Cambridge A Level Mathematics 9709 — 2025 Feb/March Paper 5 · Variant 2

1 Jacob throws three coins at the same time. The first coin is biased so that the probability of obtaining a head when it is thrown is 1. 3 The second coin is biased so that the probability of obtaining a head when it is thrown is 1. 4 The third coin is biased so that the probability of obtaining a head when it is thrown is 1. 5 The random variable X is the number of heads obtained. (a) Show that P ( X = 2) = 3 . 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(b) Draw up the probability distribution table for X. 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(c) Given that E( X ) = 47 , find Var ( X ) . 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Mark scheme: Question Answer Marks Guidance 1(a) 1 1 4 1 3 1 2 1 1  9  3 B1 Order of coins must be consistent with question if not   +   +   = = AG   identified. 3 4 5 3 4 5 3 4 5  60  20 1 1(b) B1 Table with correct values of x and at least two correct non- x 0 1 2 3 zero probabilities. P(X = x) 24 8 2 26 13 9 3 1 , , , , , , , 60 20 5 60 30 60 20 60 B1 One more correct non-zero probability linked with correct 0.4 0.433 0.15 0.0167 x value, need not be in table if clearly identified, accept unsimplified (total of 3 correct probabilities). B1 4 correct probabilities linked with the correct outcomes, may not be in table. Decimals correct to at least 3SF. SC1 for 4 or more probabilities summing to 1 placed in a probability distribution table. 3 1(c) 2 2 2 2 2 M1 Appropriate variance formula using (E(X))2 value. FT (0  24+ ) 1  26 + 2 +9 3  1  47  [Var(X) =] −  their table with 3 or more probabilities (0 < p < 1) which 60  60  need not sum to 1, with an expression no more evaluated 1  26 + 4 +9 9  1  47  2 than in bold. FT acceptable at the bold partially evaluated = −  60  60  stage with their probabilities. 2051 A1 0.5695 < Var(X) ⩽ 0.570. = , 0.570 3600 If M0 scored, SC1 for 2051, 0.570 WWW 3600 Note: 0.57 without more accurate previous value penalised as 2SF. 2

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2 Last year, an online store sold a large number of computers. 55% of the computers were made by company F, 30% were made by company G and 15% were made by company H. A random sample of 3 customers who each bought a computer from this store is chosen. (a) Find the probability that the 3 customers bought computers all made by different companies. 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A random sample of 12 customers who each bought a computer from this store is chosen. (b) Find the probability that fewer than 10 of these customers bought a computer made by company F. 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A random sample of 140 customers who each bought a computer from this store is chosen. (c) Use a suitable approximation to find the probability that more than 24 of these customers bought a computer made by company H. 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Mark scheme: 2(a) 297 B1 148500  0.55  0.3  0.15  3! =  0.1485, Accept , condone 0.149. 2000 1000000 1 2(b) Method 1 [1 – P(10, 11, 12) = ] M1 x 12 − x One term of the form 12Cx ( p ) (1 − p ) , 1 – {12C10 0.5510 0.452 + 12C11 0.5511 0.45 + 0.5512} = 0 < p < 1, x ≠ 0 or 12. [1 – (0.0338529 + 0.0075229 + 0.0007662) =] A1 Correct unsimplified expression, no terms omitted leading to final answer. Condone omission of last bracket ‘}’ only. = 0.958 B1 0.9575 < p ≤ 0.958. Method 2 [P(0,1,2,3,4,5,6,7,8,9) =] M1 x 12 − x One term of the form 12Cx ( p ) (1 − p ) , 0 < p < 1, x ≠ 0 0.4512 +12C1 0.551 0.4511 + … + 12C9 0.5590.453 or 12. A1 Correct unsimplified expression, no more than 7 ‘middle’ terms omitted leading to final answer. = 0.958 B1 0.9575 < p ⩽ 0.958. 3 2(c) [Mean = 140 0.15 =] 21 B1 17 21 and 17.85 (or 17 ) seen, allow unsimplified. [Variance = 14 0  0.15  0.85 =]17.85 20 May be in standardisation formula. ( = 17.85, 4.224926  to at least 4SF implies correct variance). Withhold mark if variance clearly identified as standard deviation, condone N(21, 17.85 ) if standardisation formula correct or variance/standard deviation correctly stated as well.  24.5 − 21  M1 Substituting their µ and their σ into the ± standardisation P(X 24) = P  Z   formula (any number for 24.5), allow σ2 or √σ.  17.85  M1 Use continuity correction 23.5 or 24.5 in their standardisation formula.  3.5   3.5  Note: If no working    or   seen gains  17.85   4.225  M2 BOD. [P( Z  0.8284 ) = 1 − Φ ( 0.8284 ) ] M1 Appropriate area Φ, from final process, must be a probability. 1 – 0.7961 May be implied by a sketch of the required probability area. Note: correct final answer implies this M1. Expect final answer < 0.5. = 0.204 A1 Final answer AWRT. 5

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3 The lengths of 250 leaves of a certain type of plant are measured, correct to the nearest centimetre. The results are summarised in the table below. Length (cm) 5 – 9 10 – 14 15 – 19 20 – 24 25 – 29 30 – 39 Frequency 18 28 60 72 48 24 (a) On the grid below, draw a cumulative frequency graph to illustrate this information. [4] (b) 38% of these leaves are of length k cm or more. Use your graph to find an estimate for k. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Calculate an estimate for the mean length of these 250 leaves. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) B1 At least 4 of 46, 106, 178, 226, 250 cumulative < 9.5 < 14.5 < 19.5 < 24.5 < 29.5 < 39.5 frequencies correct. May be implied by accurate plotting if scale suitable. CF 18 46 106 178 226 250 May be by data table. B1 Linearly scaled axes correctly labelled cumulative frequency (cf) (from 0 to 250), length (oe) and cm (from 5 to 39.5) – or a suitable title, with at least 3 values identified on each. Axes can be the other way round. Axes must be more than 50% of grid. M1 At least 4 points correctly plotted at class upper end points (9.5, 14.5, 19.5, 24.5, 29.5, 39.5) on scaled axes. A1 All points plotted correct, curve drawn (within tolerance) and joined to (4.5,0) and not going above 250 vertically within range. A0 if straight line segments used. 4 3(b) [250 × 0.62 = 155] M1 Clear indication of use of graph at cf 155 is required. Line drawn from 155 on cf axis to meet graph at l = 23 A1FT Must be an increasing cf graph. Expect an answer in the range 22.5 ⩽ l ⩽ 23.5 from correct graph. 2 3(c) Midpoints 7, 12, 17, 22, 27, 34.5 B1 At least 5 correct midpoints seen, may be unsimplified, may be in calculation, may be by data table. 7  18 + 12  28 + 17  60 + 22  72 + 27  48 + 34.5  24 M1 Correct unsimplified mean formula using their 6 Mean = midpoints (not upper bound, lower bound, upper limits, 250 lower limits, cw, fd, f or cf and must be within class) condone 1 error. 5190 If midpoints correct accept or 250 126 + 336 + 1020 + 1584 + 1296 + 828 . 250 = 20.76 A1 19 38 76 Accept 20 , 20 or 20 , not improper fraction 25 50 100 If M1 withheld, SC1 for 20.76 oe WWW. 3

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4 Eddie has 16 toy cars, of which 8 are white, 5 are black and 3 are silver. He places all the cars in a bag and selects three of them at random, without replacement. (a) Find the probability that all three cars are the same colour. 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(b) Find the probability that, when the 3 cars are selected, at least one car is white and at least one car is black. 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Mark scheme: 4(a) 8 7 6  336 1  8 C3 M1 1 outcome seen as the product of 3 fractions with P(WWW) =   = , or   16 8, 7, 6 or 5, 4, 3 or 3, 2, 1 as numerators 16 15 14  3360 10  C3 and 16, 15, 14 or 16, 16, 16 as denominators 5 4 3  60 1  5 C3 Or P(BBB) =   = , or   16 16 15 14  3360 56  C3 1 outcome correct in terms of combinations. 3 2 6 1  1  3 C3 M1 Sum of 3 correct identified scenarios (may be identified by   = , P(SSS) = or   16 16 15 14  3360 560  C3 correct unsimplified numerator values). 402 67 A1 0.1195 < p ⩽ 0.120. , ,0.120 3360 560 3 4(b) 3 cases to consider: WBW, WBB, WBS M1 One outcome seen as the product of 3 fractions with 8 5 7  840 1  correct numerators and n, (n - 1), (n - 2) only as P(WBW) =    3  = ,  denominator, where 8 ⩽ n ⩽ 16 (condone omission of ×3 16 15 14  3360 4  or ×6) 8 5 4  480 5  Or P(WBB) =    3  = ,  16 15 14  3360 42  1 outcome correct in terms of combinations with m C 3 as 8 5 3  720 3  denominator where 8 ⩽ m ⩽ 16. P(WBS) =    6  = ,  16 15 14  3360 14  Must be a probability, no additional ‘divisions’ leading to Or 8 5 final answer. C 2  C1  140  P(WBW) = = 16   C3  560  A1 1 identified outcome fully correct (accept unsimplified). 8 C1  5 C 2  80  P(WBB) = =  M1 Sum of 3 correctly identified scenarios (may be identified 16  C3  560  by correct unsimplified numerator values). 8 C1  5 C1  3 C1  120  P(WBS) = = 16   C3  560  2040 17 A1 2040 17 , ,0.607 If 1 or more M mark not scored, SC1 for , ,0.607 3360 28 3360 28 WWW. Method 2 1 – {P(WSS)+P(WWS)+P(WWW)+P(BSS)+P(BBS)+P(BBB)+P(SSS)} 8 3 2  144 3  M1 Two outcomes seen as the product of 3 fractions with P(WSS) =    3  = ,  correct numerators and n, (n - 1), (n - 2) only as 16 15 14  3360 70  denominator, where 8 ⩽ n ⩽ 16 (condone omission of ×3). 8 7 3  504 3  Attempt at 1 – p must be present. P(WWS) =    3  = ,  16 15 14  3360 20  Must be a probability, no additional ‘divisions’ leading to final answer. 8 7 6  336 1  P(WWW) =    = ,  16 15 14  3360 10  A1 2 identified outcomes fully correct (accept unsimplified). 4(b) 5 3 2  90 3  M1 1 – sum of 7 correctly identified scenarios (may be P(BSS) =    3  = ,  identified by correct unsimplified numerator values). 16 15 14  3360 112  5 4 3  180 3  P(BBS) =    3  = ,  16 15 14  3360 56  5 4 3  60 1  P(BBB) =    = ,  16 15 14  3360 56  3 2 1  6 1  P(SSS) =    = ,  16 15 14  3360 560  2040 17 A1 2040 17 , If 1 or more M mark not scored, SC1 for , ,0.607 3360 28 3360 28 WWW. 4

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5 The mass of peaches sold per day in a supermarket is normally distributed with mean 65.8 kg and standard deviation 9.6 kg. (a) Find the probability that the mass of peaches sold on any given day is between 56 kg and 75 kg. 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The mass of cherries sold per day in a supermarket is normally distributed with mean 72.4 kg and standard deviation v kg. It is known that on 10% of days less than 59.1 kg of cherries are sold. (b) Find the value of v. 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The supermarket is open 7 days a week. (c) Find the probability that, in a randomly chosen week, the first day on which less than 59.1 kg of cherries are sold is the fifth day of the week. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) Find the probability that, in a randomly chosen week, the first day on which less than 59.1 kg of cherries are sold is before the fifth day of the week. 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Mark scheme: 5(a) 56 − 65.8 75 − M1 Use of ±standardisation formula once with 65.8, 9.6 and [P(56 < X < 75) =] P(  Z  65.8) 2 9.6 9.6 either 56 or 75. No continuity correction, not , not . =  P ( − 1.0208  Z  0.9583)  [ Φ ( 0.9583 ) + Φ (1.0208 ) −]1 M1 Appropriate probability area Φ, from final process. Must be a probability. (expect > 0.5). = 0.8309 + 0.8463 – 1 or 0.8309 – (1 – 0.8463) or 0.8309 – 0.1537 or (0.8309 – 0.5) + (0.8463 – 0.5) or 0.3309 + 0.3463 = 0.677 A1 AWRT. If 1 or more M mark not awarded, SC1 for final answer 0.677 AWRT WWW. 3 5(b)   59.1 − 72.4   B1 1.282 or – 1.282 seen cao (critical value). P Z  = 0.10          M1 ±standardisation formula with 59.1, 72.4, σ equating to a z-value (not 0.1, 0.9, 0.5398, 0.4602, 0.8159, 0.1841). 59.1 − 72.4 = −1.282 Condone continuity correction of ±0.05, not σ2 and not √σ.  13.3 Condone  = −1.282 .  = 10.4 A1 AWRT. Signs must be consistent throughout. If M1 not awarded, SC1 = 10.4 WWW. 3 5(c)  ( 0.9 ) 4 ( 0.1) =  0.0656 1 B1   1 5(d) Method 1 4 M1 d [P(X < 5) =] 1 − ( 0.9 ) 1 − ( 0.9 ) d = 4, 5. = 0.344 A1 0.3439. Method 2 [P(X < 5) =] 0.1 + ( 0.1)( 0.9 ) + ( 0.1)( 0.9 ) 2 + ( 0.1)( 0.9 )3 M1 0.1 + ( 0.1)( 0.9 ) + ( 0.1)( 0.9 ) 2 + ( 0.1)( 0.9 )3  + ( 0.1)( 0.9 ) 4    or or 1 – ( ( 0.1)( 0.9 ) 4 + ( 0.1)( 0.9 )5 + ( 0.1)( 0.9 )6 + ( 0.9 )7 ) 1 – (  ( 0.1)( 0.9 ) 4 + ( 0.1)( 0.9 ) 5 + ( 0.1)( 0.9 ) 6 + ( 0.9 ) 7 ) .   = 0.344 A1 0.3439. 2

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6 Alissa has 10 different books from the series Squares and Circles. The books look similar except for their colour. There are 3 blue books, 2 red books, 2 yellow books, 1 orange book, 1 purple book and 1 green book. Alissa places the books in a row on her shelf. She is only interested in the arrangement of the colours. (a) How many different colour arrangements are there of the 10 books? 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(b) How many different colour arrangements are there of the 10 books in which the 3 blue books are together, but the 2 yellow books are not next to each other? 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(c) How many different colour arrangements are there of the 10 books with exactly 4 books between the 2 yellow books? 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Alissa selects 4 books from her 10 different books from the series Squares and Circles. (d) Find the number of different selections if the 4 books include at least 1 red book, at most 1 blue book and exactly 1 yellow book. 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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: 6(a)  10!  B1 CAO. = 151200    2!3!2!  1 6(b) Method 1 Total – Ys together 8! 7! M1 8! 7! − = − , a = 1, 2 b = 1, 2 c = 1, 3. 2!2! 2! 2!a !c ! a !b !c ! [10080 – 2520 =] 7560 A1 2 Method 2 3 Bs treated as a single unit ^ ^ ^ ^ BBB ^ and Ys inserted 6! 7 M1 6! 7 6! 7 6! 7 P2  2  C 2 or  P2 or 2! C d ! d ! d ! 2 or 6! 7  6 6! 6 + 5 + 4 + 3 + 2 + 1) 6! 7  6 or  or (  d ! e d ! 2! 2 d = 1, 2, 3 e = 1, 2. or 6! ( 6 + 5 + 4 + 3 + 2 + 1) 2! [360 × 21 =] 7560 A1 2 6(c) Method 1 8! *M1 8! 5 , b = 1,2, c = 1,3 and b ≠ c, d ≥ 1. 2!3! b !c !d DM1 Multiply by 5. 16 800 A1 Method 2 8 P4  5! *M1 8 P4 , b = 1,2, c = 1,3 and b ≠ c, d ≥ 1. 2!3! b !c !d DM1 Multiply by 5!. 16 800 A1 3 6(d) Method 1 Y R B _ 2 2 3C1  3C1 = 36 M1 One correct identified unsimplified expression Y R R B 2 1 3C1 = 6 (3C1 ≠ 3C2). Y R _ _ 2 2 3C2 = 12 Y R R _ 2 1 3C1 = 6 B1 Correct outcome/value for 2 clearly identified scenarios, accept unsimplified WWW. M1 Sum of 4 correct identified scenarios. 60 A1 Method 2 Y R _ _ 5C3 × 3C1 × 2C1 M1 5C3 seen with YR^^ identified. M1 5C3 × a, a = 2, 3, 6. B1 5C3 × 3C1 × 2C1 or 5C3 × 3 × 2. 60 A1 4

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