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Mathematics 9709/52 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_529709_w25_ms_529709/52/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
9709/52/O/N/25
9709_w25_ms_52.pdf

9709/52 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 13 marks

Question

A coin is biased so that the probability of obtaining a head when it is thrown is 0.4. The coin is thrown repeatedly until the first head is obtained. (a) Find the probability that the first head is obtained on the 5th throw. [1] (b) Find the probability that the first head is obtained after the 6th throw. [2] * 0000800000003 * DFD 3

Mark scheme

1(a) 162 B1 CAO. [( 0.6 ) ( 0.4 ) =] 0.05184, 4 3125 1 1(b) ( ) 1 − 1 − 0.66 = 0.66 M1 0.6k , k = 5, 6, 7 . Or Or 0.4 + 0.4 × 0.6 + 0.4 × 0.62 + 0.4 × 0.63 + 0.4 × 0.64 + 1 − (0.4 + 0.4 × 0.6 + 0.4 × 0.62 + 0.4 × 0.63 + 0.4 × 0.64 + 1− . 0.4 × 0.65 + 0.4 × 0.66 0.4 × 0.65 ) The blue terms may be omitted or included. Condone omission of final bracket only. Condone omission of both brackets if recovered by 1 – 0.953344… or final answer 0.046656 rounded to at least 4SF. 729 A1 0.046656 rounded to at least 3SF. = 0.0467, 15625 2

Question 27 marks

Question

Kayla has a bag containing 3 red marbles, 1 blue marble and 2 green marbles. She selects one marble from the bag at random and does not replace it in the bag. She repeats this process until she obtains a green marble. The random variable X is the number of marbles that she needs to select until she obtains a green marble. (a) Draw up the probability distribution table for X. [4] (b) Find Var(X ). [3] * 0000800000004 * DFD 4

Mark scheme

2(a) B1 Table with correct x values and at least 1 correct probability. x 1 2 3 4 5 B1 A second correct probability correctly linked to the correct x value, P(X = x) 5 4 3 2 1 need not be in table, accept un-simplified. 15 15 15 15 15 B1 Two more correct probabilities correctly linked to the correct x 40 32 24 16 8 values, need not be in table, accept un-simplified. 120 120 120 120 120 B1 5 correct probabilities linked with correct x values. 1 1 3 5 SCB1 5 non-zero probabilities (not all 1/5) summing to 1 placed in a probability distribution table with correct x values if B1B1 0.3333 0.2667 0.2[000] 0.1333 0.06667 max scored. SCB2 for All decimals correct to at least 3SF x 0 1 2 3 4 P(X = x) 5 4 3 2 1 15 15 15 15 15 OE. 4 2(b) [E(X) =] M1 Accept un-simplified expression. May be calculated in variance. 5 4 3 2 1 1 8 3 8 1 1× + 2 × + 3 × + 4 × + 5 × Accept + + + + OE for the M mark. 15 15 15 15 15 3 15 5 15 3 5 + 8 + 9 + 8 + 5 35 7 FT their table with 5 or 6 probabilities summing to 1 (0 < p < 1). = , 15 15 3 [Var(X) =] M1 Appropriate variance formula using their (E(X))2 value. 5 4 3 2 1 FT their table with 4 or more probabilities (0 < p < 1) which need 12 × + 22 × + 32 × + 42 × + 52 × not sum to 1 or with an expression no more evaluated than shown 15 15 15 15 15 2 in bold. 35 − their 15 [1×] 5 + 4 × 4 + 9 × 3 + 16 × 2 + 25[×1] 49 − 15 9 14 A1 AWRT. = , 1.56 9 WWW but allow from truncation error (e.g. 0.266 rather than 0.267). 14 Note: also comes from SCB2 but scores M1M1A0 max. 9 3

Question 37 marks

Question

The heights of the 124 Senior members of the Giraffes basketball club are normally distributed with mean 187.4 cm and standard deviation 6.4 cm. (a) How many members of the club would you expect to have heights within 5 cm of the mean? [4] * 0000800000005 * DFD 5 The heights of the Junior members of the Giraffes club are normally distributed with mean 172.7 cm and standard deviation v cm. 23% of these members have height less than 170.3 cm. (b) Find the value of v . [3] * 0000800000006 * DFD 6

Mark scheme

3(a) 182.4 − 187.4 192.4 − 187.4 M1 Substituting 187.4 and 6.4 and either 182.4 or 192.4 appropriately P <Z < 6.4 6.4 into one ±standardising formula, allow σ2 or σ . 5 5 Note − or scores M1. 6.4 6.4 [= P( − 0.78125 < Z < 0.78125) = 2Φ ( 0.7813) − 1 ] M1 Appropriate area Φ, from final process. Must be a probability. There may be small variations in the probability values used – the = 2 × 0.7826 −1 values here are from the tables. Or 0.7826 − (1 − 0.7826 ) Expect final answer > 0.5. Or 0.7826 − 0.2174 Condone omission of brackets if recovered for their values. Or ( 0.7826 − 0.5) + ( 0.7826 − 0.5) Or 2 × 0.2826 0.5652 A1 0.565 ⩽ p < 0.5655 SOI. If one or both M marks not awarded, SCB1 for 0.565 ⩽ p < 0.5655 SOI. [Expected number = 124 × 0.5652 = 70.08, ] B1FT Strict FT their at least 4 figure probability × 124 (Check with 70 calculator) One integer answer, expect 70 or 71. No indication of ‘approximation’, e.g. ≈, , about, 2SF . 4 3(b) 170.3 − 172.7 B1 0.7385 < z < 0.7395 or –0.7395 < z < −0.7385 seen. [P(X < 170.3) = 0.23, P(Z > ) = 0.77] σ M1 ±standardisation formula with 170.3, 172.7, σ equated to a z-value 170.3 − 172.7 (not 0.23, 0.77, 0.261, 0.591, 0.409, 0.7794, 0.2206, 1 – their z- = −0.739 value …). σ or Condone continuity correction ±0.05. 172.7 − 170.3 = 0.739 Do not allow σ2 or σ . σ 2.4 Condone ± = 0.739 . σ σ = 3.25 A1 3.245 ⩽ σ ⩽ 3.25. Do not award for improper fractions. There must be consistency with signs in the solution, e.g. 172.7 − 170.3 = −0.739, so σ = 3.25 is not acceptable, A0. σ 3

Question 47 marks

Question

Gio has a pack of 18 cards. Ivy has a pack of x cards. Each card has a picture of a bus or a car or a train. The number of cards with each picture in the two packs is shown in the table. Bus Car Train Gio’s pack 6 10 2 Ivy’s pack x - 12 9 3 One card is chosen at random from each pack. The probability that the two cards have pictures of buses on them is equal to twice the probability that the two cards have pictures of cars on them. (a) Write down an equation in terms of x and hence find the value of x. [4] * 0000800000007 * DFD 7 (b) Find the probability that the two cards have pictures of the same type of vehicle on them. [3] * 0000800000008 * DFD 8

Mark scheme

4(a) [P(BB) = 2 × P(CC)] B1 6 x − 12 Either × . 18 x 6 x − 12 10 9 × = 2× × 18 x 18 x 10 9 Or × seen. 18 x x − 12 10 3x = x OE. B1 Correct equation formed. Fractions may be simplified. x 2 − 12 x = 30 x M1 Rearrange probabilities to form quadratic equation and solve to or find a value for x. x 2 − 42 x = 0 Condone elimination of x on denominators: or 6 10 x ( x − 42 ) = 0 × ( x − 12 ) = 2 × × 9 OE. 18 18 Must be an equation throughout. [ x =] 42 A1 If M1 not awarded, SCB1 for [x =] 42 WWW. Note x = 42 must be selected if x = 0 is present. 4 4(b) M1 Two identified un-simplified outcomes with their x substituted. BB 6 30 6 ( x − 12 ) 10 , 0.238 × , × 18 42 18 x 42 Correct values linked to identified outcomes acceptable (using x = 42). CC 10 9 10 9 5 × , × , 0.119 18 42 18 x 42 M1 Add probabilities, 0 < p < 1, for 3 correct scenarios, no incorrect/repeated scenarios. TT 2 3 2 3 1 × , × , 0.00794 Identification can be implied by un-simplified expressions. 18 42 18 x 126 23 A1 0.365079… rounded to at least 3SF. = 0.365 , 63 OE. 10 5 1 23 SCB1 + + = OE. 42 42 126 63 3

Question 57 marks

Question

The times of 240 competitors taking part in an event are recorded correct to the nearest minute. The results are summarised in the table. Time (minutes) 1 – 10 11 – 20 21 – 25 26 – 30 31 – 50 Frequency 12 38 68 76 46 (a) Draw a histogram to represent this information. [4] * 0000800000009 * DFD 9 (b) Calculate an estimate of the mean time taken by the 240 competitors. [3] * 0000800000010 * DFD 10

Mark scheme

5(a) M1 At least 4 correct frequency densities (F/CW, e.g. CW 10 10 5 5 20 12 38 68 76 46 , , , , ). FD 1.2 3.8 13.6 15.2 2.3 10 10 5 5 20 Accept un-simplified. May be read from graph if scale sufficiently accurate. A1 All heights correct on graph. Daylight rule applied full width of bar. Minimum scale: FD axis uses at least ½ the grid. B1 Bar ends at 0.5, 10.5, 20.5, 25.5, 30.5, 50.5, with a linear scale, 0.5 to 50.5, and at least 3 values indicated ‘linearly’. Daylight rule applied 0 cm ⩽ FD ⩽ 1 cm vertically. B1 Axes labelled frequency density (OE e.g. fd) and time, minutes (OE e.g. t, min). FD scale starts at 0 with a linear scale and at least 3 values indicated ‘linearly’. (condone 0.5 on time scale and no value on FD scale). Minimum: time scale use at least ½ the grid. Axes can be reversed. 4 5(b) 5.5 × 12 + 15.5 × 38 + 23 × 68 + 28 × 76 + 40.5 × 46 M1 At least 4 correct midpoints Mean = 240 Midpoints 5.5 15.5 23 28 40.5 May be seen by data table, accept un-simplified. M1 Correct mean formula using their 5 midpoints (must be within class, not upper bound, not lower bound), condone 1 error. = 25.875 A1 207 7 Accept or 25 or 25.88 or 25.9 WWW. 8 8 207 7 If 1 or more M not scored, SCB1 for , 25 or 25.875 WWW. 8 8 3

Question 69 marks

Question

For a randomly chosen person, their next birthday is equally likely to occur on any day of the week, independently of any other person’s birthday. (a) Find the probability that, out of 10 randomly chosen people, none of them will have their next birthday on a Saturday or Sunday. [1] (b) Find the probability that, out of 10 randomly chosen people, fewer than 3 will have their next birthday on a Wednesday. [3] * 0000800000011 * DFD 11 (c) Use a suitable approximation to find the probability that, out of 392 randomly chosen people, more than 65 will have their next birthday on a Friday. [5] * 0000800000012 * DFD 12

Mark scheme

6(a) 5 10 9765625 B1 0.03457… [ =] 0.0346, 7 282475249 1 6(b) Method 1 6 1 8 2 6 1 6 9 10 M1 One term 10 Cx ( p )x (1 − p )10− x . [P(0, 1, 2) =] 10 C2 + 10 C1 + 7 7 7 7 7 With 0 < p < 1, x ≠ 0 or 10. [ = 0.2675729 + 0.3567639 + 0.2140583 =] A1 Correct un-simplified expression. Allow 10 for 10 C1 . 0.838 B1 0.838 ⩽ p ⩽ 0.839. M1 One term 10 C p x 1 − p 10− x . x ( ) ( ) 10 9 2 8 1 6 1 6 1 [P(0,1,2) =] 1 –{ + 10 C9 + 10 C1 + 10 C1 7 7 7 7 7 With 0 < p < 1, x ≠ 0 or 10. 3 7 4 6 5 5 6 1 6 1 6 1 + C1 + C1 + C1 10 10 10 A1 Correct un-simplified expression. Allow 10 for 10 C1 . 7 7 7 7 7 7 4 6 3 7 Condone omission of up to 5 of the middle 6 terms. 6 1 6 1 Condone omission of last bracket only. + C1 } 10 7 7 7 7 If both brackets omitted in un-simplified expression allow recovery for final stated calculation of 1 – 0.1616 or final answer WRT to 0.8384. 0.838 B1 0.838 ⩽ p ⩽ 0.8385. 3 6(c) 1 B1 56 and 48 seen, allow un-simplified, may be seen in the [Mean = 392 × =] 56 7 standardisation formula. 1 6 [Variance = 392 × × =] 48 ( σ = 48, 4 3, 6.928 σ 6.9283 implies correct variance. 7 7 Condone N(30, 48 ) if standardisation formula is correct or variance/standard deviation correctly stated as well. 65.5 − 56 M1 Substituting their µ and positive σ into the ± standardising formula [P(X > 65) = P( Z >] ) (any number for 65.5), allow σ2 or √σ. 48 M1 Use continuity correction 64.5 or 65.5 in ±standardisation formula 9.5 9.5 Note: If no standardisation formula seen ± or ± 6.928 48 scores M2. [= 1 − Φ (1.3712 ) ] M1 Appropriate area Φ, from final process, must be a probability. = 1 – 0.9149 May be implied by a sketch of the required probability area. Note: correct final answer implies this M1. Expect final answer < 0.5. = 0.0851 final answer A1 Final answer Accept 0.08505 ⩽ p ⩽ 0.0852. 5

Question 710 marks

Question

(a) Find the number of different arrangements of the 10 letters in the word ZOOLOGICAL in which the three Os are together and the two Ls are not next to each other. [4] (b) Find the number of different arrangements of the 10 letters in the word ZOOLOGICAL in which there are exactly 5 letters between the two Ls. [3] * 0000800000013 * DFD 13 Two letters are chosen at random from the 10 letters in the word ZOOLOGICAL. (c) Find the probability that these two letters are different. [3] * 0000800000014 * DFD 14 Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. DFD 15 * 0000800000016 * DFD 16

Mark scheme

7(a) Method 1 Total arrangements with 3 Os together – total arrangements with 3 Os together and 2 Ls together. 8! B1 8! − 7! seen alone (not multiplied/divided). 2! 2! B1 b − 7!, 5040 < b . M1 8! 7! − , c = 1, 2 d = 1, 3 . c ! c !d ! = 15120 A1 CAO. Method 2 ^ ^ OOO ^ ^ ^ , Arrangements with OOOs together and no Ls, Ls inserted separately. 7×6 B1 6!× e,1e 42 . 6! × 2 B1 7×6 7 P2 f× , 1 < f accept 7 C2 or . 2 2 M1 6! 7 × 6 × , g = 1,3 h = 1, 2,3 . g! h = 15120 A1 CAO. 4 7(b) Method 1 L _ _ _ _ _ L _ _ _ 8! B1 8! ×4 × i, i 1 . 3! 3! M1 8! × 4, j = 1, 2, 3 . j! 26880 A1 CAO. Method 2 8 4! B1 8 P5 × k , k 1 . P5 × 3! M1 8 4! Pm × or 8 Pm × 4, m = 3, 4, 5. 3! 4! or 8 C5 × or 8 C5 × 4. 3! 26880 A1 CAO. 3 7(c) Method 1 1 – P(2 letters the same) 3 2 2 1 M1 3 2 2 1 1− × + × × + × seen. 10 9 10 9 10 9 10 9 M1 3 2 2 1 1 − × + × d = 9 or 10. d d d d 6 3 2 1 2 1 Accept 1 − or or or + or . 90 45 30 15 90 45 41 A1 41 = , 0.911 If one or more M mark not awarded, SCB1 for , 0.911 WWW. 45 45 Method 2 P(O O) + P(L L) + P( OL) 3 7 2 8 5 9 × + × + × 10 9 10 9 10 9 M1 3 7 2 8 5 9 × + × + × , d = 9 or 10. d d d d d d Accept 21 7 16 8 45 9 1 or + or + or or . 90 30 90 45 90 18 2 41 A1 41 = , 0.911 If one or more M mark not awarded, SCB1 for , 0.911 WWW. 45 45 7(c) Method 3 Combination approach using OO and LL 3 C2 + 2 C2 M1 3 C + 2 C seen. 2 2 [Probability =] 1 − 10 C2 M1 f 1 − 10 , 1 ⩽ f < 45. C2 41 A1 41 = , 0.911 If one or more M mark not awarded, SCB1 for , 0.911 WWW. 45 45 Method 4 Combinations with scenarios OL, OL, OL, OL M1 3 C × 2 C + 3 C × 5 C + 5 C × 2 C + 5 C seen 1 1 1 1 1 1 2 O and L 3C × 2C 1 1 [6] M1 g 10 ,1 g < 45 O and not L 3C × 5C 1 1 [15] C2 Not O and L 5C × 2C 1 1 [10] Not O and not L 5C 2 [10] 6 + 15 + 10 + 10 [Probability = ] 10 C2 41 A1 41 = , 0.911 If one or more M mark not awarded, SCB1 for , 0.911 WWW. 45 45 3

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On this page, free: the mark scheme for AS & A Level Mathematics 9709/52, October/November 2025 (9709_w25_ms_52), and the question paper (9709_w25_qp_52), to read or download, with each question's marking points written out underneath.

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Mathematics 9709/52 for October/November 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.

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