Free until 14 Nov · 80% off for the first 100 · Pricing
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w21_qp_539709_w21_ms_539709/53/O/N/21
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
The 26 members of the local sports club include Mr and Mrs Khan and their son Abad. The club is holding a party to celebrate Abad’s birthday, but there is only room for 20 people to attend. In how many ways can the 20 people be chosen from the 26 members of the club, given that Mr and Mrs Khan and Abad must be included? [2]
Mark scheme
1 C17 M1 Cx or yC17 or zC6 , x, y or z are integers no +, –, × or ÷. 100947 A1 CAO 2
Question
Lakeview and Riverside are two schools. The pupils at both schools took part in a competition to see how far they could throw a ball. The distances thrown, to the nearest metre, by 11 pupils from each school are shown in the following table. Lakeview 10 14 19 22 26 27 28 30 32 33 41 Riverside 23 36 21 18 37 25 18 20 24 30 25 (a) Draw a back-to-back stem-and-leaf diagram to represent this information, with Lakeview on the left-hand side. [4] (b) Find the interquartile range of the distances thrown by the 11 pupils at Lakeview school. [2]
Mark scheme
2(a) Lakeview Riverside B1 Correct stem, ignore extra values. 9 4 0 1 8 8 B1 Correct Lakeview labelled on left, leaves in order from 8 7 6 2 2 0 1 3 4 5 5 right to left and lined up vertically, no commas. 3 2 0 3 0 6 7 1 4 B1 Correct Riverside labelled on same diagram, leaves in order and lined up vertically, no commas. Key: 6|2|3 means 26m for Lakeview and 23m for Riverside B1 Correct key for their diagram, need both teams identified and ‘m’ stated at least once here or in leaf headings or title. SC If 2 separate diagrams drawn: SC B1 if both keys meet these criteria. 4 2(b) UQ = 32, LQ = 19 M1 (30 ⩽ UQ ⩽ 33) – (14 ⩽ LQ ⩽ 22) IQR = 32 – 19 = 13 A1 WWW 2
Question
The times taken, in minutes, by 360 employees at a large company to travel from home to work are summarised in the following table. Time, t minutes 0≤t<5 5 ≤ t < 10 10 ≤ t < 20 20 ≤ t < 30 30 ≤ t < 50 Frequency 23 102 135 76 24 (a) Draw a histogram to represent this information. [4] (b) Calculate an estimate of the mean time taken by an employee to travel to work. [2]
Mark scheme
3(a) Cw: 5 5 10 10 20 M1 At least 4 frequency densities calculated (f/cw), accept unsimplified and class widths ±1 of true values. May be implied by graph. Fd: 4.6 20.4 13.5 7.6 1.2 A1 All heights correct on graph NOT FT B1 Bar ends at 0, 5, 10, 20, 30, 50 clear intention not to draw at 4.5 or 5.5 etc. B1 Axes labelled: Frequency density (fd), time (t) and mins (or appropriate title). Linear scales between 0 and 20.4 or above on vertical axis, and 0 and 50 or above on the horizontal axis. (Axes may be reversed.) 4 3(b) 2.5 × 23 + 7.5 × 102 + 15 × 135 + 25 × 76 + 40 × 24 M1 Uses at least 4 midpoint attempts (e.g. 2.5 ± 0.5) in 360 correct formula, accept unsimplified expression, denominator either correct or their Σfrequencies . 5707.5 41 A1 Evaluated. 360 = 15.9, 15 48 2
Question
Raj wants to improve his fitness, so every day he goes for a run. The times, in minutes, of his runs have a normal distribution with mean 41.2 and standard deviation 3.6. (a) Find the probability that on a randomly chosen day Raj runs for more than 43.2 minutes. [3] (b) Find an estimate for the number of days in a year (365 days) on which Raj runs for less than 43.2 minutes. [2] (c) On 95% of days, Raj runs for more than t minutes. Find the value of t. [3]
Mark scheme
4(a) 43.2 − 41.2 M1 Use of ±Standardisation formula once, allow continuity P ( X > 43.2 ) = P Z > = P ( Z > 0.5556) correction, not σ2, √ σ. 3.6 1 – Φ ( 0.5556 ) = 1 − 0.7108 M1 Appropriate area Φ, from final process, must be probability. 0.289 A1 AWRT 3 4(b) Probability = 1 – their (a) = 1 – 0.2892 = 0.7108 B1FT 1 – their (a) or correct. 0.7108 × 365 = 259.4 B1FT FT their 4SF (or better) probability, final answer must be 259, 260 positive integer. 2 4(c) z = ±1.645 B1 CAO, critical z value. t − 41.2 M1 Use of ±standardisation formula with µ, σ equated to a = −1.645 z-value, no continuity correction, allow σ2, √ σ. 3.6 t = 35.3 A1 3 5
Question
A security code consists of 2 letters followed by a 4-digit number. The letters are chosen from {A, B, C, D, E} and the digits are chosen from {1, 2, 3, 4, 5, 6, 7}. No letter or digit may appear more than once. An example of a code is BE3216. (a) How many different codes can be formed? [2] (b) Find the number of different codes that include the letter A or the digit 5 or both. [3] A security code is formed at random. (c) Find the probability that the code is DE followed by a number between 4500 and 5000. [3]
Mark scheme
5(a) 5 P2 × 7P4 or 5 × 4 × 7 × 6 × 5 × 4 M1 Px × 7Py, 1⩽ x ⩽ 4, 1 ⩽ y ⩽ 6 16 800 A1 2 5(b) Method 1 [Identify scenarios] With A and no 5: 8 × 6P4 or (1 × 4 × 6 ×5 × 4 × 3) ×2 or 4C1 × 2! × 6P4 = M1 One number of ways correct, accept unsimplified. 2880 With 5 and no A: 4P2 × 4 × 6P3 or (4 × 3 × 1 × 6 × 5 × 4) × 4 or 4P2 × 6C3 × M1 Add 2 or 3 identified correct scenarios only, accept 4! = 5760 unsimplified. With A and 5: 8× 4 × 6P3 or (4 × 1 × 1 × 6 × 5 × 4) × 8 or 4C1 × 2! × 6C3 × 4! = 3840 [Total =] 12 480 A1 CAO Method 2 [total number of codes – number of codes with no A or 5] 4 No A or 5 : (4 × 3 ) ×( 6 × 5 × 4 × 3) = 4320 M1 P2 × 6P4 or 4C2 × 6C4 seen, accept unsimplified. Required number = their (a) – their 4320 M1 Their 5(a) (or correct) – their (No A or 5) value. 12 480 A1 Method 3 [subtracting double counting] With A 4P1 × 7P4 × 2 or 4C1 × 2 × 7C4 × 4! = 6720 M1 One outcome correct, accept unsimplified. With 5 5P2 × 6P3 × 4 or 5C2 × 2 × 6C3 × 4! = 9600 With A and 5 = 4P1 × 6P3 × 8or 4C1 × 2! × 6C3 × 4! × 8 = 3840 Required number = 6720 + 9600 – 3840 M1 Adding ‘with a’ to ‘with 5’ and subtracting ‘A and 5’. 12 480 A1 CAO 3 5(c) Method 1 – number of successful codes divided by total (1 ×) 3 × 5P2 M1 3 × 5Pn, n = 2, 3. Condone 3 × 5C2, no + or –. their 3 × 5 P 2 M1 their 60 Probability = Probability = . their 16 800 their 16 800 1 A1 , 0.00357 280 Method 2 – product of probabilities of each part of code 1 1 1 3 5 4 1 1 3 × 5P 2 M1 1 1 × × × × × or × × × × k where 0 < k < 1 for considering letters. 5 4 7 6 5 4 5 4 7 P4 5 4 M1 1 3 3 × 5P 2 t × × or t × where 0 < t < 1 . 7 6 7 P4 1 A1 CAO 280 3
Question
In a game, Jim throws three darts at a board. This is called a ‘turn’. The centre of the board is called the bull’s-eye. The random variable X is the number of darts in a turn that hit the bull’s-eye. The probability distribution of X is given in the following table. x 0 1 2 3 P X = x 0.6 p q 0.05 It is given that E X = 0.55. (a) Find the values of p and q. [4] (b) Find Var X . [2] Jim is practising for a competition and he repeatedly throws three darts at the board. (c) Find the probability that X = 1 in at least 3 of 12 randomly chosen turns. [3] (d) Find the probability that Jim first succeeds in hitting the bull’s-eye with all three darts on his 9th turn. [1]
Mark scheme
6(a) p + q + 0.65 = 1 B1 Sum of probabilities = 1. p + 2q + 0.15 = 0.55 B1 Use given information. Solve 2 linear equations M1 Either a single expression with one variable eliminated formed or two expressions with both variables on the same side seen with at least one variable value stated. 3 1 A1 CAO, both WWW p = 0.3, , q = 0.05, If M0 with correct answers SC B1. 10 20 4 6(b) Var (X) = their 0.3 + 4 × their 0.05 + 9 × 0.05 − 0.552 M1 Appropriate variance formula including (E(X))2, accept unsimplified. é 259 ù A1 CAO (must be exact). 0.6475 ê ú êë 400 úû 2 6(c) 1 – P(0, 1, 2) = 1 – (12C0 0.30 0.712 + 12C1 0.31 0.711 + 12C2 0.32 0.710) M1 One correct term: 12Cx px (1 – p)12–x for 0 < x < 12, 0 < p <1. 1 – (0.01384 + 0.07118 + 0.16779) A1FT Correct unsimplified expression, or better in final answer. Unsimplified expression must be seen to FT their p from 6(a) or correct. 0.747 A1 3 6(d) ( 0.95) × 0.05 = 0.0332 or 0.958 − 0.959 = 0.0332 8 B1 Evaluated. 1
Question
Box A contains 6 red balls and 4 blue balls. Box B contains x red balls and 9 blue balls. A ball is chosen at random from box A and placed in box B. A ball is then chosen at random from box B. (a) Complete the tree diagram below, giving the remaining four probabilities in terms of x. [3] Box A Box B Red Red 6 10 Blue Red 4 10 Blue Blue 4 (b) Show that the probability that both balls chosen are blue is . [2] x + 10 It is given that the probability that both balls chosen are blue is 61 . (c) Find the probability, correct to 3 significant figures, that the ball chosen from box A is red given that the ball chosen from box B is red. [5] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) x +1 9 x 10 B1 One probability correct in correct position. Probabilities: , , , x + 10 x + 10 x + 10 x + 10 B1 Another probability correct in correct position. B1 Other two probabilities correct in correct positions. 3 7(b) 4 10 M1 Method consistent with their tree diagram. × their 10 x + 10 4 A1 AG x + 10 2 7(c) 4 1 B1 Find value of x. Can be implied by correct probabilities in = calculation. x + 10 6 x + 10 = 24, x = 14 P(ARed|BRed) = P(ARed ∩ BRed) ÷ P(BRed) B1 FT 6 x +1 × their as numerator or denominator of 10 x + 10 6 x +1 6 15 3 fraction. × their × 10 x + 10 = 10 24 = 8 6 x +1 4 x 6 15 4 14 73 M1 6 x +1 4 x × their + × their × + × × their + × their seen anywhere. 10 x + 10 10 x + 10 10 24 10 24 120 10 x + 10 10 x + 10 A1 FT Seen as denominator of fraction. 45 A1 If B0 M0: , 0.616[4…] 73 3 0.375 45 SC B1 for 8 or SC B1 or 0.616. 73 0.6083 73 120 5
Checked your answers against these? Let AI mark the whole paper, on your own working.
Mark my 9709/53Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Question papers and mark schemes © Cambridge University Press & Assessment. CIE Buddy is independent and not connected to Cambridge.
On this page, free: the mark scheme for AS & A Level Mathematics 9709/53, October/November 2021 (9709_w21_ms_53), and the question paper (9709_w21_qp_53), to read or download, with each question's marking points written out underneath.
Mathematics 9709/53 for October/November 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
Yes. Upload your answers to 9709/53/O/N/21 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.