Cambridge A Level Mathematics 9709 — 2017 May/June Paper 6 · Variant 3

1 A biased die has faces numbered 1 to 6. The probabilities of the die landing on 1, 3 or 5 are each equal to 0.1. The probabilities of the die landing on 2 or 4 are each equal to 0.2. The die is thrown twice. Find the probability that the sum of the numbers it lands on is 9. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 P(6) = 0.3 B1 SOI P(sum is 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) M1 Identifying the four ways of summing to 9 (3,6), (6,3) (4,5) and (5,4) = (0.03 + 0.02) × 2 M1 Mult 2 probs together to find one correct prob of (3,6), (6,3) (4,5) or (5,4) unsimplified = 0.1 A1 OE Total: 4 np = 270 × 1/3 = 90, npq = 270 × 1/3 × 2/3 = 60

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2 The probability that George goes swimming on any day is 3.1 Use an approximation to calculate the probability that in 270 days George goes swimming at least 100 times. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 B1 Correct unsimplified np and npq, SOI P( ) 100 > x = P 99.5 90 60 −   >     z = P(z > 1.2264) M1 M1 ±Standardising using 100 need sq rt Continuity correction, 99.5 or 100.5 used = 1 – 0.8899 M1 Correct area 1 – Φ implied by final prob. < 0.5 = 0.110 A1 Total: 5 P(S) = 0.65 × 0.6 + 0.35 × 0.75 M1 Summing two 2-factor probs or 1 – (sum of two 2-factor probs)

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3 A shop sells two makes of coffee, Caf´e Premium and Caf´e Standard. Both coffees come in two sizes, large jars and small jars. Of the jars on sale, 65% are Caf´e Premium and 35% are Caf´e Standard. Of the Caf´e Premium, 40% of the jars are large and of the Caf´e Standard, 25% of the jars are large. A jar is chosen at random. (i) Find the probability that the jar is small. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that the jar is Caf´e Standard given that it is large. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) = 0.653 (261/400) A1 Total: 2 Question Answer Marks Guidance 3(ii) P( ) Std L = ( ) ( ) ∩ P Std L P L = 0.35 0.25 1 0.6525 × − = 0.0875/0.3475 M1 M1 ‘P(Std)’ × ‘P(L/Std)’as num of a fraction. Could be from tree diagram in 3(i). Denominator (1 - their (i)) or their (i) or 0.65 × 0.4(or 0.6) + 0.35 × 0.25(or 0.75) = 0.26+0.0875 or P(L) from their tree diagram = 0.252 (35/139) A1 Total: 3 ±Standardising, in terms of µ and/or σ with 0 - …. in numerator,

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5 Hebe attempts a crossword puzzle every day. The number of puzzles she completes in a week (7 days) is denoted by X. (i) State two conditions that are required for X to have a binomial distribution. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ On average, Hebe completes 7 out of 10 of these puzzles. (ii) Use a binomial distribution to find the probability that Hebe completes at least 5 puzzles in a week. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Use a binomial distribution to find the probability that, over the next 10 weeks, Hebe completes 4 or fewer puzzles in exactly 3 of the 10 weeks. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) constant probability (of completing) B1 Any one condition of these two independent trials/events B1 The other condition Totals: 2 5(ii) P(5, 6, 7) = 7C5(0.7)5(0.3)2 + 7C6(0.7)6(0.3)1 + (0.7)7 M1 A1 Bin term 7Cx(0.7)x(0.3)7-x , x ≠ 0, 7 Correct unsimplified answer (sum) OE = 0.647 A1 Total: 3 5(iii) P(0, 1, 2, 3, 4) = 1 – their ‘0.6471’ = 0.3529 M1 Find P( 4 - ) either by subtracting their (ii) from 1 or from adding Probs of 0,1,2,3,4 with n=7 (or 10) and p = 0.7 P(3) = 10C3(0.3529)3(0.6471)7 M1 10C3 (their 0.353)3(1 – their 0.353)7 on its own = 0.251 A1 First digit in 2 ways. 2 × 4 × 3 × 2 or 2 × 4P3 1, 2 or 3 × 4P3 OE as final answer

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6 (a) Find how many numbers between 3000 and 5000 can be formed from the digits 1, 2, 3, 4 and 5, (i) if digits are not repeated, [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) if digits can be repeated and the number formed is odd. [3] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) A box of 20 biscuits contains 4 different chocolate biscuits, 2 different oatmeal biscuits and 14 different ginger biscuits. 6 biscuits are selected from the box at random. (i) Find the number of different selections that include the 2 oatmeal biscuits. [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) Find the probability that fewer than 3 chocolate biscuits are selected. [4] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................

Mark scheme: 6(a)(i) M1 Total = 48 ways A1 Total: 2 6(a)(ii) 2 × 5 × 5 × 3 M1 M1 Seeing 52 mult; this mark is for correctly considering the middle two digits with replacement Mult by 6; this mark is for correctly considering the first and last digits = 150 ways A1 Totals: 3 Question Answer Marks Guidance 6(b)(i) OO**** in 18C4 ways M1 18Cx or the sum of five 2-factor products with n = 14 and 4, may be × by 2C2: 4C0 × 14C4 + 4C1 × 14C3 + 4C2 × 14C2 + 4C3 × 14C1 + 4C4 (× 14C0) = 3060 A1 Totals: 2 Question Answer Marks Guidance 6(b)(ii) Choc Not Choc 0 6= 1 × 16C6 = 8008 0.2066 1 5= 4C1 × 16C5 = 17472 0.4508 2 4= 4C2 × 16C4 = 10920 0.2817 OR Choc Oats Ginger 0 0 6 0 1 5 0 2 4 1 0 5 1 1 4 1 2 3 2 0 4 2 1 3 2 2 2 B1 The correct number of ways with one of 0, 1 or 2 chocs , unsimplified or any three correct number of ways of combining choc/oat/ginger, unsimplified Total = 36400 ways M1 sum the number of ways with 0, 1 and 2 chocs and two must be totally correct, unsimplified OR sum the nine combinations of choc, ginger, oats, six must be totally correct, unsimplified Probability = 36400/ 20 C6 M1 dividing by 20C6 (38760) oe = 0.939 (910/969) A1 Totals: 4 freq = fd × cw 10, 40, 120, 30

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7 The following histogram represents the lengths of worms in a garden. 12 8 density 4 Frequency 0 0 5 10 15 20 25 Length (cm) (i) Calculate the frequencies represented by each of the four histogram columns. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) On the grid on the next page, draw a cumulative frequency graph to represent the lengths of worms in the garden. [4] (iii) Use your graph to estimate the median and interquartile range of the lengths of worms in the garden. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ [Question 7 (iv) is printed on the next page.] (iv) Calculate an estimate of the mean length of worms in the garden. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) M1 A1 Attempt to multiply at least 3 fds by their ‘class widths’ Totals: 2 Question Answer Marks Guidance 7(ii) length < 5 < 10 < 20 < 25 cf 10 50 170 200 B1 B1 M1 A1 3 or more correct cfs heights on graph 10, 50, 170, 200 Labels correct cf and length(cm), linear scales from zero (allow 0.5 on horizontal axis) Attempt (at least three) at plotting at upper end points (either 5 or 5.5, 10 or 10.5 etc.) Starting at (0, 0) polygon or smooth curve increasing with plotted points at lengths 5, 10, 20 and 25 Totals: 4 7(iii) median = 14.2 B1 Median (accept 13.2 – 15.2) ‘18.5’ – ‘10’ M1 Subt their LQ from their UQ if reasonable from their graph IQ range = 8.5 A1FT Correct FT using LQ = 10 and UQ between 17.5 and 19.5 Totals: 3 7(iv) mean = (2.5×10 + 7.5×40 + 15×120 + 22.5×30) / 200 M1 Using mid points (± 0.5) and their frequencies from 7(i) in correct formula = 14 A1 Totals: 2 cf 200 150 100 50 0 5 10 15 20 25 length (cm)

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