Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 5 · Variant 1

1 Nicola throws an ordinary fair six-sided dice. The random variable X is the number of throws that she takes to obtain a 6. (a) Find P ( X 1 8) . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the probability that Nicola obtains a 6 for the second time on her 8th throw. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(b) 6 2 M1 6 2  5   1   5   1    7   d d an integer ≥ 1, no inappropriate addition.          6   6   6   6  0.0651 A1 0.0651 ⩽ p < 0.06512. 2

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2 The random variable X takes the values -2, -1, 0, 2, 3. It is given that P ( X = x) = k ( x 2 + 2 ) , where k is a positive constant. (a) Draw up the probability distribution table for X, giving the probabilities as numerical fractions. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the value of Var( X ). [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) [Probs 6k , 3k , 2k , 6k , 11k so 28k = 1, ] B1 k must be identified 1 k = 28 M1 Table with correct outcomes and 2 correct probabilities. x –2 –1 0 2 3 FT substituting their k correctly into formula, with 0 < p < 1. No additional x values unless probability 0. P(X = x) 6 3 2 6 11 6 k 28 28 28 28 28 Condone in terms of k of the form or 6k. 28 0.2143 0.1071 0.07143 0.2143 0.3929 A1 Fully correct. Decimal answers to at least 3 sig figures, condone not summing exactly to 1. 3 2(b)  6 3  2  6 11  M1 Accept unsimplified expression. May be calculated in the variance. FT their table with 5 probabilities 0 < p < 1 that E(X) =  −2 + −1 +  0   + 2  + 3    28 28  28  28 28  sum to 1. 1  15  ( −12 −+3 12 + 33) =   28  14  2 2 2 2 M1 Appropriate variance formula using their (E(X))2 value. FT 6 −( 2 ) + 3 −( 1) + 6  2 + 11 3 Var(X) = their table with at least 4 probabilities 0 < p < 1, that may not 28 sum to 1. 15  2 −their    14  41 A1 825 = 4.21, 4196 Condone 196 . If one or both M marks not awarded, SC B1 for correct answer WWW. 3

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3 The time taken, in minutes, to walk to school was recorded for 200 pupils at a certain school. These times are summarised in the following table. Time taken t G 15 t G 25 t G 30 t G 40 t G 50 t G 70 (t minutes) Cumulative 18 46 88 140 176 200 frequency (a) Draw a cumulative frequency graph to illustrate the data. [2] (b) Use your graph to estimate the median and the interquartile range of the data. 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(c) Calculate an estimate for the mean value of the times taken by the 200 pupils to walk to school. 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Mark scheme: 3(a) M1 At least 4 points plotted within tolerance at upper bounds. Linear cf scale 0 ⩽ cf ⩽ 200 and linear time scale 0 ⩽ t ⩽ 70, with at least 3 values identified on each. Minimum scale uses at least ½ the grid. A1 All points plotted correctly. Curve drawn and joined to (0, 0). Axes labelled cumulative frequency (cf), time (t) and minutes (min) – or a suitable title. 2 3(b) Median = 33 B1 FT Must be identified. Evidence of use of graph must be seen. Strict FT ± ½ square on time axis. [IQR = ] 42 – 26 M1 41 ⩽ UQ ⩽ 43 − 25 < LQ ⩽ 27 . If outside of range FT ± ½ square on time axis. 16 A1 FT 3 3(c) B1 At least 5 correct midpoints Midpoint 7.5 20 27.5 35 45 60 or 5 correct frequencies seen. Frequency 18 28 42 52 36 24 18  7.5 + 28  20 + 42  27.5 + 52  35 + 36  45 + 24  60 M1 Correct mean formula using their 6 midpoints (must be Mean = within class, not upper bound, not lower bound) condone 1 200 error and their 6 frequencies (not cumulative frequencies). 13 A1 673 = 33.65, 33 Accept 33.7, not . 20 20 3

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4 Rahul has two bags, X and Y. Bag X contains 4 red marbles and 2 blue marbles. Bag Y contains 3 red marbles and 4 blue marbles. Rahul also has a coin which is biased so that the probability of obtaining a head when it is thrown is 1. 4 Rahul throws the coin. • If he obtains a head, he chooses at random a marble from bag X. He notes the colour and replaces the marble in bag X. He then chooses at random a second marble from bag X. • If he obtains a tail, he chooses at random a marble from bag Y. He notes the colour and discards the marble. He then chooses at random a second marble from bag Y. (a) Find the probability that the two marbles that Rahul chooses are the same colour. 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(b) Find the probability that the two marbles that Rahul chooses are both from bag Y given that both marbles are blue. 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Mark scheme: 4(a) 1 4 4 16 4 B1 2 clearly identified unsimplified probabilities from P(HRR), P(HRR) =   = , P(TRR), P(HBB) and P(TBB) correct. 4 6 6 144 36 3 3 2 18 3 P(TRR) =   = , 4 7 6 168 28 1 2 2 4 1 P(HBB) =   = , 4 6 6 144 36 3 4 3 36 6 3 P(TBB) =   = , , 4 7 6 168 28 14 4 3 1 6 M1 Sum of 4 correct scenarios, may be identified by the + + + unsimplified probability calculations. 36 28 36 28 29 A1 (0.460317… to at least 3SF). = or 0.460 63 3 4(b)  3 4 3  M1 3 4 3 36 3     , oe, , 0.2142857  seen as numerator of a  P ( T  BB )   4 7 6  4 7 6 168 14  P ( T|BB ) =  =   fraction, accept unsimplified, FT their P(TBB) from 4(a)  P ( BB )   1 + 6   36 28  M1 1 6 their + their FT from 4(a) or correct, 0.24206…, 3 36 28 3 61 seen as denominator of a fraction, accept unsimplified. 14  or 14 252 61 252 54 A1 Accept 0.8852589… rounded to at least 3SF. or 0.885 If one or both Ms not awarded, SC B1 for correct final 61 answer WWW. 3

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5 The weights of the green apples sold by a shop are normally distributed with mean 90 grams and standard deviation 8 grams. (a) Find the probability that a randomly chosen green apple weighs between 83 grams and 95 grams. 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(b) The shop also sells red apples. 60% of the red apples sold by the shop weigh more than 80 grams. 160 red apples are chosen at random from the shop. Use a suitable approximation to find the probability that fewer than 105 of the chosen red apples weigh more than 80 grams. 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Mark scheme: 5(a) 83 − 90 95 − 90 M1 Using ± standardisation formula with 90, 8 and either 83 or P(83 < X < 95) = P(  Z  ) 2 95. Not  , not σ, no continuity correction. 8 8 = P( −0.875  Z  0.625) A1 Both ±0.875 OE and ±0.625 OE seen. If M0 scored, SC B1 for both ±0.875 and ±0.625 seen M1 Calculating the appropriate probability area, leading to their [ Φ ( 0.625 ) + Φ ( 0.875 ) − 1 ] final probability. Expect final answer > 0.5. = 0.7340 + 0.8092 – 1 = 0.543 A1 0.5432, 0.543 ⩽ p < 0.5435. Only dependent on the 2nd M mark. 4 5(b) [Mean =160 0.6 =]96 B1 96 and 38.4 seen, allow unsimplified. May be seen in the standardisation formula. [Var =160 0.6  0.4 =]38.4  8 15   , 6.19677  to at least 4 SF   5     implies correct variance  Withold mark if variance clearly identified as standard deviation, condone N(96, 38.4 ) if standardisation formula correct or variance/standard deviation correctly stated as well. 104.5 − M1 Substituting their 96 and their 38.4 into the ±standardising P(X < 105) = P( Z  96) formula (any number for 104.5), condone σ2 or √σ. 38.4 M1 Use continuity correction 104.5 or 105.5 in their [P( Z  1.372) = Φ (1.372 ) ] standardisation formula.  8.5 8.5 Note: or seen gains M2 BOD. 38.4 6.197 M1 Appropriate area Φ, from final process, must be a probability. Expect final answer > 0.5. = 0.915[0] A1 0.9149 ⩽ p ⩽ 0.915. If one or more M marks not scored, SC B1 for 0.9149 ⩽ p ⩽ 0.915.

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6 The heights of the female students at Breven college are normally distributed: • 90% of the female students have heights less than 182.7 cm. • 40% of the female students have heights less than 162.5 cm. (a) Find the mean and the standard deviation of the heights of the female students at Breven college. 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Ten female students are chosen at random from those at Breven college. (b) Find the probability that fewer than 8 of these 10 students have heights more than 162.5 cm. 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Mark scheme: 6(a) 182.7 −  B1 1.282 or – 1.282 seen, CAO (critical value). = 1.282  B1 −0.2535 < z ⩽ −0.253 or 0.253 ⩽ z < 0.2535 seen. 162.5 −  = −0.253  M1 One standardisation formula, not σ2, or √σ, with 182.7 or 162.5 substituted correctly equated to a z value (not 0.9, 0.1, 0.8159, 0.5398, 0.4, 0.6, 0.6554, 0.7257, …). Solve, obtaining values for and σ 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. = 165.8, σ = 13.2 A1 Answers must be to at least 1 DP (context). 5 6(b) Method 1 8 2 M1 x 10 − x [P(X < 8) = 1 – P(8, 9, 10) =] 1 – (10C8 ( 0.6 ) ( 0.4 ) + 10C9 One term 10Cx ( p ) (1 − p ) . With 0  p  1, x  0 or 10. ( 0.6 )9 ( 0.4 )1 + ( 0.6 )10 ) A1 Correct unsimplified expression. Allow 10 for 10C9. Condone omission of last bracket only. [= 1 – (0.12093 + 0.040311 + 0.0060466)] If both brackets omitted in unsimplified expression allow recovery for final stated calculation of 1 – 0.1673 or final answer WRT 0.8327. = 0.833 B1 0.8327 < p ⩽ 0.833. Method 2 ( 0.4 )10 +10C1 ( 0.6 )1 ( 0.4 )9 +10C2 ( 0.6 ) 2 ( 0.4 )8 +10C3 ( 0.6 )3 ( 0.4 )7 + M1 One term 10Cx ( p ) x (1 − p )10 − x . With 0  p  1, x  0 or 10. 10C4 ( 0.6 ) 4 ( 0.4 )6 +10C5 ( 0.6 )5 ( 0.4 )5 + 10C6 ( 0.6 )6 ( 0.4 ) 4 + 10C7 A1 Correct unsimplified expression. ( 0.6 )7 ( 0.4 )3 1.0486  10 −4 + 1.5729 10 −3 ++ 0.21499    = 0.833 B1 0.8327 < p ⩽ 0.833. 3

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7 (a) How many different arrangements are there of the 9 letters in the word INTELLECT in which the two Ts are together? 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(b) How many different arrangements are there of the 9 letters in the word INTELLECT in which there is a T at each end and the two Es are not next to each other? 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Four letters are selected at random from the 9 letters in the word INTELLECT. (c) Find the percentage of the possible selections which contain at least one E and exactly one T. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) 8! M1 k ! k = 7 or 8, m = 1, 2. 2!2! 2! m ! = 10080 A1 7(b) Method 1 Number of ways with no restriction on Es – ways with Es together 7! 6! M1 7! − – r, r integer > 1. 2!2! 2! 2!2! [= 1260 – 360] M1 6! s − , s integer > 360. 2! = 900 A1 Method 2 T ^ ^ ^ ^ ^ T with Es inserted in gaps 5! 6  5 5! 6 M1 6  5 6  or  C 2 t  or t  C2 , t an integer > 1. 2! 2 2! 2 [= 60 × 15] M1 5!  u , u an integer > 1. 2! =900 A1 3 7(c) Method 1 – addition T E _ _ = 2C1 2C1  5C2 = 40 B1 Either identified or correct unsimplified expression, either alone or in an addition. T E E _ = 2C1 2C2  5C1 = 10 B1 Either identified or correct unsimplified expression, either alone or in an addition. M1 a ( 40 + 10 ) , a an integer < 126. Probability 9 9 C 4 C 4 Denominator value must be seen as 9 C 4 somewhere.  50  A1 39.68 ⩽ percentage ⩽ 39.7. Percentage =  100 = 39.7%    126  Method 2 – subtraction (total arrangements with 1 T – number of arrangements with 1T 0 E) T ^ ^ ^ = 2C1  7C3 = 70 B1 Either identified or correct unsimplified expression, either alone or in a subtraction. T * * * = 2C1  5C3 = 20 B1 Either identified or correct unsimplified expression, either alone or in a subtraction. M1 a ( 70 − 20 ) , a an integer < 126. Probability 9 9 C 4 C 4 Denominator value must be seen as 9 C 4 somewhere.  50  A1 39.68 ⩽ percentage ⩽ 39.7. Percentage =  100 = 39.7%    126  4

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