Discrete random variables — Paper 5 — A Level Mathematics 9709 topical questions

Question 1

In a certain large college, 22% of students own a car. (a) 3 students from the college are chosen at random. Find the probability that all …

Accepted Answer

2(a) 1 2(b) P(2, 3, 4) = 16C2 2 14 16 3 13 16 4 12 3 4 0.22 0.78 0.22 0.78 0.22 0.78 C C + + M1 0.179205 + 0.235877 + 0.216221 A1 0.631 A1 3

Question 2

The random variable X takes each of the values 1, 2, 3, 4 with probability 1 Two independent values 4. of X are chosen at random. If the tw…

Accepted Answer

4(a) y 1 2 3 4 prob 7 16 5 16 3 16 1 16 B1 1 2 3 4 1 1 1 2 3 2 1 2 1 2 3 2 1 3 1 4 3 2 1 4 Probability distribution table with correct scores with at least one probability, allow extra score values if probability of zero stated’ B1 One probability (linked with correct score) correct B1 2 more probs (linked with correct scores) correct B1 FT 4th prob correct, FT sum of 3 or 4 terms = 1 4 Question Answer Marks Guidance 4(b) P(2|even) = 5 16 6 16 M1 ( ) ( ) ( ) P 2 P 2 P 4 their their their + seen or correct outcome space. 5 6 or 0·833 A1 2

Question 3

A fair six-sided die, with faces marked 1, 2, 3, 4, 5, 6, is thrown repeatedly until a 4 is obtained. (a) Find the probability that obtaini…

Accepted Answer

1(a) 5 5 1 6   −    or 2 3 4 1 5 1 5 1 5 1 5 1 6 6 6 6 6 6 6 6 6       + × + × + × + ×             or p + pq + pq2+pq3+ pq4 (+ pq5) 0 < p < 1, p + q = 1, 0·598, 4651 7776 A1 2 1(b) (1 – P(0, 1, 2)) 1 – 10 5 6        +10C1 9 1 5 6 6   +       10C2 2 8 1 5 6 6              M1 10Cx ( ) 10 1 , − − x x p p 0 < p < 1, any p, x ≠ 0,10 1 – (0·1615056 + 0·3230111 + 0·290710) A1 Correct expression, accept unsimplified, condone omission of final bracket 0·225 A1 0·2247 < p ≤ 0·225, WWW 3

Question 4

An ordinary fair die is thrown until a 6 is obtained. (a) Find the probability that obtaining a 6 takes more than 8 throws. [2] ...........…

Accepted Answer

2(a) 8 5 6       0.233 A1 2 2(b) 36 B1 1 2(c) P(X =10) + P(X=11) = 9 10 35 1 35 1 36 36 36 36     +         M1 OE, unsimplified expression in form 9 10 + p q p q , p + q = 1, no × 0.0425 A1 2

Question 5

The 13 00 train from Jahor to Keman runs every day. The probability that the train arrives late in Keman is 0.35. (a) For a random sample o…

Accepted Answer

4(a) 7 0.65 + 7C1 6 1 0.65 0.35 + 7C2 5 2 0.65 0.35 M1 Binomial term of form 7Cx ( ) 7 1 , − − x x p p 0 < p < 1, any p, x ≠ 0, 7 0.049022 + 0.184776 + 0.29848 A1 Correct unsimplified answer 0.532 A1 3 4(b) Mean = 142 0.35 49.7 × = Variance = 142 0.35 0.65 32.305 × × = B1 Correct unsimplified np and npq (condone σ = 5.684 evaluated) P(X > 40) = P( 40.5 49.7) 32.305 − > z M1 Substituting their µ and σ (no 2 or σ σ ) into ±standardisation formula with a numerical value for '40.5' P( 1.619) > − z M1 Using either 40.5 or 39.5 within a ±standardisation formula M1 Appropriate area Φ , from standardisation formula P(z >…) in final solution, must be probability 0.947 A1 Correct final answer 5

Question 6

Every day Richard takes a ﬂight between Astan and Bejin. On any day, the probability that the ﬂight arrives early is 0.15, the probability …

Accepted Answer

5(a) ( ) 3 [ 0.7 ] 0.343 = Alternative method for Question 5(a) [(0.15)3 + 3C1(0.15)2(0.55) + 3C2(0.15)(0.55)2 + (0.55)3 =] 0.343 B1 Evaluated WWW 1 5(b) 1 – (0.859 + 9C1 0.151 0.858 + 9C2 0.152 0.857) [1 – (0.231617 + 0.367862 + 0.259667)] M1 One term: 9Cx px (1 – p)9-x for 0 < x < 9, any 0 < p < 1 A1 Correct expression, accept unsimplified. 0.141 A1 0.1408 ⩽ ans ⩽ 0.141, award at most accurate value. Alternative method for Question 5(b) 9C3 0.153 0.856 + 9C4 0.154 0.855 + 9C5 0.155 0.854 + 9C6 0.156 0.853 + 9C7 0.157 0.852 + 9C8 0.158 0.85 + 0.159 M1 One term: 9Cx px (1 – p)9-x for 0 < x < 9, any 0 < p < 1 A1 Correct expression, accept unsimplified. 0.141 A1 0.1408 ⩽ ans ⩽ 0.141, award at most accurate value. 3 Question Answer Marks Guidance 5(c) Mean [ ] 60 0.15 9 = × = Variance [ ] 60 0.15 0.85 7.65 = × × = B1 Correct mean and variance, allow unsimplified. (2.765 ≤ σ ≤ 2.77 imply correct variance) ( ) 11.5 9 12 7.65 −    ≥ = >       X P Z M1 Substituting their mean and variance into ±standardisation formula (any number for 11.5), not σ2 or √ σ M1 Using continuity correction 11.5 or 12.5 in their standardisation formula. ( ) 1 0.9039 1 0.8169 −Φ = − M1 Appropriate area Φ, from final process, must be probability. 0.183 A1 Final AWRT 5

Question 7

In a large college, 28% of the students do not play any musical instrument, 52% play exactly one musical instrument and the remainder play …

Accepted Answer

5(a) [P(10, 11, 12) =] 12C10 10 2 0.72 0.28 + 12C11 11 1 0.72 0.28 + 12C12 12 0 0.72 0.28 M1 One term 12Cx 12 1 x x p p   , for 0 < x < 12, 0 < p < 1. = 0.193725 + 0.0905726 + 0.0194084 A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.304 B1 Final answer 0.3036 < p ⩽ 0.304. Alternative method for question 5(a) [1 – P(0,1,2,3,4,5,6,7,8,9) =] 1 – (12C0 0 12 0.72 0.28 + 12C1 1 11 0.72 0.28 + 12C2 2 10 0.72 0.28 + 12C3 3 9 0.72 0.28 + 12C4 4 8 0.72 0.28 + 12C5 5 7 0.72 0.28 + 12C6 6 6 0.72 0.28 + 12C7 7 5 0.72 0.28 + 12C8 8 4 0.72 0.28 + 12C9 9 3 0.72 0.28 ) M1 One term 12Cx   12 1 x x p p   , for 0 < x < 12, 0 < p < 1. A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.304 B1 Final answer 0.3036 < p ⩽ 0.304. 3 5(b) Mean = [   0.52 90 46.8, var 0.52 0.48 90] 22.464       B1 46.8 and 22.464 or 22.46 seen, allow unsimplified, (4.739 < σ ⩽ 4.740 imply correct variance). [P(X < 40) =] P 39.5 46.8 22.464 z         M1 Substituting their mean and their variance into ±standardisation formula (any number for 39.5), not σ2, √ σ. M1 Using continuity correction 39.5 or 40.5 in their standardisation formula. = [P( 1.540)] 1 0. Z   9382 M1 Appropriate area Φ, from final process, must be probability. 0.0618 A1 0.06175 ⩽ p ⩽ 0.0618 5

Question 8

The times taken, in minutes, to complete a word processing task by 250 employees at a particular company are summarised in the table. Time …

Accepted Answer

4(a) Cw 20 20 10 10 40 M1 At least 4 frequency densities calculated Fd 1.6 2.3 9.6 5.2 0.6 f 32  f  eg  condone if unsimplified  , accept unsimplified, cw 20  cw  0.5  may be read from graph using their scale no lower than 1 cm = fd 1 A1 All bar heights correct on graph, using their suitable linear scale with at least 3 values indicated, no lower than 1 cm = fd 2. B1 Bar ends at [0,] 20, 40, 50, 60, 100 (at axis), 5 bars drawn 0 ⩽ time axis ⩽ 100, linear scale with at least 3 values indicated. B1 Axes labelled frequency density (fd), time (t) and minutes (mins, m) or appropriate title. (Axes may be reversed). 4 4(b) Midpoints 10 30 45 55 80 B1 At least 4 correct midpoints seen (check data table). [Mean = 43.2 given] M1 Appropriate variance formula with their 5 midpoints (not upper 32  10 2 + 46  30 2 + 96  45 2 + 52  55 2 + 24  80 2 2 bound, lower bound, class width, frequency density, frequency or [Var =] − 43.2 cumulative frequency). 250 Condone 1 frequency error. Or 2 2 2 If correct midpoints seen accept 32 (10 − 43.2 ) + 46 ( 30 − 43.2 ) + 96 ( 45 − 43.2 )  3200 + 41400 + 194400 + 157300 + 153600 549900  2 2  or  +52 ( 55 − 43.2 ) + 24 ( 80 − 43.2 )  250 250  250 −{43.2 2 or 1866.24} .  549900 2  A1 www, final answer 18.25814887 to at least 3SF. = − 43.2 = 333.36   If M0 earned SC B1 for final answer 18.25814887 to at least 3SF.  250  Sd = 18.3 3 5(a) Method 1: Scenarios identified ignoring unbiased coin 1 3 3 M1 All 3 different calculations seen unsimplified. P(BH1 BT2) =  = 4 4 16 3 1 3 P(BT1 BH2) =  = 4 4 16 1 1 1 P(BH1 BH2) =  = 4 4 16 3 3 1 7 A1 Clear identification of all scenarios, linked probabilities and sum. + + = AG 16 16 16 16 5(a) Method 2: Scenarios identified with all 3 coins 1 1 3 3 M1 All 6 different calculations seen unsimplified. P(H BH1 BT2) =   = 2 4 4 32 1 1 3 3 P(T BH1 BT2) =   = 2 4 4 32 1 3 1 3 P(H BT1 BH2) =   = 2 4 4 32 1 3 1 3 P(T BT1 BH2) =   = 2 4 4 32 1 1 1 1 P(H BH1 BH2) =   = 2 4 4 32 1 1 1 1 P(T BH1 BH2) =   = 2 4 4 32 1 + 3 + 3 + 1 + 3 + 3 14 7 A1 Clear identification of all scenarios, linked probabilities and sum. P(B) = = = AG 32 32 16 Method 3: 1- P(BT1 BT2) ignoring unbiased coin 2 M1 Calculation seen unsimplified  3  1 – P(BT1 BT2) = 1 −  and 1 – probability seen.  4  7 A1 Clear identification of scenario used, linked probability and = calculation. AG 16

Question 9

George has a fair 5-sided spinner with sides labelled 1, 2, 3, 4, 5. He spins the spinner and notes the number on the side on which the spi…

Accepted Answer

2(a) Method 1: [P(5) = 0.2] M1 1 – 0.8n, n = 6, 7. [P(X < 7) =] 1 − 0.86 11529 A1 0.737856 to at least 3SF. = 0.738, 15625 Method 2: [P(X < 7) =] 0.2 + 0.2  0.8 + 0.2  0.82 + 0.2  0.83 + 0.2  0.8 4 + 0.2  0.85 M1 0.2 + 0.2  0.8 + 0.2  0.8 2 + 0.2  0.83 + 0.2  0.8 4 + 0.2  0.85 +0.2  0.86 ( ) 11529 A1 0.737856 to at least 3SF. = 0.738, 15625 2 2(b) Method 1: [P(5, 6, 7) = ] M1 One term: 5 5 6 4 7 3 x 10 − x 10C5 ( 0.2 ) ( 0.8 ) +10C6 ( 0.2 ) ( 0.8 ) + 10C7 ( 0.2 ) ( 0.8 ) 10Cx ( p ) (1 − p ) , 0  p  1, x  0,10. [0.02642 + 5.505 × 10-3 + 7.864 × 10-4] A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. = 0.0327 B1 awrt Method 2: [P(X < 8) – P(X ⩽ 4) = 1 – P(X ⩾ 8) – P(X ⩽ 4) =] M1 One term: x 10 − x 10Cx ( p ) (1 − p ) , 0  p  1, x  0,10. 1 – {10C8(0.2)8(0.8)2 + 10C9(0.2)90.8 + (0.2)10} – {(0.8)10 +10C1(0.2)(0.8)9 +10C2(0.2)2(0.8)8 +10C3(0.2)3(0.8)7 +10C4(0.2)4(0.8)6} A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. [1 − {7.373  10 −5 + 4.096  10 −6 + 1.024  10 −7 } − 0.1074 + 0.2684 + 0.3020 + 0.2013 + 0.08808 = 0.0327 B1 awrt 3

Question 10

Harry has three coins: • One coin is biased so that the probability of obtaining a head when it is thrown is 1.3 • The second coin is biase…

Accepted Answer

6(a) x 0 1 2 3 P(X = x) 24 60 26 60 9 60 1 60 2 5 13 30 3 20 1 60 0.4 0.433 0.15 0.0167 B1 Table with correct X values and at least one probability. Values need not be in order, lines may not be drawn, may be vertical, X and P(X) may be omitted. Condone any additional X values if probability stated as 0. B1 P(X = 1) or P(X = 2) correct and identified, need not be in table, accept unsimplified. B1 Two more correct and identified probabilities, need not be in table, accept unsimplified. B1 4 correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3sf. SC B1 for four probabilities summing to 1 placed in a probability distribution table with the correct x values. 4 Question Answer Marks Guidance 6(b) [P(Y = 0) =]   2 2 3 4 1 ; 3 4 5    p [P(Y = 5) =] 2 1 1 1 3 4 5   p B1 Either   2 , 2 3 4 1 3 4 5 p     not   2 2 1 5 ; p   or 2, 1 1 1 3 4 5 p    not 2 1 60 . p    2 2 2 3 4 1 1 1 1 6 3 4 5 3 4 5          p p   2 2 24 1 6        p p 2 3 8 4 0    p p M1 Equating and forming a 3 term quadratic equation. Their P(Y = 0) = 6 × their P(Y = 5). 2 3  p A1 Not dependent on B1. A0 if p = 2 seen and not clearly rejected. SC B1 if 2 3  p obtained from a correct quadratic with more than three terms. If p = 2 seen and not clearly rejected, SC B0. 3

Question 11

Jasmine has one $5 coin, two $2 coins and two $1 coins. She selects two of these coins at random. The random variable X is the total value,…

Accepted Answer

5(a) [$7 =] [$]5 + [$]2 [Probability =] 1 2 1 2 0.2 5 4 5     Or [Probability =] 0.2 × 0.5 × 2 = 0.2 B1 AG Must include [$7], 5, 2 and link the probabilities to the appropriate value 1 2 1 1 5 2 C C 0.2. C      1 2 2 1 1 2 1 2 , not 5 4 5 4 5 4 5 4       unless 5 and 2 and 2 and 5 seen in solution. If all possibilities identified (e.g. outcome table), must be clearly labelled and terms fulfilling the condition identified. 1 5(b) x 2 3 4 6 7 P(X = x) 0.1 0.4 0.1 0.2 0.2 B1 Table with correct x values and at least one further non-zero probability correct. Condone extra x values if probability stated as 0. B1 Two more correct non-zero probabilities linked with correct outcomes. Accept probabilities not in table if clearly identified. B1 All five probabilities correct. Accept probabilities not in table if clearly identified. SC B1 for four further non-zero probabilities adding to 0.8 if B1 max scored. 3 Question Answer Marks Guidance 5(c) [E(X) = 0.1 × 2 + 0.4 × 3 + 0.1 × 4 + 0.2 × 6 + 0.2 × 7] 0.2 + 1.2 + 0.4 + 1.2 + 1.4 [ = 4.4] M1 Accept unsimplified expression. May be calculated in the variance, FT their table with at least 5 probabilities, 0 < p < 1, that sum to 1. FT acceptable at the bold partially evaluated stage.   2 2 2 2 2 2 [Var 0.1 2 0.4 3 0.1 4 0.2 6 0.2 7 4.4            X 2 0.1 4 0.4 9 0.1 16 0.2 36 0.2 49 4.4         M1 Appropriate variance formula using their (E(X))2 value. FT their table with at least 4 probabilities, 0 < p < 1, that may not sum to 1. FT acceptable at the bold partially evaluated stage. Note: if table is correct, 22.6 – (4.42 or 19.36) implies this M1. = 3.24 A1 CAO 81 6 , 3 25 25 scores A0. Only dependent upon previous M1 (M0 M1 A1 possible). If M0 M0 scored, SC B1 for 3.24 WWW. 3

Question 12

The residents of Mahjing were asked to classify their local bus service: • 25% of residents classified their service as good. • 60% of resi…

Accepted Answer

6(a) Mean [=110 0.25] 27.5 Variance [=110 165 0.25 0.75] 20.625, 8    B1 27.5 and 20.625 (CAO) seen, allow unsimplified. May be in standardisation formula (4.541475… to at least 4sf or 165 330 or 8 4 implies correct variance). Penalise incorrect identification, condone no identification. P(X < 22) = P 21.5 27.5 20.625 Z         M1 Substituting their 27.5 and their 20.625 into the ± standardising formula (any number for 21.5), not 2,  not .  M1 Using continuity correction 21.5 or 22.5 in their standardisation formula. [P( 1.3212)  Z =   1 Φ 1.3212 ]  1 – 0.9068 = M1 Appropriate probability area, from final process, must be a probability. May be implied by a sketch of the required probability area. Expect final answer < 0.5. 0.0932 A1 0.0932 ⩽ p < 0.09325 If either M1 M1 not awarded for standardisation and/or M1 not awarded for finding probability area, SC B1 0.0932 ⩽ p < 0.09325 WWW. 5 Question Answer Marks Guidance 6(b) Method 1 [1 – P(8, 9, 10) = ] 1 – (10C8 0.858 0.152 + 10C9 0.8590.151 + 0.8510) [ = 1 – (0.275897 + 0.347425 + 0.196874)] M1 One term 10Cx   10 , 1 x x p p   0 1, 0 p x    or 10. A1 Correct unsimplified expression. Condone omission of last bracket only. = 0.180 B1 0.1795 < p ⩽ 0.180 Method 2 [P(0, 1, 2, 3, 4, 5, 6, 7) = ] 0.1510 + 10C1 0.85×0.159 + … + 10C7 0.8570.153 (M1) One term 10Cx   10 , 1 x x p p   0 1, 0 p x    or 10. (A1) Correct unsimplified expression. = 0.180 (B1) 0.1795 < p ⩽ 0.180 3 6(c) 0.25 0.6 0.15 6    M1 0.25 0.6 0.15 ,k    k an integer > 1. 0.135, 27 200 A1 2

Question 13

The numbers on the faces of a fair six-sided dice are 1, 2, 2, 3, 3, 3. The random variable X is the total score when the dice is rolled tw…

Accepted Answer

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

Question 14

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) D…

Accepted Answer

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

Question 15

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 pro…

Accepted Answer

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.

Question 16

30% of the residents of Wimfield own an electric car. Three residents are chosen at random. (a) Find the probability that either all three …

Accepted Answer

Question Answer Marks Guidance 1(a) 0.33 + 0.7 3 M1 p 3 + q 3, p + q = 1, p , q  0 or or 1 – (3  0.32  0.7 + 3  0.3  0.72 ) = 1 − 0.63 1 − (3  p 2  q + 3  p  q 2 ), p + q = 1, p, q  0 . 0.37[0] A1 37 . 100 2 1(b) [Mean = 125  0.3 =] 37.5 B1 37.5 or 37½ and 26.25, 26¼ seen, allow unsimplified. May be seen in [Variance = 125  0.3  0.7 =] 26.25 standardisation formula. (=  5.12, 105 implies correct 2 variance). 45.5 − 37.5 M1 Substituting their mean and their positive P(X > 45) = P( Z  ) standard deviation into the ±standardising 26.25 formula (any number for 45.5), not their σ2, not their . M1 Use continuity corrections 44.5 or 45.5 in their standardisation formula 8 8 Note: or seen gains M2 26.25 5.123 BOD [1 − Φ ( their 1 .5614 ) ] = 1 – their 0.9407 M1 Appropriate area Φ, from final process, must be a probability. Expect final answer < 0.5. Note: appropriate final answer implies this M1. 0.0593 A1 0.0592 ⩽ p ⩽ 0.0593. 5

Question 17

A red fair six-sided dice has faces labelled 1, 1, 1, 2, 2, 2. A blue fair six-sided dice has faces labelled 1, 1, 2, 2, 3, 3. Both dice ar…

Accepted Answer

2(a) B1 Table with correct x values and at least one correct probability linked with the correct x- x 1 2 3 4 6 value. Values need not be in order, lines may not 6 12 6 6 6 be drawn, may be vertical, P(X = x) 36 36 36 36 36 x and P(X) may be omitted. Condone any additional x values if probability stated as 0. 1 1 1 1 1 6 3 6 6 6 B1 4 correct probabilities linked with the correct x-values, need not be in table, 0.167 0.333 0.167 0.167 0.167 accept unsimplified. B1 5 correct probabilities linked with correct x- values, may not be in table. Decimals correct to at least 3 SF. SC B1 4 or 5 probabilities summing to 1 placed in a probability distribution table with 4 or 5 x-values between 1 and 6 inclusive. 3 2(b) 1 B1 FT FT their table with 4 or 5 probabilities (0 < [E(X) = ( 6 + 24 + 18 + 24 + 36 ) =] 3 p < 1) summing to 1. 36 1

Question 18

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 T…

Accepted Answer

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

Question 19

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…

Accepted Answer

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

Question 20

A bag contains 10 marbles, of which 4 are red and 6 are blue. Four marbles are selected from the bag at random, without replacement. The ra…

Accepted Answer

6(a) Method 1 6 5 4 3 4! 3 M1 AG. P(X = 2) =     = 10 9 8 7 2!2! 7 6 5 4 3 4 3 6 5     k or     k for k an integer, k > 1. 10 9 8 7 10 9 8 7 A1 4! 4C2 may be seen for 2!2!. Method 2 6 C 2  4 C 2  15  6  3 M1 AG. = = 6 4   10C4  210  7 C 2  C 2 seen as the numerator of a fraction. Condone use of permutations if used consistently. A1 2 6(b) B1 Table with correct values of x and at least one further non-zero probability correct. x 0 1 2 3 4 Condone extra x values if probability stated as 0. P(X = x) 1 4 3 8 1 , , , , B1 Third probability correct. 210 35 7 21 14 Accept probabilities not in table if clearly identified. 0.00476 0.114 0.381 0.0714 B1 Fourth probability correct. Accept probabilities not in table if clearly identified. B1 Fifth probability correct. Accept probabilities not in table if clearly identified. 4 SCB1 for 4 further non-zero probabilities adding to , 0.5712 if 7 B2 max scored. 4 6(c) 3 8 M1 3 8 + + their seen as the numerator of a fraction. 7 21 7 21 [P(2B, 3B | 3B1R or 2B2R or 1B 3R) =] 4 3 8 + + 35 7 21 B1 FT 4 3 8 their + + their seen as the denominator of a fraction. 35 7 21  17  A1 Accept 0.87628…to at least 3SF.  21 =  = 170 , 85 , 0.876 97 194 97    105  3

Question 21

In Millford, 70% of the residents own a bicycle. A random sample of 160 residents is selected. Use a suitable approximation to find the pro…

Accepted Answer

2 Mean = 160 0.7 = 112 B1 112 and 33.6 (CAO) seen, allow un-simplified. Variance = 160  0.7  0.3 = 33.6 May be in standardisation formula. ( = 33.6, 5.79655 to at least 4SF implies correct variance). Withold mark if variance clearly identified as standard deviation. Condone N 112, 33.6 if standardisation formula ( ) correct or variance/standard deviation stated correctly as well. 120.5 − M1 Substituting their 112 and their 33.6 into the P(X 120 ) = P( Z  112) ±standardising formula (any number for 120.5), 33.6 allow σ2 or √σ. M1 Use continuity correction 119.5 or 120.5 in their standardisation formula. Note: If no standardisation formula seen  8.5   8.5     or   scores M2 BOD.  33.6   5.797  [P( Z  1.466 ) = 1 − Φ (1.466 )] M1 Appropriate area Φ , from final process, must be a probability. = 1 – 0.9287 May be implied by a sketch of the required probability area. Note: correct final answer implies this M1. Expect final answer < 0.5. = 0.0713 final answer A1 Final answer AWRT. 5

Question 22

Two fair 6-sided dice with faces labelled 1, 2, 3, 4, 5, 6 are thrown. The two scores are noted. The random variable X is defined as follow…

Accepted Answer

Question Answer Marks Guidance 1(a) B1 Table with correct X values and at least one X 0 1 2 3 4 5 probability correct. Values need not be in order, lines may not be drawn, may be vertical, X and P(X=x) P(X=x) 6 10 8 6 4 2 may be omitted. 36 36 36 36 36 36 Condone any additional X values if probability stated as 0. 1 5 2 1 1 1 B1 Total of four correct probabilities linked with correct 6 18 9 6 9 18 outcomes, may not be in table. 0.167 0.278 0.222 0.167 0.111 0.056 B1 All six probabilities are correct and linked with correct outcomes, may not be in table. If decimals are used, condone correct rounding (which will not sum to 1) or one value rounded inaccurately to sum to 1. 3 SCB1 for six probs linked to X-values 0 – 5 summing to 1 with no more than 3 correct. 1(b)   0  6  + 10 +1 8  2 + 6 +3 4  4 + 2  5  10 + 16 + 18 + 16 + 10 M1 May be implied by use in Variance, accept un-  E ( X ) = =  simplified expression.  36  36 FT their table if their 4 or more non-zero probabilities sum to 1 or 0∙999. 2 2 2 2 2 M1 Appropriate variance formula using their (E(X))2 10 +1 8  2 + 6  3 + 4  4 + 2  5  their 35  [Var =] −  value. FT their table even if their 4 or more non-zero 36  18  probabilities not summing to 1  210 1225 665  210 2 = − = = 2.05   Note: If table is correct, − ( their E ( X ) ) is M1.  36 324 324  36 35 665 17 A1 OE. E(X) = , 1.94 , Var(X) = , 2 , 2.05 Answers for E(X) and Var(X) must be identified. 18 324 324 Accept Var = 2.052469… to 3sf or better. 3

Question 23

The random variable X takes the value x with probability kx2, where k is a constant and x takes the values - 2 , 1, 2, 3 only. (a) Draw up …

Accepted Answer

Question Answer Marks Guidance 1(a) 1 B1 Using sum of probabilities =1 to form an equation in k or value of 4 k + k + 4 k + 9 k = 1, k = k stated. 18 B1 Table with at least 2 correctly linked probabilities accurate. May x –2 1 2 3 be in terms of k. P(X = x) 4 1 4 9 X –2 1 2 3 18 18 18 18 P(X = x) 4k k 4k 9k Condone extra X values if probability stated as 0. B1 4 correctly linked probabilities accurate. May not be in a table. 3 1(b) B1FT FT 28 × their k or correct with 0  p .1 [E(X) = 28k =] 14, 1.56 9 1 1(c) 10 M1 their P (1) + P ( 3 ) . P ( X  2| X  0) = 18 their P (1) + P ( 2 ) + P ( 3 ) 14 May be in terms of k. With 0  p .1 18 10 5 If table correct, accept 18 or 9 . 14 7 18 9 5 A1 = , 0.714 7 2

Question 24

Kayla has a bag containing 3 red marbles, 1 blue marble and 2 green marbles. She selects one marble from the bag at random and does not rep…

Accepted Answer

2(a) B1 Table with correct x values and at least 1 correct probability. x 1 2 3 4 5 B1 A second correct probability correctly linked to the correct x value, P(X = x) 5 4 3 2 1 need not be in table, accept un-simplified. 15 15 15 15 15 B1 Two more correct probabilities correctly linked to the correct x 40 32 24 16 8 values, need not be in table, accept un-simplified. 120 120 120 120 120 B1 5 correct probabilities linked with correct x values. 1 1 SCB1 5 non-zero probabilities (not all 1/5) summing to 1 placed 3 5 in a probability distribution table with correct x values if B1B1 max scored. 0.3333 0.2667 0.2[000] 0.1333 0.06667 SCB2 for x 0 1 2 3 4 All decimals correct to at least 3SF P(X = x) 5 4 3 2 1 15 15 15 15 15 OE. 4 2(b) [E(X) =] M1 Accept un-simplified expression. May be calculated in variance. 5 4 3 2 1 1 8 3 8 1 1  + 2  + 3  + 4  + 5  Accept + + + + OE for the M mark. 15 15 15 15 15 3 15 5 15 3  5 + 8 + 9 + 8 + 5 35 7  FT their table with 5 or 6 probabilities summing to 1 (0 < p < 1). = ,    15 15 3  [Var(X) =] M1 Appropriate variance formula using their (E(X))2 value. 2 5 2 4 2 3 2 2 2 1 FT their table with 4 or more probabilities (0 < p < 1) which need 1  + 2  + 3  + 4  + 5  not sum to 1 or with an expression no more evaluated than shown 15 15 15 15 15 2 in bold.  35  − their   15   1  5 + 4 +4 9 +3 16 +2 25 1 49   −   15 9  A1 AWRT. = 14, 1.56 WWW but allow from truncation error (e.g. 0.266 rather than 9 0.267). 14 Note: also comes from SCB2 but scores M1M1A0 max. 9 3

Question 25

For a randomly chosen person, their next birthday is equally likely to occur on any day of the week, independently of any other person’s bi…

Accepted Answer

6(a) 10 B1 0.03457…  5  9765625 [ =] 0.0346,    7  282475249 1 6(b) Method 1 8 2 9 10 M1 x 10 − x  6  1   6  1   6  One term 10Cx ( p ) (1 − p ) . [P(0, 1, 2) =] 10C2    + 10C1    +    7  7   7  7   7  With 0  p  1, x  0 or 10.  = 0.2675729 + 0.3567639 + 0.2140583 =  A1 Correct un-simplified expression. Allow 10 for 10C1. 0.838 B1 0.838 ⩽ p ⩽ 0.839. 10 9 2 8 M1 x 10 − x  1   6  1   6  1  One term 10Cx ( p ) (1 − p ) . [P(0,1,2) =] 1 –{   + 10C9    + 10C1    + 10C1  7   7  7   7  7  With 0  p  1, x  0 or 10.  6 3 1  7  6 4 1  6  6 5 1 5    + 10C1    + 10C1    + 10C1 A1 Correct un-simplified expression. Allow 10 for 10C1.  7  7   7  7   7  7  Condone omission of up to 5 of the middle 6 terms.  6 4 1  6  6 3 1  7 Condone omission of last bracket only.    + 10C1    }  7  7   7  7  If both brackets omitted in un-simplified expression allow recovery for final stated calculation of 1 – 0.1616 or final answer WRT to 0.8384. 0.838 B1 0.838 ⩽ p ⩽ 0.8385. 3 6(c) 1 B1 56 and 48 seen, allow un-simplified, may be seen in the [Mean = 392  =] 56 standardisation formula. 7 1 6 [Variance = 392   = ] 48 (  = 48,4 3, 6.928  6.9283 implies correct variance. 7 7 Condone N(30, 48 ) if standardisation formula is correct or variance/standard deviation correctly stated as well. 65.5 − 56 M1 Substituting their µ and positive σ into the ± standardising formula [P(X > 65) = P( Z  ] ) (any number for 65.5), allow σ2 or √σ. 48 M1 Use continuity correction 64.5 or 65.5 in ±standardisation formula  9.5   9.5  Note: If no standardisation formula seen   or    48   6.928  scores M2. [= 1 − Φ (1.3712 ) ] M1 Appropriate area Φ, from final process, must be a probability. = 1 – 0.9149 May be implied by a sketch of the required probability area. Note: correct final answer implies this M1. Expect final answer < 0.5. = 0.0851 final answer A1 Final answer Accept 0.08505 ⩽ p ⩽ 0.0852. 5 7(a) Method 1 Total arrangements with 3 Os together – total arrangements with 3 Os together and 2 Ls together. 8! B1 8! 2!− 7! 2! seen alone (not multiplied/divided). B1 b − 7!, 5040  b . M1 8! 7! − , c = 1, 2 d = 1, 3 . c ! c ! d ! = 15120 A1 CAO. Method 2 ^ ^ OOO ^ ^ ^ , Arrangements with OOOs together and no Ls, Ls inserted separately. 7  6 B1 6!e,1 e 42 . 6!  2 B1 7  7 2P f  6, 1  f accept 7C2 or . 2 2 M1 6! 7   6, g = 1,3 h = 1,2,3 . g ! h = 15120 A1 CAO. 4 7(b) Method 1 L _ _ _ _ _ L _ _ _ 8! B1 8! 3! 4 3!  i , i  1 . M1 8!  4, j = 1, 2, 3 . j ! 26880 A1 CAO. Method 2 4! B1 8 5P  k , k  1 . 8 P5  3! M1 8 4! 8 Pm  or Pm  4, m = 3, 4, 5. 3! 8 4! 8 or C5  or C5  4. 3! 26880 A1 CAO. 3

Question 26

There are a large number of students at Greenfield college. Each student travels to college by car, by bus or on foot, independently of any…

Accepted Answer

Question Answer Marks Guidance 1(a) 3 27 B1 OE. [( 0.6 ) =] 0.216,125 1 1(b) Method 1 [P(X < 9) = 1 – P(9, 10, 11) =] 1 – {11C9(0.75)9 (0.25)2 + 11C10 (0.75)10 M1 x 11− x One term of form 11Cx ( p ) (1 − p ) with 0  p  1, x  0 or (0.25) + (0.75)11} 11. [= 1 – (0.258104 + 0.154862 + 0.042235)] A1 Correct un-simplified expression, no terms omitted leading to final answer. Condone omission of final bracket ‘}’ If other brackets omitted, allow recovery if 1 – 0.455 (or better) seen. 11 11 x 11− x Accept C x ( p ) (1 − p ) with p = 0.75.  9 = 0.545 B1 0.5445 p 0.545 . Method 2 [P(X < 9) = P(0,1,2,3,4,5,6,7,8) =] (0.25)11 + 11C1(0.75)(0.25)10 + … + M1 One term of form 11Cx ( p ) x (1 − p )11− x with 0  p  1, x  0 or 11C7(0.75)7(0.25)4 + 11C8(0.75)8(0.25)3 11. A1 Correct un-simplified expression, no more than 6 ‘middle’ terms omitted leading to final answer. 8 11 x 11− x Accept C x ( p ) (1 − p ) with p = 0.75.  0 = 0.545 B1 0.5445  p  0.545 . 3

Question 27

A fair red spinner has 4 sides, numbered 1, 2, 3, 4. A fair blue spinner has 4 sides, numbered 0, 1, 2, 3. When a spinner is spun, the scor…

Accepted Answer

4(a) B1 Table with correct x values and one correct probability. x 0 1 2 3 4 B1 2 more probabilities correct linked to the correct x value, need P(X = x) 3 1 3 5 4 not be in table, accept un-simplified. 16 16 16 16 16 B1 5 correct probabilities linked with correct outcomes. 0.1875 0.0625 0.1875 0.3125 0.25 SCB1 5 non-zero probabilities (not all 1/5) summing to 1 Accept 0.188 0.0625 0.188 0.313 0.25 placed in a probability distribution table with correct x values if only one of the first two B marks are scored. 3 4(b) 1 3 5 4 M1FT Accept un-simplified or bold expression with probabilities (0 < [E(X) = ] 0 + 1  + 2  + 3  + 4  p < 1) that add to 1. May be calculated in variance. 16 16 16 16 FT their table with 4 or 5 probabilities adding to 1.  0 + 1 + 6 + 15 + 16   =   16   38 19  = =    16 8  2 2 2 2 M1FT Appropriate variance formula using their (E(X))2 value. 1 + 3  2 + 5  3 + 4  4  19  Var(X) = −  FT their table with 4 or more probabilities (0 < p < 1) which 16  8  need not sum to 1 or with an expression no more evaluated than  1 + 3  4 + 5  9 + 4  16 361  shown. = −   16 64      122  19  2   = −     16  8   127 63 A1 Accept 1.984375 to 3 or more SF. = , 1 , 1.98 SCB1 for 1.98 if M0 M1 scored. 64 64 3 4(c) Method 1 7 M1 n  3   3  1 −  1 −  with n = 6, 7, 8.  4   4  = 0.867 A1 = 0.8665 to 4SF. Method 2 2 3 4 M1 7 6 1  3  1   3  1   3  1   3  1   3  1   3  1  + + + + + Condone extra term or the term missing.                   4  4  4   4  4   4  4   4  4   4  4   4  4  [P(X < 8) =]  3 5 1   3 6 1  +        4  4   4  4  = 0.867 A1 2

Question 28

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 …

Accepted Answer

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

Question 29

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 …

Accepted Answer

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

Question	Answer	Marks	From
1	see sheet	4	9709/53 May/June 2020
2	see sheet	6	9709/51 Oct/Nov 2020
3	see sheet	5	9709/52 Oct/Nov 2020
4	see sheet	5	9709/53 Oct/Nov 2020
5	see sheet	8	9709/53 Oct/Nov 2020
6	see sheet	9	9709/52 May/June 2021
7	see sheet	8	9709/52 May/June 2022
8	see sheet	7	9709/52 Oct/Nov 2022
9	see sheet	5	9709/52 Oct/Nov 2023
10	see sheet	7	9709/51 May/June 2024
11	see sheet	7	9709/52 May/June 2024
12	see sheet	10	9709/52 May/June 2024
13	see sheet	8	9709/53 May/June 2024
14	see sheet	6	9709/51 Oct/Nov 2024
15	see sheet	9	9709/51 Oct/Nov 2024
16	see sheet	7	9709/53 Oct/Nov 2024
17	see sheet	4	9709/53 Oct/Nov 2024
18	see sheet	6	9709/52 Feb/March 2025
19	see sheet	9	9709/52 Feb/March 2025
20	see sheet	9	9709/51 May/June 2025
21	see sheet	5	9709/52 May/June 2025
22	see sheet	6	9709/55 May/June 2025
23	see sheet	6	9709/51 Oct/Nov 2025
24	see sheet	7	9709/52 Oct/Nov 2025
25	see sheet	9	9709/52 Oct/Nov 2025
26	see sheet	4	9709/53 Oct/Nov 2025
27	see sheet	8	9709/53 Oct/Nov 2025
28	see sheet	7	9709/55 Oct/Nov 2025
29	see sheet	11	9709/55 Oct/Nov 2025

Discrete random variables — Paper 5 · A Level Mathematics 9709

Questions44 pages

Mark scheme29 answers

Another paper, or another topic

Questions as text

Q1 · In a certain large college, 22% of students own a car 9709/53 May/June 2020

Q2 · The random variable X takes each of the values 1, 2, 3, 4 with probability 1 Two… 9709/51 Oct/Nov 2020

Q3 · A fair six-sided die, with faces marked 1, 2, 3, 4, 5, 6, is thrown repeatedly until a 4… 9709/52 Oct/Nov 2020

Q4 · An ordinary fair die is thrown until a 6 is obtained 9709/53 Oct/Nov 2020

Q5 · The 13 00 train from Jahor to Keman runs every day 9709/53 Oct/Nov 2020

Q6 · Every day Richard takes a ﬂight between Astan and Bejin 9709/52 May/June 2021

Q7 · In a large college, 28% of the students do not play any musical instrument, 52% play… 9709/52 May/June 2022

Q8 · The times taken, in minutes, to complete a word processing task by 250 employees at a… 9709/52 Oct/Nov 2022

Q9 · George has a fair 5-sided spinner with sides labelled 1, 2, 3, 4, 5 9709/52 Oct/Nov 2023

Q10 · Harry has three coins: • One coin is biased so that the probability of obtaining a head… 9709/51 May/June 2024

Q11 · Jasmine has one $5 coin, two $2 coins and two $1 coins 9709/52 May/June 2024

Q12 · The residents of Mahjing were asked to classify their local bus service: • 25% of… 9709/52 May/June 2024

Q13 · The numbers on the faces of a fair six-sided dice are 1, 2, 2, 3, 3, 3 9709/53 May/June 2024

Q14 · The random variable X takes the values -2, -1, 0, 2, 3 9709/51 Oct/Nov 2024

Q15 · The weights of the green apples sold by a shop are normally distributed with mean 90… 9709/51 Oct/Nov 2024

Q16 · 30% of the residents of Wimfield own an electric car 9709/53 Oct/Nov 2024

Q17 · A red fair six-sided dice has faces labelled 1, 1, 1, 2, 2, 2 9709/53 Oct/Nov 2024

Q18 · Jacob throws three coins at the same time 9709/52 Feb/March 2025

Q19 · Last year, an online store sold a large number of computers 9709/52 Feb/March 2025

Q20 · A bag contains 10 marbles, of which 4 are red and 6 are blue 9709/51 May/June 2025

Q21 · In Millford, 70% of the residents own a bicycle 9709/52 May/June 2025

Q22 · Two fair 6-sided dice with faces labelled 1, 2, 3, 4, 5, 6 are thrown 9709/55 May/June 2025

Q23 · The random variable X takes the value x with probability kx2, where k is a constant and x… 9709/51 Oct/Nov 2025

Q24 · Kayla has a bag containing 3 red marbles, 1 blue marble and 2 green marbles 9709/52 Oct/Nov 2025

Q25 · For a randomly chosen person, their next birthday is equally likely to occur on any day… 9709/52 Oct/Nov 2025

Q26 · There are a large number of students at Greenfield college 9709/53 Oct/Nov 2025

Q27 · A fair red spinner has 4 sides, numbered 1, 2, 3, 4 9709/53 Oct/Nov 2025

Q28 · Kai has a spinner with four sides, labelled 1, 2, 3, 4 9709/55 Oct/Nov 2025

Q29 · A large number of runners took part in two charity runs to raise money for a new… 9709/55 Oct/Nov 2025