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Mathematics 9709/53 May/June 2025: question paper and mark scheme

Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_s25_qp_539709_s25_ms_539709/53/M/J/25

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Paper
Paper 5 Probability & Statistics 1
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/53/M/J/25
9709_s25_ms_53.pdf

9709/53 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 14 marks

Question

For a set of 40 values of x, it is found that / (x - k) = 836.0, / (x - k) 2 = 25410.8 , where k is a constant. (a) Given that the mean of these 40 values is 124.0, find the value of k. [2] (b) Find the standard deviation of these 40 values of x. [2]

Mark scheme

1(a) 4960 − 40k = 836 B1 or Also accept 124 – k = 20.9 or 40k = 4124 OE. 4960 836 –k= 40 40 k =103.1 B1 2 1(b) Method 1 25410.8 836 2 M1 2 [sd =] − 40 40 [sd2 = 198.46, ] A1 14.08758…to at least 3SF. sd = 14.1 If M not awarded SCB1 for correct answer WWW. Method 2 622978.4 M1 622978.4 comes from expansion of [sd2 =] −1242 ∑(x – k)2 = 25410.8 substituting in relevant values. 40 [sd2 = 198.46, ] A1 14.08758…to at least 3SF. sd = 14.1 If M not awarded SCB1 for correct answer WWW. 2

Question 25 marks

Question

At a large college, all students who study Science also study exactly one of Art or Drama or Music. 20% of these students study Art, 45% study Drama and 35% study Music. (a) 3 students are selected at random from the students who study Science. Find the probability that at least 1 of these students studies Drama. [2] (b) 10 students are selected at random from the students who study Science. Find the probability that more than 7 study Art or Music. [3]

Mark scheme

2(a) Method 1 1 − ( 0.55) 3 M1 Might also be seen as 1 – 3C0(0.55)3[(0.45)0]. 6669 A1 AWRT. = 0.834, 8000 Method 2 3 C1 (0.45)(0.55)2 + 3C2 (0.45)2(0.55) + (0.45)3[0.55]0 M1 6669 A1 AWRT. = 0.834, 8000 2 2(b) Method 1 [P(8, 9, 10) =] 10C8 0.558 0.452 + 10C9 0.559 0.451 + 0.5510 M1 One term of the form 10Cx ( p ) x (1 − p )10− x , 0 < p < 1, x ≠ 0 or 10. [= 0.0763026 + 0.0207241 + 0.00253295] A1 Correct un-simplified expression. = 0.0996 B1 0.09955 < p ≤ 0.0996. Method 2 [1 – P(0,1,2,3,4,5,6,7)] M1 One term of the form 10Cx ( p ) x (1 − p )10− x , 0 < p < 1, x ≠ 0 or 10. 1 – (0.4510 + 10C1 0.55 0.459 + … + 10C7 0.557 0.453) A1 Correct un-simplified expression. Condone omission of up to 5 of the 6 middle terms. Condone omission of last bracket only. If both brackets omitted in un-simplified expression, allow recovery for final stated calculation of 1 – 0.90044… or better. = 0.0996 B1 0.09955 < p ⩽ 0.0996. 3

Question 36 marks

Question

A fair six-sided dice with faces labelled 1, 2, 3, 4, 5, 6 is thrown repeatedly until a 3 is obtained. The number of throws taken is denoted by the random variable X. (a) Find P(X = 8) . [1] (b) Find P(X 1 9) . [2] (c) Find the probability that a 3 is obtained for the second time before the 6th throw. [3]

Mark scheme

3(a) B1 1 5 7 78125 P ( X = 8 ) = = 0.0465, 6 6 1679616 1 3(b) Method 1 5 8 M1 5 d [P(X < 9) =] 1 − 1 − , d = 8, 9. 6 6 = 0.767 A1 AWRT. Method 2 [P(X < 9) =] M1 2 3 4 5 1 5 1 5 1 5 1 5 1 5 1 2 3 4 5 1 5 1 5 1 5 1 5 1 5 1 + + + + + + + + + + + + 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 5 1 5 1 5 1 6 7 8 6 5 1 5 1 7 + + + 6 6 6 6 6 6 6 6 6 6 = 0.767 A1 AWRT. 2 3(c) Method 1 1 2 1 B1 Two identified scenarios correct, accept un-simplified 2 throws calculations. 6 36 2 1 5 10 M1 Four correctly identified scenarios added (here they may be 3 throws: × 2 C1 identified by correct un-simplified calculations). Condone 2C1 = 2 6 6 216 2 2 etc. 1 5 75 4 throws: × 3 C1 6 6 1296 2 3 1 5 500 5 throws: × 4 C1 6 6 7776 763 A1 0.196 ⩽ p < 0.1963 0.196, SCB1 for correct answer if M mark not awarded 3888 Method 2 5 5 3125 B1 1 – 1 identified scenario correct, accept un-simplified calculations. No threes in 5 throws 6 7776 M1 1 – 2 correctly identified scenarios (here they may be identified by 4 1 5 3125 correct un-simplified calculations). Condone 5C1 = 5. One three in 5 throws × 5 C1 6 6 7776 5 4 5 1 5 1 – – { × 5 C1 } 6 6 6 763 A1 0.196 ⩽ p < 0.1963. 0.196, SCB1 for correct answer if M mark not awarded. 3888 3

Question 45 marks

Question

84 people attempt a particular puzzle. The times taken, in minutes, to complete the puzzle are recorded. These times are represented in the cumulative frequency graph below. 100 90 80 70 Cumulative frequency 60 50 40 30 20 10 0 0 1 2 3 4 5 6 7 8 9 10 11 12 Time in minutes (a) Use the graph to estimate how many people took between 4 and 7.5 minutes to complete the puzzle. [1] (b) On the grid below, draw a box-and-whisker plot to summarise the information in the cumulative frequency graph. [4]

Mark scheme

4(a) Accept 43 or 44 B1 1 4(b) B1 Plotted on box. Median = 4.8 LQ = 3.7 or 3.8, UQ = 6.5 or 6.6 B1 Plotted on box. Correct end points for whiskers and whiskers drawn B1 Correct end points of whiskers (2 and 12). Whiskers not through box, not drawn at corners of box. Accept end points of whiskers as dots or without vertical line. Linear scale and labelled Time in minutes B1 OE. Linear scale from 2 to 12, at least 3 equally spaces values marked using at least half of the grid. Label to include time and minutes. 4

Question 58 marks

Question

Bag A contains 6 red marbles, 5 blue marbles and 1 green marble. Bag B contains 5 red marbles and 3 blue marbles. A marble is chosen at random from bag A and placed in bag B. A marble is now chosen at random from bag B. (a) Draw a tree diagram to represent this information, giving the probability on each branch. [3] (b) Find the probability that both marbles chosen are the same colour. [2] (c) Find the probability that the marble chosen from bag A is blue, given that the marble chosen from bag B is blue. [3]

Mark scheme

5(a) B1 Correct structure and probabilities for Bag A branches which must be the first set of branches, including labels R, B, G. B1 Correct structure and probabilities for one Bag B branch, including labels R, B, [G]. Condone inclusion of branches with 0 probability. B1 Correct structure and probabilities for remaining Bag B branches, including labels R, B, [G]. Condone inclusion of branches with 0 probability. Additional branches without a probability of zero lose this mark. 6 5 1 Probabilities for Bag A selection R,B,G ( , , ) 12 12 12 6 3 5 4 5 3 1 Probabilities for Bag B selection: , ,[ 0] , , [ 0] , , 9 9 9 9 9 9 9 3 5(b) 6 6 5 4 1 1 M1 Correct or FT tree diagram probabilities with all 3 (R, B and G) [P(RR) + P(BB) + P(GG) =] × + × + × branches. 12 9 12 9 12 9 Note: working may be by tree diagram. 57 19 A1 Accept 0.5278, 0.528. = , 108 36 2 5(c) 5 4 M1 5 4 20 5 × × or or or their P(BB) from Q5(b) seen as a P(A blue | B blue) = 12 9 12 9 108 27 6 3 5 4 1 3 numerator of a single fraction. × + × + × 12 9 12 9 12 9 FT tree diagram with values from only 2 bags. B1 FT 6 3 5 4 1 3 × + × + × seen as the denominator of a single 12 9 12 9 12 9 fraction. 5 5 4 or their P(BB) may be seen from Q5b in place of × 27 12 9 FT tree diagram with values from only 2 bags. 20 A1 Accept 0.488, 0.4878. = 20 41 SCB1 if either M or B mark not scored or 0.4878. 41 3

Question 610 marks

Question

A company sells bags of pasta. The masses of large bags of pasta are normally distributed with mean 2.50 kg and standard deviation 0.12 kg. (a) Find the probability that the mass of pasta in a randomly chosen large bag is less than 2.65 kg. [2] A restaurant manager buys 160 of these large bags of pasta. (b) Find the number of bags for which you would expect the mass of pasta to be more than 1.65 standard deviations above the mean. [3] The masses of small bags of pasta sold by the company are normally distributed with mean n kg and standard deviation v kg. Tests show that 77% of these bags have masses greater than 1.26 kg, and 44% have masses less than 1.35 kg. (c) Find, in either order, the value of n and the value of v. [5]

Mark scheme

6(a) 2.65 − 2.50 M1 Substituting 2.65, 2.50 and 0.12 into the ±standardisation formula [P(X < 2.65) =] P(Z < ) [ = P(Z < 1.25)] allow σ2 or √σ. Condone continuity correction ±0.005. 0.12 = 0.894 A1 0.894 ⩽ p ⩽ 0.8944. 2 6(b) 2.50 + 1.65 × 0.12 − 2.50 M1 Seeing 2.50 + 1.65 × 0.12 or 2.698 and 2.50 and 0.12 substituted P(Z > ) = P(Z > 1.65) into the ±standardisation formula with a direct attempt to find the 0.12 probability area. Accept numerator of 1.65 × 0.12. Accept Z linked to 1.65 with a direct attempt to find the probability area. Do not allow σ2 or √σ. No continuity correction. = 1 – 0.9505 = 0.0495 A1 AWRT. Expected number = 0.0495 ×160 = 7.92 , so 7 or 8 B1 FT Strict FT their at least 4-DP calculated probability, seen anywhere (not a z-value). Final answer must be a single positive integer value, no approximation or rounding stated. 3 6(c) 1.26 − μ B1 0.738 < z1 < 0.740 or −0.740 < z1 < −0.738 seen. = −0.739 σ B1 0.150 < z2 < 0.152 or −0.152 < z2 < −0.150 seen. 1.35 − μ = −0.151 M1 Use of the ±standardisation formula once with µ, σ equating to a z- σ value (not 0.77, 0.23, 0.44, 0.56, 0.7794, 0.5910, 0.6700, 0.7123, 0.2206, 0.4090, 0.3300, 0.2877). No continuity correction, not σ2, not σ . Solve, obtaining values for µ and σ M1 Solve 2 equations in µ and σ using the elimination method, substitution method or other appropriate approach to obtain values μ = 1.37, σ = 0.153 for both µ and σ. A1 AWRT μ = 1.37, σ = 0.153 . If one or both the M marks have been withheld, SCB1 for both correct. 5

Question 712 marks

Question

A set of friends consists of 7 men and 4 women. Three of the men are brothers: Ali, Ben and Charlie. (a) Find the number of different arrangements of the 7 men in a line in which Ali and Ben do not stand next to each other. [3] (b) Find the number of different arrangements of the 7 men and 4 women in a line in which all the men stand together and all the women stand together. [3] (c) In how many ways can the 7 men and 4 women be divided into a group of 6, a group of 3 and a group of 2 if there are no restrictions? [2] (d) The 7 men and 4 women are divided at random into a group of 6, a group of 3 and a group of 2. Find the probability that Ali, Ben and Charlie are all in the same group. [4] Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown.

Mark scheme

7(a) Method 1 Total – A and B together 7! – 6! ×2 M1 7! – k, k an integer ⩾ 1. M1 m – 6! × n : n = 1, 2 where m an integer > 6! n = 1 can be implied. = 3600 A1 Method 2 Arrangements of 5 people then Ali and Ben placed 5! × 6 × 5 M1 5! × p where p an integer > 1. M1 6×5 q × 6 × 5, q × 6C2, q × 6P2, q × , where q an integer > 1. 2 = 3600 A1 3 7(b) 7! × 4! × 2 M1 7! × r, where r is an integer, r > 1 or 4! × s, where s is an integer, s > 1. M1 7! × 4! × t, where t is an integer, t ⩾ 1, t = 1 can be implied. 241920 A1 CAO. 3 7(c) 11 C6 × 5C3 × [2C2] M1 11 Cv × 11-vCu × [11-v-uCw], u, v, w = 6, 3, 2 u ≠ v ≠ w. = 4620 A1 2 7(d) Method 1 Ali, Ben and Charlie must be in the group of 6 or the group of 3. group of 6: 8 C3 × 5C3 × (2C2) = 560 B1 560 seen, accept un-simplified. group of 3: 8 C6 ( × 2C2 × 3C3 ) = 28 B1 28 seen accept un-simplified. 560 + 28 M1 their ( 560 + 28) their ( 560 + 28 ) Probability they are in same group = or 4620 their (c) 4620 588 7 A1 = , , 0.127 4620 55 Method 2 Ali, Ben and Charlie must be in the group of 6 or the group of 3. 6 5 4 3 2 1 8 B1 4 Group of 6: × × × × × × C3 , 0.1212… seen, accept un-simplified. 11 10 9 8 7 6 33 1 4 = × 56 = 462 33 3 2 1 1 B1 1 Group of 3: × × = , 0.00606 ( 06) seen, accept un-simplified. 11 10 9 165 165 4 1 M1 4 1 Probability they are in same group = + their + their . 33 165 33 165 7 A1 = ,0.127 55 4

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Mathematics 9709/53 for May/June 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.

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