Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 5 · Variant 3

1 30% of the residents of Wimfield own an electric car. Three residents are chosen at random. (a) Find the probability that either all three own an electric car or none of them owns an electric car. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ A random sample of 125 of the residents of Wimfield is selected. (b) Use a suitable approximation to find the probability that more than 45 of these residents own an electric car. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 0.33 + 0.7 3 M1 p 3 + q 3, p + q = 1, p , q  0 or or 1 – (3  0.32  0.7 + 3  0.3  0.72 ) = 1 − 0.63 1 − (3  p 2  q + 3  p  q 2 ), p + q = 1, p, q  0 . 0.37[0] A1 37 . 100 2 1(b) [Mean = 125  0.3 =] 37.5 B1 37.5 or 37½ and 26.25, 26¼ seen, allow unsimplified. May be seen in [Variance = 125  0.3  0.7 =] 26.25 standardisation formula. (=  5.12, 105 implies correct 2 variance). 45.5 − 37.5 M1 Substituting their mean and their positive P(X > 45) = P( Z  ) standard deviation into the ±standardising 26.25 formula (any number for 45.5), not their σ2, not their . M1 Use continuity corrections 44.5 or 45.5 in their standardisation formula 8 8 Note: or seen gains M2 26.25 5.123 BOD [1 − Φ ( their 1 .5614 ) ] = 1 – their 0.9407 M1 Appropriate area Φ, from final process, must be a probability. Expect final answer < 0.5. Note: appropriate final answer implies this M1. 0.0593 A1 0.0592 ⩽ p ⩽ 0.0593. 5

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2 A red fair six-sided dice has faces labelled 1, 1, 1, 2, 2, 2. A blue fair six-sided dice has faces labelled 1, 1, 2, 2, 3, 3. Both dice are thrown. The random variable X is the product of the scores on the two dice. (a) Draw up the probability distribution table for X. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find E(X ). [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) B1 Table with correct x values and at least one correct probability linked with the correct x- x 1 2 3 4 6 value. Values need not be in order, lines may not 6 12 6 6 6 be drawn, may be vertical, P(X = x) 36 36 36 36 36 x and P(X) may be omitted. Condone any additional x values if probability stated as 0. 1 1 1 1 1 6 3 6 6 6 B1 4 correct probabilities linked with the correct x-values, need not be in table, 0.167 0.333 0.167 0.167 0.167 accept unsimplified. B1 5 correct probabilities linked with correct x- values, may not be in table. Decimals correct to at least 3 SF. SC B1 4 or 5 probabilities summing to 1 placed in a probability distribution table with 4 or 5 x-values between 1 and 6 inclusive. 3 2(b) 1 B1 FT FT their table with 4 or 5 probabilities (0 < [E(X) = ( 6 + 24 + 18 + 24 + 36 ) =] 3 p < 1) summing to 1. 36 1

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3 In Molimba, the heights, in cm, of adult males are normally distributed with mean 176 cm and standard deviation 4.8 cm. (a) Find the probability that a randomly chosen adult male in Molimba has a height greater than 170 cm. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 60% of adult males in Molimba have a height between 170 cm and k cm, where k is greater than 170. (b) Find the value of k, giving your answer correct to 1 decimal place. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 170 − 176 M1 Using ± standardisation formula with 170, [P(X > 170) = P( Z ] ) 176 and 4.8 substituted appropriately. 4.8 Condone 2 and . No continuity correction. = 0.894  M1 Appropriate area Φ, from final process, Φ ( 1.25 )  must be a probability. Expect final answer > 0.5. A1 0.894 or 0.89435 ⩽ p ⩽ 0.8944. If A0 scored, SC B1 for 0.894 or 0.89435 ⩽ p ⩽ 0.8944, WWW. 3 3(b) P(h < 170) = 1 – 0.8944 = 0.1056 M1 1 – their 3(a) seen or implied by 0.7056 or 0.2944 k − 176 −1 B1 0.540 < z ⩽0.541 or −0.541 ⩽ z < 0.540  = 0.541 = Φ ( 0.1056 + 0.6 )  seen.   4.8 M1 Use of ±standardisation formula with k, 176, 4.8 equated to a z-value (not 1.25, 0.7601, 0.2399, 0.7056, 0.7257, 0.8313, 0.253 0.894, 0.6, 0.4), not 4.82, not 4.8 , no continuity correction. k = 178.6 A1 CAO (answer required to 1 dp). 4

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4 On a certain day, the heights of 150 sunflower plants grown by children at a local school are measured, correct to the nearest cm. These heights are summarised in the following table. Height 10–19 20–29 30–39 40–44 45–49 50–54 55–59 (cm) Frequency 10 18 32 42 28 14 6 (a) Draw a cumulative frequency graph to illustrate the data. [4] (b) Use your graph to estimate the 30th percentile of the heights of the sunflower plants. [2] ............................................................................................................................................................ ............................................................................................................................................................ (c) Calculate estimates for the mean and the standard deviation of the heights of the 150 sunflower plants. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) B1 Cf values 28, 60, 102, 130, 144, 150 seen, H(cm) 10–19 20–29 30–39 40–44 45–49 50–54 55–59 Condone omission of 10. May be implied by accurate plotting (scale UB 19.5 29.5 39.5 44.5 49.5 54.5 59.5 no less than 1cm = 10). May be by data table. cf [10] 28 60 102 130 144 150 B1 Linearly scaled axes correctly labelled cumulative frequency (cf) (from 0 to 150) and height (h) and centimetres (cm) (from 9.5 to 59.5) with at least 3 values identified on each. Axes can be the other way round. M1 At least 4 points plotted at upper boundary ± 0.5, (e.g. allow (19, 19.5 or 20, 10) etc.) on correctly scaled axes. (9.5,0), (19.5,10), (29.5, 28), (39.5, 60), (44.5,102), (49.5, 130), (54.5, 144), (59.5,150). A1 All points plotted correctly, curve drawn (within tolerance), joined to (9.5, 0) and not going beyond above 150 vertically. A0 if straight line segments used. 4 4(b) [150 × 0.3 = 45] M1 Use of graph must be seen. Line drawn from 45 on cf axis to meet graph at h = 36 A1 FT Must be an increasing cf graph. Expect an answer in range 35 ⩽ h ⩽ 37 for a correct graph. 2 4(c) Midpoints 14.5, 24.5, 34.5, 42, 47, 52, 57 B1 At least 6 correct midpoints seen, may be unsimplified, may be in calculation, may be by data table. 10  14.5 + 18  24.5 + 32  34.5 + 42  42 + 28  47 + 14  52 + 6  57 M1 Correct unsimplified mean formula with Mean = their midpoints (not ub, lb, upper limits, 150 lower limits, cw, fd, f or cf and must be  145 + 441 + 1104 + 1764 + 1316 + 728 + 342  = within class).    150  If midpoints correct, accept partially evaluated. 5840 584 14 A1 Accept answers wrt 38.9 WWW = , , 38 , 38.9 If M1 withheld, SC B1 for 150 15 15 5840 584 14 , ,38 , 38.9. 150 15 15 sd2 = M1 Correct unsimplified variance formula with 2 2 2 2 2 2 2 2 their midpoint (not ub, lb, upper limits, 10  14.5 + 18  24.5 + 32  34.5 +42  42 + 28  47 + 14  52 + 6  57  5840  − their  lower limits, cw, fd, f or cf and must be 150  150  within class). 2 If midpoints correct, accept partially 244285  5840  [ = − their  ] evaluated 150  150  [= 112.76] A1 AWRT 10.6 WWW. standard deviation  112.76  = 10.6 If second M1 withheld, SC B1 for 10.6   WWW. 5 5(a) Method 1 [P(0, 1, 2) =] 8C2(0.75)6 (0.25)2 + 8C1(0.75)7(0.25)1 + (0.75)8 M1 One term 8Cx . ( p ) x (1 − p )8− x , 0  p  1, 0  x  8 . [= 0.31146 + 0.26697 + 0.10011] = A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. = 0.679 B1 AWRT. 3 Method 2 [P(0, 1, 2) = 1 – P(3, 4, 5, 6, 7, 8) = ] 1 – {8C3(0.75)5 (0.25)3 + 8C4(0.75)4 (0.25)4 + M1 x 8 − x One term 8Cx ( p ) (1 − p ) , 8C5(0.75)3 (0.25)5 + 8C6(0.75)2 (0.25)6 + 8C7(0.75) (0.25)7 + (0.25)8} 0  p  1, 0  x  8 . A1 Correct expression, accept unsimplified, condone omission of up to 3 ‘middle’ terms, leading to final answer. = 0.679 B1 AWRT. 3 5(b) Method 1 1 − 0.756 M1 1 − 0.75 n , n = 6,7 . 3367 A1 0.82202148… to at least 3SF. = 0.822, 4096 Method 2 0.25 + 0.25  0.75 + 0.25  0.752 + 0.25  0.753 + 0.25  0.75 4 + 0.25  0.755 M1 Summing 6 or 7 terms – condone extra term 0.25  0.756 . =0.822 A1 Method 3 1 - 0.757 − 0.25  0.756 M1 Correct expression. =0.822 A1 2 5(c) Method 1 P(2nd gold 5th is first unwrapped). R G RorG RorG U 0.25  0.3  0.55  0.55  0.45  = 0.01021 M1 a  0.3 b c 0.45 0 < a, b, c < 1. a ≠ 0.3, 0.45 b, c ≠ 0.45. on its own or as a numerator multiplied in that order, or correct. A1 5 correct probabilities multiplied. Method 2 P(2nd gold 5th is first unwrapped).4 possible scenarios M1 a  0.3 b c 0.45 0 < a, b, c < 1. a ≠ 0.3, 0.45 b, c ≠ 0.45. R G R G U 0.25  0.3  0.25  0.3  0.45 [= 0.00253125] 4 terms in this form seen added on their own or as a numerator. R G R R U 0.25  0.3  0.25  0.25  0.45 [= 0.002109375] A1 All probabilities correct and attempt to sum R G G R U 0.25  0.3  0.3  0.25  0.45 [= 0.00253125] the 4 scenarios. R G G G U 0.25  0.3  0.3  0.3  0.45 [= 0.0030375] [Total 0.010209375] For either approach 4 B1 [P(5th is first unwrapped) =] ( 0.55 ) ( 0.45 ) = 0.041178 0.25  0.3  0.55  0.55  0.45 M1 their P ( 2nd gold  5th is first unwrapped ) [P(2nd is first gold | 5th is first unwrapped) =] 4 their P ( 5th is first unwrapped ) . ( 0.55 ) ( 0.45 ) Their probabilities must be clearly  0.010209375  = identified if incorrect.    0.0411778125  30 A1 0.24793… = 0.248, 121

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6 (a) Find the number of different arrangements of the 9 letters in the word HAPPINESS. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the number of different arrangements of the 9 letters in the word HAPPINESS in which the first and last letters are not the same as each other. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the number of different arrangements of the 9 letters in the word HAPPINESS in which the two Ps are together and there are exactly two letters between the two Ss. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The 9 letters in the word HAPPINESS are divided at random into a group of 5 and a group of 4. (d) Find the probability that both Ps are in one group and both Ss are in the other group. 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Mark scheme: 6(a)  9!  B1 = 90720    2!2!  1 6(b) Method 1 Total arrangements – arrangements with repeated letters at ends 9! 7! M1 7! −  2 a −  b a = their 6(a) or correct, b = 1,2. 2!2! 2! 2! M1 7! a −  2 a = their 6(a) or correct, c = 1,2. c ! 85680 A1 FT ft their 6(a) – 5040. Method 2 Adding no of different ways P and S at ends 2  7! = 10080 M1 Finding correct number of ways for one of these correctly identified scenarios. 7! P or S at one end only 4 5 = 50400 2! M1 Adding no of ways for 3 correctly identified 7! scenarios. Neither P nor S at an end 5 4 = 25200 2!2! Total 85680 A1 Method 3 7! M1 Finding correct number of ways for one of P at beginning 7  = 17640 these correctly identified scenarios. 2! 7! S at beginning 7  = 17640 M1 Adding no of ways for 3 correctly identified 2! scenarios. 8! Neither P nor S at beginning 5  = 50400 2!2! Total 85680 A1 3 6(c) Method 1 arrangements with PP between Ss { S P P S ^ ^ ^ ^ ^ } add arrangements with PP not between Ss { (S ^ ^ S) ^ P P ^ } 6!+ 5!5 4 M1 6! + d, d an integer ≥ 1, may be implied. 6!+ 5!5 4 M1 e + 5! f , e, f integers ≥ 1, may be implied. , e an integer ≥ 1, g = M1 e + g ! ( 5  4 or 5 P2 ) 4,5,6. [Total ]= 3120 A1 Method 2 - considers the 6 positions for S ^^S Positions 1 and 6 there are 5 5! ways M1 Identifying no of ways if S^^S is in position 1 or 6. Positions 2, 3, 4 and 5 there are 4 5! ways M1 Identifying no of ways if S^^S is in position 2, 3, 4 or 5. 2 +5 5! 4 4 5! M1 Adding no of ways for 6 scenarios ( or 26 × 5!). [Total] = 3120 A1 SC B1 for 3120 if any method marks are withheld. 4 6(d) Method 1 Either PP in the group of 5 or PP in the group of 4 5 C3 5 C2 M1 a  5 C 2 , a  5 C 3 , or 5 C 2 + 5 C 3 seen as a + , 9 C5 9 C5 numerator of one or two fractions where a is 1 or 2, no extra terms. 5 C3 + 5 C 2 M1 9 9 9 C 5 or C 4 seen (no addition, C5 multiplication) as a denominator of one or two fractions. 20 10 A1 Probability = , , 0.159 126 63 Method 2 Considering the positions of P and then S  5 4 4 3   5 4 4 3  M1 a × 5 × 4 × 4 × 3 seen as a numerator of a      +      fraction.  9 8 7 6   7 6 9 8  where a = 1 or 2.  5 4 4 3   5 4 4 3  M1 9 × 8 × 7 × 6 seen as a denominator of a      +      fraction.  9 8 7 6   7 6 9 8  10 A1 = 63 3

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