of the diagram and Brecons on the right-hand side. [4] (ii) Find the median and the interquartile range for the heights of the Anvils. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The heights of the 11 members of the Anvils are denoted by x cm. It is given that Σx = 1923 and Σx2 = 337 221. The Anvils are joined by 3 new members whose heights are 166 cm, 172 cm and 182 cm. (iii) Find the standard deviation of the heights of all 14 members of the Anvils. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) Anvils Brecons 8 15 9 5 16 6 5 3 2 0 17 0 1 2 2 8 4 1 0 18 1 2 3 3 6 19 2 Key: 5|16|6 means 165 cm for Anvils and 166 cm for Brecons B1 Correct Anvils labelled on left, leaves in order from right to left and lined up vertically, no commas B1 Correct Brecons labelled on same diagram on right hand side in order from left to right and lined up vertically, no commas B1 Correct key, not split, both teams, at least one with cm 4 7(ii) Median = 173 B1 Correct median (or Q2) LQ = 169; UQ = 181 IQR = 181 – 169 M1 Either UQ = 181 ± 4, or LQ = 169 ± 4 and evaluating UQ – LQ = 12 A1 Correct answer from 181 and 169 only 3 Question Answer Marks Guidance 7(iii) Σx = 1923 + 166 + 172 + 182 (= 2443) 2 x ∑ = 337221 + 1662 + 1722 + 1822 (= 427485) M1 Correct unsimplified expression for x ∑ and 2 x ∑ , may be implied Mean = 2443 14 14 x ∑ = = 174.5 M1 Correct unsimplified mean Variance = 2 2 2 427485 2443 14 14 14 14 x x ∑ ∑     − = −         M1 Correct unsimplified variance using 14, their Σx and their Σx2, not using 1923 and/or 337221 S d = 9.19 A1 Correct answer 4

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