Cambridge A Level Mathematics 9709 — 2024 May/June Paper 5 · Variant 3

1 The numbers on the faces of a fair six-sided dice are 1, 2, 2, 3, 3, 3. The random variable X is the total score when the dice is rolled twice. (a) Draw up the probability distribution table for X. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the value of Var(X ). [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the probability that X is even given that X 2 3 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) x 2 3 4 5 6 P(X = x) 1 36 4 36 10 36 12 36 9 36 1 36 1 9 5 18 1 3 1 4 Decimal equivalent 3sf: 0.0278, 0.111, 0.278, 0.333, 0.25 associated with the correct X value. Values need not be in order, lines may not be drawn, may be vertical, X and P(X) may be omitted. Condone any additional X values if probability stated as 0. B1 Three other probabilities associated with correct x values, need not be in table, accept unsimplified. B1 Five correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3sf. SC B1 for five probabilities summing to 1 placed in a probability distribution table with the correct x values. 3 Question Answer Marks Guidance 1(b) E(X) = 1 2 4 3 10 4 12 5 9 6 36          2 12 40 60 54 14 or 4.67 36 36 36 36 36 3            M1 Accept unsimplified expression or sum of fractions seen. May be calculated in variance. FT their table with five probabilities summing to 0.999 ⩽ total ⩽ 1 (0 < p < 1). Var(X) = 2 2 2 2 2 2 1 2 4 3 10 4 12 5 9 6 14 36 3               M1 Appropriate variance formula using their (E(X))2 value. FT their table with 4 or more probabilities. (0 < p < 1) which need not sum to 1. Note: If table is correct, then 2 824 206 196 14 or or 22.89 or 21.78 or 36 9 9 3                     implies M1. [= 824 196 22.89 21.78] 36 9    = 10 9 A1  1 1 , 1.11 1 , 9 1.1 3 1(c) P(X even | X > 3) = 10 9 36 36 31 36  M1      P 4 P 6 , P 4 P 5 P 6 their their their    all probabilities (0 < p < 1). If sample space seen in any part of the question, then M1 1 9. 31 their their = 19 31 A1 0.613 2

More questions on Probability

2 In a certain country, the heights of the adult population are normally distributed with mean 1.64 m and standard deviation 0.25 m. (a) Find the probability that an adult chosen at random from this country will have height greater than 1.93 m. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ In another country, the heights of the adult population are also normally distributed. 33% of the adult population have height less than 1.56 m. 25% of the adult population have height greater than 1.86 m. (b) Find the mean and the standard deviation of this distribution. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) P 1.93 1.64 0.25 Z          P(Z > 1.16) substituted, not 2,  not ,  no continuity correction. 1 – 0.8770 M1 Appropriate area Φ resulting from a standardisation, from final process, must be probability. Note: the appropriate probability answer implies this M1. 0.123 A1 0.123 ⩽ p < 0.12303 If M0 M0, SC B1 if no standardisation shown. 3 2(b) 1.56 0.44     1.86 0.674     B1 −0.441 < z1 < −0.439 or 0.439 < z1 < 0.441 seen. B1 z2 = 0.674 or z2 = – 0.674 seen, CAO, critical value. M1 Use of the ±standardisation formula once with μ, σ equating to a z- value (not 0.33, 0.67. 0.25, 0.75, 0.6293, 0.5987, 0.7486, 0.7734, (1 – 0.44), (1 – 0.674)). Condone continuity correct ± 0.005, not 2, .   Solve, obtaining values for and σ  1.68, 0.269     M1 Solve two equations in μ and σ using the elimination method, substitution method or other appropriate approach to obtain values for both μ and σ. A1 AWRT 1.68, 0.269.     If one or both of the M marks have not been awarded, SC B1 for both correct. 5

More questions on The normal distribution

3 Box A contains 6 green balls and 3 yellow balls. Box B contains 4 green balls and x yellow 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) Draw a tree diagram to represent this information, showing the probability on each of the branches. [4] 8 The probability that both the balls chosen are the same colour is . 15 (b) Find the value of x. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) B1 Correct structure and probabilities for Box A branches. B1 Completely correct structure and one correct probability for a Box B branch including label for G or Y. B1 Completely correct structure and second correct probability on a Box B branch including label G or Y. B1 Completely correct structure and remaining two probabilities correct on Box B branches, including labels for G or Y. SC B1 if correct shape diagram but only four correct algebraic probs for GG, GY, YG and YY. 4 3(b) P(same colour) = 6 5 3 1 9 5 9 5 x x x       M1 P(GG) + P(YY) = 6 3 5 3 6 1 or or 9 9 5 9 9 5 x their their x x                               6 5 3 1 8 9 5 9 5 15 x x x        and arrange as a linear equation M1     15 11 24 5 x x    OE Accept sum of their products equated to 8 15 and rearranged to form a linear equation. Solve: 5 x  A1 3 Box A Box B

More questions on Probability

4 The times taken, in seconds, by 15 members of each of two swimming clubs, the Penguins and the Dolphins, to swim 50 metres are shown in the following table. Penguins 35 39 42 44 45 45 48 50 56 58 59 61 66 68 72 Dolphins 36 41 43 48 49 49 50 51 54 56 56 60 61 64 71 (a) Draw a back-to-back stem-and-leaf diagram to represent this information, with Penguins on the left-hand side. [4] The diagram shows a box-and-whisker plot representing the times for the Penguins. (b) On the same diagram, draw a box-and-whisker plot to represent the times for the Dolphins. [3] Penguins 30 35 40 45 50 55 60 65 70 75 Times in seconds (c) Hence state one difference between the distributions of the times for the Penguins and the Dolphins. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) Penguins Dolphins 9 5 3 6 8 5 5 4 2 4 1 3 8 9 9 9 8 6 0 5 0 1 4 6 6 8 6 1 6 0 1 4 2 7 1 Key: 2 | 4 | 1 means 42 seconds for Penguins and 41 seconds for Dolphins If a split stem-and-leaf plot is used (i.e. stem values are repeated), the remaining B marks are available. B1 Correct Penguins labelled on left, leaves in order from right to left and lined up vertically (less than halfway to next column), no commas or other punctuation. B1 Correct Dolphins labelled on same diagram, leaves in order and lined up vertically (less than halfway to next column), no commas or other punctuation. If the correct data for Penguins & Dolphins is transposed, treat as a single error in Penguins and condone in Dolphins. B1 Correct key for their diagram, need both clubs labelled and ‘sec’ or ‘s’ stated at least once here, or in leaf headings or title. If two separate diagrams drawn: SC B1 if both keys meet these criteria. 4 Question Answer Marks Guidance 4(b) For Dolphins, median is 51 B1 Plotted on box. LQ = 48, UQ = 60 B1 Plotted on box. Correct end points for whiskers and diagram labelled Dolphins B1 Correct end points of whiskers (36 and 71). Whiskers not through box, not drawn at corners of boxes, diagram labelled. 3 4(c) Dolphins have more consistent times than Penguins or Penguins are faster (have faster times) than Dolphins B1 Reason given in context. Can be reference to either the central tendency or spread. 1

More questions on Representation of data

6 (a) How many different arrangements are there of the 9 letters in the word RECORDERS? [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) How many different arrangements are there of the 9 letters in the word RECORDERS in which there is an E at the beginning, an E at the end and the three Rs are not all together? [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The 9 letters of the word RECORDERS are divided at random into two groups: a group of 5 letters and a group of 4 letters. (c) Find the probability that the three Rs are in the same group. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) 9! 30240 2!3!        1 6(b) Method 1: Number of arrangements with E at each end – Number of arrangements with E at each end and the three Rs together 7! 5! 3!   B1 7! 3! e  , 7P4 – e, e a positive integer. M1 5! ! f r  , f > 120, r = 1, 2 720 A1 If no marks scored SC B1 for 840 – 120 = 720. Method 2: Number of arrangements with E at each end and no Rs together + No of arrangements with E at each end and two Rs together 5C3 × 4! + 4C1 × 5! or 3 5P 4! 3!  + 5P2 4!  (B1) One of 5C3 × 4!, 3 5P 4! , 3!  4C1 × 5! or 5P2 4!  seen. (M1) a × 4! + b × 5! where a and b are integers between 1 and 10 inclusive, or c × 4! + d × 4! where c and d are integers between 1 and 20 inclusive. 240 + 480 = 720 (A1) 3 Question Answer Marks Guidance 6(c) Method 1 Group of 5 3 Rs 2 Es = 1 3 Rs 1 E = 2C1 × 4C1 = 8 3 Rs 0 Es = 4C2 = 6 Group of 4 3 Rs 1 E = 2C1 = 2 3 Rs 0 Es = 4C1 = 4 B1 Correct no of ways for two correct identified scenarios other than three Rs two Es. [Total =] 21 M1 No of ways for five correct identified scenarios added or correct. [Number of ways of splitting into the two groups =] 9C5 (= 126) seen as a denominator M1 Accept evaluated, accept 9C4. Probability = 21 1 126 6  (0.167) A1 Question Answer Marks Guidance 6(c) Method 2 3Rs in Group of 5 = 6C2 = 15 3Rs in Group of 4 = 6C1 = 6 (B1) One correct case evaluated accurately and linked with correct scenario. [Total =] 21 (M1) No of ways for two correct scenarios added or correct. [Number of ways of splitting into the two groups =] 9C5 (= 126) seen as a denominator (M1) Accept evaluated, accept 9C4. Probability = 21 1 126 6  (0.167) (A1) Method 3: Considering the possible positions of R within the groups 3Rs in Group of 5 5 4 3 9 8 7   = 15 126 3Rs in Group of 4 4 3 2 9 8 7   6 126  (B1) For one correct product unsimplified and linked with correct scenario. (M1) For second correct product. 15 126 + 6 126 (M1) For adding probabilities of two correct scenarios or correct. Probability = 21 1 126 6  (0.167) (A1) Question Answer Marks Guidance 6(c) Method 4: Probability method Group of 5 2Es 3 2 1 2 1 5! 1 9 8 7 6 5 3!2! 126       1E 4 3 2 1 2 5! 8 9 8 7 6 5 3! 126       0E 3 2 1 4 3 5! 6 9 8 7 6 5 3!2! 126       Group of 4 3 2 1 6 4! 6 9 8 7 6 3! 126      (B1) Two correct probabilities linked with correct scenarios, accept unsimplified. (M1) Four probabilities with denominators including a factor of 9  8  7  6  n, where n is 1 or 5. 1 8 6 6 126    (M1) Probabilities of four correct scenarios added or correct. Probability = 21 1 126 6  (0.167) (A1) 4

More questions on Permutations and combinations