Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 5 · Variant 5

1 A fair 6-sided dice with faces labelled 1, 2, 3, 4, 5, 6 is thrown repeatedly. The random variable X denotes the number of throws required to obtain a 6. (a) Find P ( X = 7) . [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find P ( 3 G X G 5) . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a)   5 6 1  15625 B1    =  0.0558,   6  6   279936 1 1(b) 2 3 4 M1 2 3 4  5  1   5  1   5  1  (1 − p ) p + (1 − p ) p + (1 − p ) p + +           6  6   6  6   6  6  Or Or 2 5 p 2 − p 5 0  p  1  5   5  −      6   6  A1 = 2275, 0.293 7776 2

More questions on The Poisson distribution

2 Kai has a spinner with four sides, labelled 1, 2, 3, 4. When the spinner is spun, the score is the number on the side on which the spinner lands. The random variable X denotes this score. The probability distribution table for X is given below. x 1 2 3 4 P ( X = x) p 0.4 2p p (a) Find the numerical value of Var ( X ) . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Kai spins his spinner 10 times. (b) Find the probability that a score of 2 is obtained fewer than 8 times. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 3 B1 May be seen used in formulae.  4 p + 0.4 = 1,  p = 0.15 , 20 [E(X) = ] 1 0.15 + 2  0.4 + 3  0.3 + 4  0.15 M1 Accept un-simplified numerical expression. May be calculated in 49 variance. [ = 0.15 + 0.8 + 0.9 + 0.6 = 2.45, ] 1  p + 2  0.4 + 3  2 p + 4  p, 0  p  1 , ∑p = 1. 20 [Var (X) =] 0.15  12 + 0.4  2 2 + 0.3  32 + 0.15  4 2 −their( 2.45) 2 M1 Appropriate variance formula using their (E(X))2 value FT their p, acceptable at the bold partially evaluated stage.  0.15 +1 0.4  4 + 0.3 +9 0.15  16 − 6.0025  Must be a numerical expression, [6.85 – 6.0025] condone ∑p .1 Accept 0.15 + 1.6 + 2.7 + 2.4 − 2.45 2 or 6.0025 . 339 A1 If either or both M marks not awarded, SCB1 for correct answer = 0.8475, WWW. 400 4 2(b) Method 1 [P(X < 8) = 1 – (P(8, 9, 10) =] M1 x 10 − x One term of form 10Cx ( p ) (1 − p ) . 1 – {10C8 ( 0.4 )8 ( 0.6 ) 2 + 10C9 ( 0.4 )9 ( 0.6 )1 + 10 C10 ( 0.4 )10 } With 0  p  1, x  0 or 10. [= 1 – (0.0106168 + 0.00157286 + 0.00010486)] A1 Correct un-simplified expression, no terms omitted, leading to final answer. Condone omission of final bracket ‘}’ . = 0.988 B1 0.9875 < p ⩽ 0.988. Method 2 [P(X < 8) = P(0, 1, 2, 3, 4, 5, 6, 7) =] M1 x 10 − x One term of form 10Cx ( p ) (1 − p ) . 10C0 ( 0.6 )10 + 10C1 ( 0.4 )1 ( 0.6 )9 + 10C2 ( 0.4 ) 2 ( 0.6 )8 + 10C3 With 0  p  1, x  0 or 10. ( 0.4 )3 ( 0.6 )7 + 10C4 ( 0.4 ) 4 ( 0.6 )6 + 10C5 ( 0.4 )5 ( 0.6 )5 + 10C6 6 4 7 3 A1 Correct un-simplified expression, no more than 5 ‘middle’ terms ( 0.4 ) ( 0.6 ) + 10C7 ( 0.4 ) ( 0.6 ) omitted, leading to final answer. = 0.988 B1 0.9875 < p ⩽ 0.988. 3

More questions on Discrete random variables

3 Priti has two bags of discs, X and Y. Bag X contains 8 red discs and 7 blue discs. Bag Y contains 6 red discs and 9 blue discs. Priti tosses a fair coin. If she obtains a head, she chooses at random and without replacement two discs from bag X. If she obtains a tail, she chooses at random and with replacement two discs from bag Y. (a) Draw a tree diagram to represent this information, showing all the probabilities. [2] (b) Find the probability that the two discs that Priti chooses are blue. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the probability that the two discs that Priti chooses come from bag X given that at least one of the discs is red. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) Tree diagram with all 2, 4, 8 branches B1 Structure, including labelling H, T, R & Bs and 4 or more correct probabilities. 1 1 8 7 6 9 7 7 8 6 6 9 6 9 B1 Remaining probabilities correctly placed. , , , , , , , , , , , , 2 2 15 15 15 15 14 14 14 14 15 15 15 15 2 3(b) 1 7 6 1 9 9 M1 OE.   +   FT P(HBB) + P(TBB) from incorrect tree diagram 2 15 14 2 15 15 or 1 a a − 1 1 b b   +   , a =8, 7 b = 6, 9 2 15 14 2 15 15 7 A1 = , 0.28 25 2 3(c) 7 18 B1 18 [P(1R or 2R) = 1 – P(BB) = 1 − =] , 0.72 FT 1 – their 3(b) or seen. 25 25 25 1 8 1 7 8  6  M1 FT from incorrect tree diagram P(X ∩ 1R or 2R) =  +1   =   or 2 15 2 15 14  15  1 a 1 15 − a a  +1   2 15 2 15 14 1 a  a − 1 15 − a  1 a − 1 a or    +  +   2 15  14 14  2 15 14 a = 8, 7. 6 M1 their P ( X  1R or 2R ) Identified values substituted into or   P ( X  1R or 2R ) 15 their P (1R or 2R ) = P(X|1R or 2R) =    P ( 1R or 2R )  18 correct. 25 5 A1 = , 0.556 9 4

More questions on Probability

4 The heights, in cm, of 15 players from each of two sports teams, Pelicans and Swans, are given in the table. Pelicans 156 160 164 165 167 170 171 173 178 182 182 184 185 186 187 Swans 170 180 183 165 174 158 170 181 162 178 174 163 191 182 174 (a) Draw a back-to-back stem-and-leaf diagram to represent the heights of the players from Pelicans and Swans, with Pelicans on the left-hand side. [4] (b) Find the median and the interquartile range of the heights of the Pelicans. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ For the Swans, the lower quartile of the heights is 165 cm, the median is 174 cm and the upper quartile is 181 cm. (c) Represent the data shown in the back-to-back stem-and-leaf diagram by a pair of box-and-whisker plots in a single diagram. [3] (d) Make one comparison between the heights of the Pelicans players and the heights of the Swans players. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) Pelicans Swans B1 Correct stem, ignore extra values (not in reverse, not split) 6 15 8 If a split stem-and-leaf plot is used (i.e. stem values are repeated) 7 5 4 0 16 2 3 5 the remaining B marks are available. 8 3 1 0 17 0 0 4 4 4 7 6 5 4 2 2 18 0 1 2 3 B1 Correct Pelicans labelled on left, leaves in order from right to left 19 1 and lined up vertically no commas or other punctuation. B1 Correct Swans labelled on same diagram, leaves in order from Key: 1|17|8 represents 171 cm for Pelicans and 178 cm for Swans right to left and lined up vertically, no commas or other punctuation. Penalise each error only once in question. E.g. commas in both sets of data. Missing or incorrect leaves can be penalised in both cases. B1 Correct key, for their diagram, need both team names and ‘cm’ at least once here, or in leaf headings or a suitable title. If 2 separate diagram drawn max marks: B1 if both stems correct, B1 if Pelicans correct to the left of the stem, B1 if both keys correct including ‘cm’ and team name. 4 4(b) Median = 173 (cm) B1 Accept Q2, must be identified. [IQR =] 184 – 165 M1 182 ⩽ UQ ⩽ 185 − 164 ⩽ LQ ⩽ 167 Implied if both quartile values are stated and appropriate IQR calculated accurately. = 19 [cm] A1 WWW. If M0 scored, SC1 for 19 WWW. 3 4(c) Box-and-whisker plot on provided grid B1 All 5 key values for Swans plotted accurately in standard format using a linear scale with 3 identified values. Labelled Swans. Swans 158 165 174 181 191 Vertical lines may be omitted at both limits. Pelican 156 165 173 184 187 Condone freehand lines B1FT All 5 key values for Pelicans, FT from 4b or correct, plotted Check accuracy in the middle of the vertical line using daylight accurately in standard format using a linear scale with 3 identified rule values. Labelled Pelicans. Vertical lines may be omitted at both limits B1 Whiskers not through box for both, not drawn at corners of boxes, single linear scale for the diagram and labelled “heights” (OE) and ‘cm’ . • If only one plot is labelled, assume other plot is as anticipated and mark as in scheme. • If neither of 2 plots is labelled, SCB1 for 1 fully ‘correct’ set of values plotted. • If no scale or incorrect linear scale stated, SCB1 if both correct relative to each other. 3 4(d) Swans are taller or Swans’ heights more consistent. B1 Appropriate comment about heights. If they say Pelicans’ heights are more consistent, they need to refer to the Range 1

More questions on Representation of data

5 In a group of 25 athletes, there are 8 sprinters, 5 hurdlers and 12 throwers. (a) Find the number of different ways in which a team of 6 athletes can be selected if it consists of at least 3 sprinters, at most 2 hurdlers and at most 1 thrower. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ A group of 8 athletes chosen from the group of 25 athletes consists of 1 sprinter, 3 hurdlers and 4 throwers. These 8 athletes stand in a row. (b) How many different arrangements of the 8 athletes are there if the 3 hurdlers do not stand all together? [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) How many different arrangements of the 8 athletes are there in which there are at least two athletes between any two hurdlers? [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) 6S 8C6 (= 28) M1 One product using 2 or 3 combinations 8Cx  5Cy  12Cz with x + 5S 1H 8C5  5C1 (= 280) y + z = 6 (one of y or z may be 0) and linked to a correct identified 5S 1T 8C5  12C1 (= 672) scenario. 4S 2H 8C4  5C2 (= 700) 4S 1H 1T 8C4  5C1  12C1 (= 4200) B1 2 correct identified outcomes, accept un-simplified. 3S 2H 1T 8C3  5C2  12C1 (= 6720) Identification can be implied by un-simplified expressions. 2nd and 3rd can be combined as 8C5  17C1 and this counts as 2 M1 Add values of 6 correct scenarios, no incorrect/repeated scenarios, scenarios for the second M1 no extra scenarios. Identification can be implied by un-simplified expressions. Total = 12600 A1 If either or both Ms not awarded, SCB1 for 12600 WWW. 4 5(b) Method 1 Total arrangements – arrangements with 3 hurdlers together [With 3 hurdlers not together ] 8! − (6!×3!) B1 6!  3! seen as a term added or subtracted. M1 8! − ( 6!  k ) , k  1 , k is an integer. = 36000 A1 Method 2 Summing no of ways with 2 hurdlers together or all separate 6C 3  3!  5! + 6C 2 2 3!  5! B1 6Cx × 5! or 6Px × 5! seen, x = 2 or 3 seen.  = 14400 + 21600  M1 ( 6C 3  5! + 6C 2  2  5! ) k , or 6 P3 +5! 6 P2  5! k k  1, k is an integer. = 36000 A1 3

More questions on Permutations and combinations

6 A large number of runners took part in two charity runs to raise money for a new community centre. In the first run, the times to complete the run were normally distributed with mean 46.3 seconds and standard deviation 6.4 seconds. (a) Find the probability that a randomly chosen runner took more than 55.1 seconds to complete the run. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ In the second run, the times to complete the run were normally distributed with mean 39.8 seconds and standard deviation v seconds. 10% of the runners took more than 48.6 seconds. (b) Find the value of v. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 150 runners are chosen at random from those who took part in the second run. (c) Use an approximation to find the probability that fewer than 20 of the 150 runners took more than 48.6 seconds to complete the run. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 6(a) 55.1 − 46.3 M1 Use of ± standardisation formula with 55.1, 46.3 and 6.4 [P(X > 55.1)] = P( Z  ) [= P(Z > 1.375)] substituted appropriately, no continuity correction, allow 6.4 2 6.4 2 or  6.4 . ( ) ( ) [= 1 − Φ (1.375 ) ] M1 Calculating the appropriate probability area (leading to the final answer, expect <0.5). = 1 − 0.9155 = 0.0845 A1 0.0845 ⩽ z ⩽ 0.0846 If one or both M marks not awarded, SCB1 for 0.0845 ⩽ z ⩽ 0.0846. 3 6(b)  48.6 − 39.8  B1 ±1.282 seen, CAO, critical value. [P(X > 4.86) = P  Z   = 0.10,     48.6 − 39.8  P  Z   = 0.90]    48.6 − 39.8 M1 ± standardisation formula with 48.6, 39.8 and σ equating to a z- = 1.282  value (not 0.10, 0.90, 0.5398,0.4602, 0.8159, 0.1841, 1 – their z- value), not ,2 condone continuity correction of ±0.05.  8.8  A1 AWRT. Not dependent on B mark being awarded. Condone 6.87.  = = 6.86    1.282  3 6(c)  Mean = 0.1  150 = 15 B1 15 and 13.5 seen, allow un-simplified. May be seen in standardisation formula.  Variance = 150  0.1  0.9 = 13.5 3 6 (  = 13.5, 3.674   3.675 implies correct variance. 2 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.  19.5 − 15  M1 Using 20 (with or without a cc) and their mean and sd found using [P(X < 20) =] P  Z   [= P(Z < 1.2247)] 2 150 in ± standardisation formula, not .  13.5  M1 Using continuity correction 19.5 or 20.5 in a standardisation formula. 4.5 4.5 Note: or gains M2. 13.5 3.674 = 0.8897 M1 Appropriate area Φ, from final process. Must be a probability.   Φ ( 1.2247 )  Note: correct final answer implies this M1. = 0.890 A1 AWRT. 5

More questions on The normal distribution