Mathematics 9709/52 February/March 2024: question paper and mark scheme

Q: Where can I find the 9709/52 February/March 2024 mark scheme?

On this page, free: the mark scheme for AS & A Level Mathematics 9709/52, February/March 2024 (9709_m24_ms_52), and the question paper (9709_m24_qp_52), to read or download, with each question's marking points written out underneath.

Q: How many marks is 9709/52 out of?

Mathematics 9709/52 for February/March 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.

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Cambridge AS & A Level Mathematics (9709), Paper 5 Probability and Statistics 1, February/March 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9709_m24_qp_529709_m24_ms_529709/52/F/M/24

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Paper: Paper 5 Probability and Statistics 1
Total marks: 50
Time: 1 hour 15 minutes
Code on the paper: 9709/52/F/M/24

9709_m24_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 14 marks

Question

A bag contains 9 blue marbles and 3 red marbles. One marble is chosen at random from the bag. If this marble is blue, it is replaced back into the bag. If this marble is red, it is not returned to the bag. A second marble is now chosen at random from the bag. (a) Find the probability that both the marbles chosen are red. [1] (b) Find the probability that the first marble chosen is blue given that the second marble chosen is red. [3]

Mark scheme

1(a) 3 2 1 B1 6 12 × 11 = 22 Accept , 0.04545... to at least three significant figures. 132 1 1(b) 9 3 M1 9 3 27 3 × × = , , 0.1875 seen as numerator or denominator P(B1 | R2) = 12 12 12 12 144 16 9 3 3 2 of a fraction. × + × 12 12 12 11 M1 9 3 3 2 27 1 Their × + their × or their + their seen as 12 12 12 11 144 22 denominator of a fraction. FT from part (a). 3 A1 Accept 4752 oe , 0.804878… rounded to at least three 16 33 5904 = 3 1 = , 0.805 significant figures. + 41 16 22 If A0, SC B1 for correct final answer www. 3

Question 28 marks

Question

Sam is a member of a soccer club. She is practising scoring goals. The probability that Sam will score a goal on any attempt is 0.7, independently of all other attempts. (a) Sam makes 10 attempts at scoring goals. Find the probability that Sam will score goals on fewer than 8 of these attempts. [3] (b) Find the probability that Sam’s first successful attempt will be before her 5th attempt. [2] (c) Wei is a member of the same soccer club. He is also practising scoring goals. The probability that Wei will score a goal on any attempt is 0.6, independently of all other attempts. Wei is going to keep making attempts until he scores 3 goals. Find the probability that he scores his third goal on his 7th attempt. [3]

Mark scheme

2(a) Method 1 [P(X < 8) = 1 – P(8, 9, 10) =] 1 – (10 C8 (0.7)8 (0.3)2 + 10 C9 (0.7)9 (0.3) + (0.7)10 ) ( )( ) M1 One term 10 Cx p x 1 − p 10− x with 0 < p < 1, x ≠ 0 or 10. A1 Correct unsimplified expression. Condone omission of last bracket only. = [1 – (0.2335 + 0.1211 + 0.0282)] = 0.617 B1 0.617 ⩽ p < 0.6175 www. Method 2 x ( ) ( ) [P(0,1,2,3,4,5,6,7) = ] M1 One term 10 C p x 1 − p 10− x with 0 < p < 1, x ≠ 0 or 10. 0.310 +10 C9 0.7 0.39 + … + 10 C3 0.77 0.33 A1 Correct unsimplified expression. [= 5.905 × 10−6 + 1.378 × 10−3 + + 0.2668] B1 0.617 ⩽ p < 0.6175 www. = 0.617 3 2(b) Method 1 [P(X < 5) = ] 1 − 0.34 M1 1 – bd ; b = 0.3, 0.7; d = 4, 5. 1 – ce – (1 – c) × ce – 1 ; c = 0.3, 0.7; e = 5, 6. 9919 A1 Condone 0.992. 0.9919, 10000 9919 If M0 scored, SC B1 for 0.9919 or only. 10000 Method 2 [P(X < 5) = ] (0.7) + (0.3)(0.7) + (0.3) 2 (0.7) + (0.3) 3 (0.7) M1 (e) + (f)(e) + (f)2 (e) + (f)3 (e) [+ (f)4 (e)] e = 0.7, 0.3; e + f = 1. 9919 A1 Condone 0.992. 0.9919, 10000 9919 If M0 scored, SC B1 for 0.9919 or only. 10000 2 2(c) (0.4)4 (0.6)2 × 0.6 × 6 C2 M1 ( 0.4)4 ( 0.6)r ; r = 2, 3. No inappropriate addition. M1 ( 0.4)a ( 0.6)b × 6 C2 ; a + b = 6, 7. 1296 A1 Accept 0.082944 correct to at least three significant figures. = 0.0829, If A0 scored, SC B1 for correct answer www. 15625 3

Question 38 marks

Question

The times taken, in minutes, by 150 students to complete a puzzle are summarised in the table. Time taken 0 G t 1 20 20 G t 1 30 30 G t 1 35 35 G t 1 40 40 G t 1 50 50 G t 1 70 (t minutes) Frequency 8 23 35 52 20 12 (a) Draw a histogram to represent this information. [4] (b) Calculate an estimate for the mean time for these students to complete the puzzle. [3] (c) In which class interval does the lower quartile of the times lie? [1]

Mark scheme

3(a) M1 8 23 Class 20 10 5 5 10 20 At least five frequency densities (f/cw), e.g. , , 20 10 width Accept unsimplified. Frequency 0.4 2.3 7.0 10.4 2.0 0.6 density A1 All heights correct on graph (no FT). B1 Bar ends at [0,] 20, 30, 35, 40, 50, 70 with a linear scale and at least three values indicated ‘linearly’. B1 Axes labelled frequency density (fd) and time (mins). Frequency density scale vertical starts at 0 with a linear scale and at least three values indicated ‘linearly’. Axes can be reversed. 4 3(b) Midpoints 10, 25, 32.5, 37.5, 45, 60 B1 At least five correct mid-points seen or used in formula. 10 × 8 + 25 × 23 + 32.5 × 35 + 37.5 × 52 + 45 × 20 + 60 ×12 M1 Correct mean formula using their 6 midpoints (must be within Mean = class, not upper bound, lower bound). Condone one error. 150 5362.5 5362.5 If correct midpoints seen, accept or = 150 150 80 + 575 + 1137.5 + 1950 + 900 + 720 . 150 3 A1 143 = 35.75, 35 Accept 35.8, not . 4 4 3 If A0 scored, SC B1 for 35.75, 35 only. 4 3 3(c) 30 t < 35 B1 Condone ‘3rd’ interval, 30 – 35. 1

Question 412 marks

Question

A company sells small and large bags of rice. The masses of the small bags of rice are normally distributed with mean 1.20 kg and standard deviation 0.16 kg. (a) In a random sample of 500 of these small bags of rice, how many would you expect to have a mass greater than 1.26 kg? [4] The masses of the large bags of rice are normally distributed with mean 2.50 kg and standard deviation v kg. 20% of these large bags of rice have a mass less than 2.40 kg. (b) Find the value of v . [3] A random sample of 80 large bags of rice is chosen. (c) Use a suitable approximation to find the probability that fewer than 22 of these large bags of rice have a mass less than 2.40 kg. [5]

Mark scheme

4(a) 1.26 − 1.20 M1 Use of ± standardisation formula with 1.26, 1.20 and 0.16, not [P(X > 1.26) = P Z >] 0.16 σ 2 , not σ , no continuity correction. P ( Z > 0.375) = = 1 – 0.6462 M1 Calculating the appropriate probability area, (leading to their final probability, expect < 0.5). = 0.354 A1 0.3538, 0.3535 < p ⩽ 0.354. Only dependent on the 2nd M mark. SC B1 for 0.3535 < p ⩽ 0.354 if M0M0A0 awarded. [500 × their 0.3538] = 176, 177 B1 FT Strict FT their at least 4-figure calculated probability, seen anywhere (not a z-value). Final answer must be a single positive integer value, no approximation or rounding stated. 4 4(b) 2.40 − 2.50 B1 –0.842 ⩽ z < –0.8415 or 0.8415 < z ⩽ 0.842 seen. P Z > σ = 0.20 M1 Use of the ± standardisation formula with 2.40, 2.50, σ and a z- value (not 0.20, 0.80, 0.158, 0.7881, 0.2119, 0.5793, 0.4207, 2.40 − 2.50 …), not σ 2 . = −0.842 Condone continuity correction of ± 0.005. σ 0.1 Condone − = −0.842 etc. for M1. σ σ = 0.119 A1 0.1185 < σ ⩽ 0.119. 3 4(c) [Mean = 80 ×0.2 =]16 B1 16 and 12.8 seen, allow unsimplified. [Variance = 8 0 × 0.2 × 0.8 =]12.8 May be seen in standardisation formula. 8 5 , 3.5777… to at least three significant figures implies 5 correct variance. Incorrect notation penalised. 21.5 − 16 M1 Substituting their 16 (not 1.2, 2.5) and their 12.8 (not 0.16, [P(X < 22) = P( Z <] ) their 0.119) in the ± standardising formula (any number for 12.8 21.5), condone σ 2 or σ . [P( Z < 1.537 ) = Φ (1.537 ) ] M1 Using continuity correction 21.5 or 22.5 in their standardisation formula. M1 Appropriate area Φ, from final process, must be a probability. 0.938 A1 0.9375 < p ⩽ 0.938. 5

Question 58 marks

Question

Anil is taking part in a tournament. In each game in this tournament, players are awarded 2 points for a win, 1 point for a draw and 0 points for a loss. For each of Anil’s games, the probabilities that he will win, draw or lose are 0.5, 0.3 and 0.2 respectively. The results of the games are all independent of each other. The random variable X is the total number of points that Anil scores in his first 3 games in the tournament. (a) Show that P (X = 2) = 0.114 . [2] (b) Complete the probability distribution table for X. [3] x 0 1 2 3 4 5 6 P (X = x) 0.114 0.207 0.285 0.125 (c) Find the value of Var (X ) . [3]

Mark scheme

5(a) P(X = 2) = P(WLL or DDL) M1 0.5 × 0.22 × 3 Ca ( or 3) + x; 0 < x < 1; a = 1, 2. = 0.5 × 0.22 × 3 C1 + 0.32 × 0.2 × 3 C1 = [0.06 + 0.054] Or 0.32 × 0.2 × 3 Cb ( or 3) + y; 0 < y < 1; b = 1, 2. Or 0.5 × 0.22 × a + 0.32 × 0.2 × b; a, b = 1, 2, 3. = 0.114 A1 AG. Fully correct solutions with outcomes identified and linked to appropriate probabilities. Condone 2 = W, 1 = D, 0 = L. Probabilities alone do not identify outcomes. If individual scenarios are identified, separate calculations must correspond to the order. 2 5(b) B1 One additional correct probability in table or clearly identified. x 0 1 2 3 4 5 6 B1 A second additional correct probability in table or clearly P(X = x) 0.008 0.036 0.114 0.207 0.285 0.225 0.125 identified. 1 9 9 125 250 40 B1 Final correct probability, all probabilities in table. If 0/3 scored, SC B1 for three additional probabilities in table that sum to 0.269 exactly. 3 5(c) [E(X) = M1 Accept unsimplified expression. May be calculated in the [0.008 × 0 +] 0.036 ×1 + 0.114 × 2 + 0.207 × 3 + 0.285 × 4 variance, FT their table with probabilities, 0 < p < 1, that sum to 1. +0.225 × 5 + 0.125 × 6 =] FT acceptable at the bold partially evaluated stage. [0] + 0.036 + 0.228 + 0.621 + 1.140 + 1.125 + 0.750 [= 3.9] OR E(X) = 3(0.5 × 2 + 0.3 × 1) [= 3.9] [Var(X) = M1 Appropriate variance formula using their (E(X))2 value. FT 0.008 × 02 + 0.036 × 12 + 0.114 × 22 + 0.207 × 32 + 0.285 × 42 their table with probabilities, 0 < p < 1, that may not sum to 1. +0.225 × 52 + 0.125 × 62 − their 3.92 =] [0.008 × 0 +] 0.036 ×1 + 0.114 × 4 + 0.207 × 9 + 0.285 ×16 +0.225 × 25 + 0.125 × 36 − their 3.92 = 17.04 − 3.92 = 1.83 A1 Cao. Condone 183 . 100 3

Question 610 marks

Question

A new village social club has 10 members of whom 6 are men and 4 are women. The club committee will consist of 5 members. (a) In how many ways can the committee of 5 members be chosen if it must include at least 2 men and at least 1 woman? [4] The 10 members of the club stand in a line for a photograph. (b) How many different arrangements are there of the 10 members if all the men stand together and all the women stand together? [2] For a second photograph, the members stand in two rows, with 6 on the back row and 4 on the front row. Olly and his sister Petra are two of the members of the club. (c) How many different arrangements are there of the 10 members in which Olly and Petra stand next to each other on the front row? [4] Additional page If you use the following page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

Mark scheme

6(a) 4M 1W: 6 C4 × 4 C1 = 60 M1 6C a × Cb with a + b = 5 seen, no extra terms. 4 3M 2W: 6 C3 × 4 C2 = 120 B1 Correct outcome/value for one clearly identified scenario. Accept unsimplified, www. 2M 3W: 6 C2 × 4 C3 = 60 Condone use of × 5 C0 . M1 Add values of three correct scenarios, no incorrect scenarios, no repeated scenarios. Condone 6 Ca × 4 Cb with a + b = 5 to identify M, W for this mark. Total 240 A1 Not dependent on B1. If A0 scored, SC B1 for 240 www. 4 6(b) 6! × 4! × 2 M1 6! × 4! × k; k = 1, 2. 1 can be implied. = 34 560 A1 Cao. If M0 scored, SC B1 for 34 560 www. 2 6(c) Method 1 – Arrangements of OP in front row, 8 remaining people arranged. 8! × 3 × 2 M1 8! × g, g an integer greater than 1. M1 h! × 3 × j; h = 7, 8, 9; j = 1, 2 (1 may be implied). Condone 3 C1 for 3. M1 h! × 3 × 2; h = 7, 8, 9. Condone 2 C1 for 2. (Condone h! × 3! For M1M1). = 241 920 A1 If A0 Scored, SC B1 for 241 920. Method 2 – Two additional people selected for front row, front row arranged, remaining 6 people arranged in back row. 8 C × 6! × 3! × 2 M1 8 C × d, a = 2,6, d an integer greater than 1. 2 a M1 6! × e, e an integer greater than 1. M1 8 C × f ! × 3! × 2 or 8 C × f ! × 6 × 2; a = 2, 6; f = 5, 6, 7. a a = 241 920 A1 If A0 Scored, SCB1 for 241 920. Method 3 – Arrangements of two additional people for front row, front row arranged, remaining 6 people arranged in back row. 8P 8P 2 × 6! × 3! M1 2 × d, d an integer greater than 1. M1 6! × e, e an integer greater than 1. M1 8P 8P 2 × h! × 3! or 2 × h! × 6; h = 5, 6, 7. = 241 920 A1 If A0 Scored, SC B1 for 241 920. 6(c) Method 4 – Arrangements of 6 people for back row, front row arranged. 8P 8P 6 × 3! × 2! M1 6 × d, d an integer greater than 1. M1 3! × e, e an integer greater than 1. M1 8P 6 × j! × 2; j = 1, 2, 3. = 241 920 A1 If A0 Scored, SC B1 for 241 920. 4

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Where can I find the 9709/52 February/March 2024 mark scheme?

On this page, free: the mark scheme for AS & A Level Mathematics 9709/52, February/March 2024 (9709_m24_ms_52), and the question paper (9709_m24_qp_52), to read or download, with each question's marking points written out underneath.

How many marks is 9709/52 out of?

Mathematics 9709/52 for February/March 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.

Can I get my 9709/52 answers marked?

Yes. Upload your answers to 9709/52/F/M/24 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.