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Mathematics 9709/51 October/November 2025: question paper and mark scheme

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

9709_w25_qp_519709_w25_ms_519709/51/O/N/25

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Paper
Paper 5 Probability & Statistics 1
Total marks
50
Time
1 hour 15 minutes
Code on the paper
9709/51/O/N/25
9709_w25_ms_51.pdf

9709/51 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 16 marks

Question

The random variable X takes the value x with probability k x 2, where k is a constant and x takes the values - 2 , 1, 2, 3 only. (a) Draw up the probability distribution table for X, giving the probabilities as numerical fractions. [3] (b) Find E(X ). [1] (c) Find P (X ! 2 X 2 0) . [2] * 0000800000004 * DFD 4

Mark scheme

1(a) 1 B1 Using sum of probabilities =1 to form an equation in k or value of 4k + k + 4k + 9k = 1, k = k stated. 18 B1 Table with at least 2 correctly linked probabilities accurate. May x –2 1 2 3 be in terms of k. P(X = x) 4 1 4 9 X –2 1 2 3 18 18 18 18 P(X = x) 4k k 4k 9k Condone extra X values if probability stated as 0. B1 4 correctly linked probabilities accurate. May not be in a table. 3 1(b) 14 B1FT FT 28 × their k or correct with 0 < p < 1 . [E(X) = 28k =] , 1.56 9 1 1(c) 10 M1 their P (1) + P ( 3) . P ( X ≠ 2| X > 0) = 18 their P (1) + P ( 2 ) + P ( 3) 14 18 May be in terms of k. With 0 < p < 1 . 10 5 If table correct, accept 18 or 9 . 14 7 18 9 5 A1 = , 0.714 7 2

Question 24 marks

Question

A fair six-sided dice with faces labelled 1, 2, 3, 4, 5, 6 is thrown repeatedly until a 6 is obtained. (a) Find the probability that a 6 is obtained for the first time on the 8th throw. [1] (b) Find the probability that a 6 is obtained for the third time on the 7th throw. [3] * 0000800000005 * DFD 5

Mark scheme

2(a) 1 5 7 B1 Accept 0.0465136… rounded to 3 or more SF. = 0.0465 78125 6 6 Accept . 1679616 1 2(b) 4 5 1 3 B1 4 5 1 3 × C2 × d, d 1. 6 6 6 6 6 M1 6 P2 p e (1 − p ) 7 −e × 6 C2 , 0 < p < 1, accept . 2 0.0335 A1 AWRT. 3125 9375 , . 93312 279936 3

Question 38 marks

Question

The back-to-back stem-and-leaf diagram shows the annual salaries, in dollars, of 27 employees at each of two companies, Browns and Greens. Browns Greens (3) 9 8 4 30 6 8 (2) (7) 8 8 5 3 3 1 0 31 2 4 5 8 (4) (7) 9 7 6 4 2 2 0 32 3 6 6 7 9 (5) (6) 8 7 5 5 3 1 33 1 1 3 7 8 8 9 9 (8) (3) 4 2 2 34 0 2 3 5 6 9 (6) (1) 7 35 3 7 (2) Key: 6|32|7 means $32 600 for Browns and $32 700 for Greens. (a) Find the median and interquartile range for the annual salaries of employees at Browns. [3] The annual salary of an employee at Browns is denoted by x thousand dollars and the annual salary of an employee at Greens is denoted by y thousand dollars. It is given that, for the 27 employees at each of the companies, / x = 878.3, / x 2 = 28 616.09, / y = 896.5, / y 2 = 29 815.63 . (b) Find the mean and standard deviation of the annual salaries of these 54 employees. [5] * 0000800000006 * DFD 6

Mark scheme

3(a) Median = [$]32400 B1 Accept Q2, must be identified. If thousands consistently omitted SCB1 for median = [$]324. [IQR = $]33500 – [$]31300 M1 335[00] ⩽ UQ ⩽ 337[00] – 313[00] ⩽ LQ ⩽ 315[00]. Implied if both quartile values are stated and an appropriate IQR calculated accurately. = [$]2200 A1 CAO. 3 3(b) 878.3 + 896.5 B1 878300 + 896500 [Mean =] [ = 32.867 ] Accept . 54 54 so [$]32900 B1 Accept [$]32866.67 to 3 or more SF, condone [$]32866, use of thousand dollars. 878.3 896.5 Condone from + ÷2 . 27 27 28616.09 + 29815.63 1774.8 2 M1 Accept un-simplified variance formula. [Variance =] − = FT their mean. 54 54 Ignore any square root leading to sd for this mark. 2 Condone 1082.068… – 1080.217… 58431.72 1774.8 − 54 54 = 1.8511 A1 Accept un-simplified, condone 1.85. σ = 1.36, so σ = [$]1360 A1 Condone 1.36 if B1B0 scored for mean. Must be identified, e.g. sd, std d, s, σ. Condone short square root signs. 5

Question 45 marks

Question

Bag A contains 8 red marbles and 3 blue marbles. Bag B contains 4 red marbles and 1 blue marble. A marble is chosen at random from bag A. If the marble chosen is red, it is discarded. If the marble chosen is blue, it is placed in bag B. A marble is then chosen at random from bag B. If the marble chosen is red, it is discarded. If the marble chosen is blue, it is placed in bag A. A marble is now chosen at random from bag A. (a) Complete the tree diagram below by entering all the remaining outcomes and probabilities. [3] Bag A Bag B Bag A Red 8 11 4 3 6 11 Blue 2 6 (b) Find the probability that all three marbles chosen are the same colour. [2] * 0000800000007 * DFD 7

Mark scheme

4(a) B1 Bag B branches completed correctly including Red/Blue. B1 2 sets of Bag A branches completely correctly including Red/Blue. B1 Final 2 sets of Bag A branches completely correctly including Red/Blue. Penalise omission of Red/Blue only once. If not identified, assume RB continues as for first Bag A. 3 4(b) 8 4 7 3 2 3 M1 Both, FT their tree diagram probabilities. [P(RRR) + P(BBB) =] × × + × × May be seen in 4(a). 11 5 10 11 6 11 224 18 1307 A1 0.4320661… to 3 or more SF. 550 + 726 = 3025 , 0.432 2

Question 58 marks

Question

On any given day, Cooper either wears a blue jumper or he wears a green jumper or he does not wear a jumper. The probability that he wears a blue jumper is 0.6 and the probability that he wears a green jumper is 0.3. Whether Cooper wears a jumper of either colour, or does not wear a jumper, on any day is independent of his choice on any other day. (a) Find the probability, that in a week (7 days), Cooper wears a blue jumper on at least 5 days. [3] (b) Use a suitable approximation to find the probability that, in any 150-day period, Cooper does not wear a jumper on fewer than 22 days. [5] * 0000800000008 * DFD 8

Mark scheme

5(a) Method 1 [P(5, 6, 7) =] 7 C5 (0.6)5 (0.4)2 + 7 C6 (0.6)6 (0.4)1 + (0.6)7 M1 One term of the form 7 Cx p x 1 − p 7− x . [= 0.261274 + 0.130637 + 0.027994] ( )( ) 0 < p < 1, x ≠ 0 or 7. A1 Correct un-simplified expression no terms omitted leading to the final answer. = 0.420 B1 0.4198 < p ⩽ 0.42[0]. Method 2 [1-P(0,1,2,3,4) =] 1 – {(0.4)7 +7 C1 (0.6) (0.4) 6 +7 C2 (0.6)2 (0.4)5 M1 One term of the form 7 Cx ( p ) x (1 − p )7− x 0 < p < 1, x ≠ 0 or 7. +7 C3 (0.6)3 (0.4)4 +7 C4 (0.6)4 (0.4)3 } A1 Correct un-simplified expression, no more than 2 ‘middle’ terms omitted leading to the final answer. Condone omission of final bracket ‘}’. If other brackets omitted, allow recovery if correct answer obtained. = 0.420 B1 0.4198 < p ⩽ 0.42[0]. 3 5(b) [Mean = 0.1×150 =]15 B1 15 and 13.5 seen, allowed un-simplified. May be seen in the [Variance = 0.1×150 × 0.9 =]13.5 standardisation formula. ( σ = 13.5, 3.674 σ 3.6742346 implies correct variance . ) Withhold mark if variance clearly identified as standard deviation, ( ) condone N 15, 13.5 if standardisation formula correct or variance/standard deviation correctly stated as well. 21.5 − 15 M1 Substituting their µ and σ into the ±standardising formula (any [P(X < 22 ) = P( Z <] ) number for 21.5), allow σ2 or √σ. 13.5 M1 Use continuity correction 21.5 or 22.5 in standardisation formula. [P( Z < 1.769 ) = Φ (1.769 ) ] M1 Appropriate area Φ, from final process, must be a probability Note: correct final answer implies this M1. 0.962 A1 AWRT 0.962. 5

Question 69 marks

Question

A factory produces chocolate bars. The weights of the bars are normally distributed with mean 155 g and standard deviation 6 g. A random sample of 350 of these bars is chosen. (a) How many of these 350 bars would you expect to weigh between 148 g and 160 g? [4] * 0000800000009 * DFD 9 A second factory also produces chocolate bars. The weights of these bars are normally distributed with mean n g and standard deviation v g. Tests show that 8% of the bars weigh more than 114.0 g and 20% weigh less than 99.5 g. (b) Find the value of n and the value of v. [5] * 0000800000010 * DFD 10

Mark scheme

6(a) 148 − 155 160 − 155 M1 Substituting 155, 6 and 148 or 160 into ± standardisation formula, P( <Z < ) 6 6 not σ 2 , not σ , no continuity correction, −7 5 Award or M1. 6 6 [= P( − 1.1667 < Z < 0.8333) M1 Calculating the appropriate probability area (leading to their final = Φ (1.1667 ) + Φ ( 0.8333) − 1 ] answer. Expect p > 0.5). = 0.7975 + 0.8784 – 1 Or (0.7975 – 0.5) + (0.8784 – 0.5) Or 0.2975 + 0.3784 = 0.6759 A1 AWRT 0.676 SOI. If one or both M marks not awarded, SCB1 for AWRT 0.676 SOI. [Expected number = 0.6759 ×350 = 236.565] B1FT Strict FT their at least 4 figure probability × 350 236 (Check with calculator). One integer answer only, either 236 or 237. No indication of ‘approximation’, e.g. ≈, , about, 3SF. Condone use of 3SF probability if more accurate probability seen in workings. 4 6(b) 114 − μ B1 1.405 < z1 < 1.407, −1.407 < z1 < −1.405 seen. = 1.406 σ B1 0.841 < z < 0.843, −0.843 < z < −0.841 seen. 99.5 − μ = −0.842 M1 Use of the ±standardisation formula once with µ, σ equating to a z- σ value (not 0.20, 0.80, 0.08, 0.92, 0.5793, 0.4207, 0.5319, 0.4681, 0.7881, 0.2119, 0.8212, 0.1788). Condone continuity correction ±0.05, not σ2 , not σ . Solve, obtaining values for µ and σ M1 Solve two equations in µ and σ using the elimination method, substitution method or other appropriate approach to obtain values μ = 105, σ = 6.45 for both µ and σ. A1 104.5 μ 105.5 . 6.44 σ 6.475 . If one or both the M marks have been withheld, SCB1 for both correct. 5

Question 710 marks

Question

(a) How many different arrangements are there of the 10 letters in the word SEYCHELLES? [1] (b) How many different arrangements are there of the 10 letters in the word SEYCHELLES in which there are exactly two letters between the Ss and one of these two letters is C? [3] * 0000800000011 * DFD 11 (c) How many different arrangements are there of the 10 letters in the word SEYCHELLES in which there is an S at the beginning, an S at the end and the three Es are not all next to each other? [3] 5 letters are selected at random from the 10 letters in the word SEYCHELLES. (d) Find the probability that these 5 letters include the three Es. [3] * 0000800000012 * DFD 12 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) 10! B1 CAO. 2!2!3! = 151200 1 7(b) Method 1[SC_ S _ _ _ _ _ _] 7! M1 7! × 2 ×7 × k 1 k , a positive integer. 2!3! 2!3! M1 Integer ×7 . = 5880 A1 3 Method 2 Considering each case separately [SC-S, S-CS] 6! M1 6! 6! 6! With Y × 2 × 7 = 840 ×l , × m and × n seen. 2!3! 2!3! 2!2! 3! 6! With H × 2 × 7 = 840 2!3! 1 l , m, n , positive integers. 6! With E × 2 × 7 = 2520 2!2! 6! With L × 2 × 7 = 1680 3! 840 + 840 + 2520 + 1680 M1 Summing 4 correct or correctly identified scenarios, oe = 5880 A1 7(b) Method 3 Considering each case separately [SC-S, S-CS] 7! M1 7! 7! 7! SCYS ×2 = 840 ×l , × m and × n seen 2!3! 2!3! 2!2! 2! 7! SCHS ×2 = 840 1 l , m, n , positive integers 2!3! 7! SCES × 2 = 2520 2!2! 7! SCLS × 2 = 1680 3! 840 + 840 + 2520 + 1680 M1 Summing 4 correct or correctly identified scenarios, OE. = 5880 A1 3 7(c) Method 1 Total arrangements with Ss at ends – arrangements with Ss at ends and Es together 8! 6! M1 8! − − q, 1 < q < 3360 . 2!3! 2! 2!3! [= 3360 – 360] M1 6! m− , m > 360 . 2! = 3000 A1 3 7(d) Method 1 [Number of ways with 3 Es =] 7 C2 (= 21) B1 7 C seen with no addition, subtraction, multiplication. 2 [Total number of ways is] 10 C5 (= 252) M1 Seen. 21 1 A1 21 1 [Probability =] , If M mark not awarded, SCB1 for , WWW. 252 12 252 12 Method 2 3 2 3 B1 3 2 3 × × × 5 C2 × × . 11 6 11 11 6 11 M1 5C × k ,0 < k < 1 . 2 Accept 5 C3 × k , 0 < k < 1 . 21 1 A1 21 1 [Probability =] , If M mark not awarded, SCB1 for , WWW. 252 12 252 12 3

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On this page, free: the mark scheme for AS & A Level Mathematics 9709/51, October/November 2025 (9709_w25_ms_51), and the question paper (9709_w25_qp_51), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/51 for October/November 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.

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