Probability — Paper 5 — A Level Mathematics 9709 topical questions

Question 1

In Greenton, 70% of the adults own a car. A random sample of 8 adults from Greenton is chosen. (a) Find the probability that the number of …

Accepted Answer

5(a) 1 – P(6, 7, 8) M1 x 8 − x One term 8Cx p (1 − p ) , 0 < p < 1, x ≠ 0 = 1 – (8C6 0.7 6 0.32 + 8C7 0.7 7 0.31 + 0.7 8 ) = 1 – 0.55177 A1 Correct unsimplified expression, or better = 0.448 A1 Alternative method for question 5(a) P(0, 1, 2, 3, 4, 5) M1 x 8 − x One term 8Cx p (1 − p ) , 0 < p < 1, x ≠ 0 = 0.38 + 8C1 0.710.37 +8C2 0.720.36 + 8C3 0.730.35 + 8C40.740.34 + 8C5 0.750.33 A1 Correct unsimplified expression, or better = 0.448 A1 3 5(b) Mean = 120 × 0.7 = 84 B1 Correct mean and variance, allow unsimplified Var = 120 × 0.7 × 0.3 = 25.2  75.5 − 84  M1 Substituting their µ and σ into the ±standardising formula (any P( more than 75) = P  z >  number), not σ2, not √σ  25.2  M1 Using continuity correction 75.5 or 74.5 P( z > −1.693) M1 Appropriate area Φ , from final process, must be a probability = 0.955 A1 Allow 0.9545 < p ⩽ 0.955 5

Question 2

Box A contains 7 red balls and 1 blue ball. Box B contains 9 red balls and 5 blue balls. A ball is chosen at random from box A and placed i…

Accepted Answer

6(a) B1 Both correct probs, box A Box A Box B B1 2 probs correct for box B Red d B1 All correct probs for box B Red Blue Red Blue Blue 3 6(b) 7 5 1 9 M1 Two 2 factor terms added, correct or FT their 6(a). × + × 8 15 8 15 44  11  A1 OE = or 0.367   120  30  2 6(c) P ( A blue ∩ B blue ) M1 1 6 P(A blue |B blue) = their × seen as numerator or denom of fraction P ( B blue ) 8 15 1 6 1 × 8 15 20 = = 7 5 1 6 41 × + × 8 15 8 15 120 M1 7 5 1 6 their × + × seen 8 15 8 15 M1 7 5 1 6 their × + × seen as denominator 8 15 8 15 6 A1 = or 0.146 41 4

Question 3

A company produces small boxes of sweets that contain 5 jellies and 3 chocolates. Jemeel chooses 3 sweets at random from a box. (a) Draw up…

Accepted Answer

3(a) x 0 1 2 3 Probability 1 56 15 56 30 56 10 56 (B1 for probability distribution table with correct outcome values) P(0) = 3 2 1 1 8 7 6 56 × × = P(1) = 5 3 2 15 3 8 7 6 56 × × × = P(2) = 5 4 3 30 3 8 7 6 56 × × × = P(3) = 5 4 3 10 8 7 6 56 × × = (M1 for denominator 8×7×6) M1 Any one probability correct (with correct outcome) A1 All probabilities correct A1 4 3(b) 1 – P(8, 9, 10) = 1 – 10 8 2 10 9 1 10 8 9 C 0.64 0.36 C 0.64 0.36 0.64   + +   M1 1 – (0.164156 + 0.064852 + 0.11529) M1 0.759 A1 3

Question 4

On Mondays, Rani cooks her evening meal. She has a pizza, a burger or a curry with probabilities 0.35, 0.44, 0.21 respectively. When she co…

Accepted Answer

5(a) Fully correct labelled tree for method of transport with correct probabilities. B1 Fully correct labelled branches with correct probabilities for lateness with either 1 branch after W or 2 branches with the prob 0 B1 2 5(b) 0.35 × 0.3 + 0.44 × 0.8 (+ 0) M1 0.457 A1 2 Question Answer Marks 5(c) P(not B|not fruit) = ( ) ( ) P B' F' P F' ∩ M1 ( ) 0.35 0.7 0.21 1 1 their × + × − b M1 0.455 0.543 (M1 for 1 – their (b) or summing three appropriate 2-factor probabilities, correct or consistent with their tree diagram as denominator) M1 0.838 or 455 543 A1 4

Question 5

A total of 500 students were asked which one of four colleges they attended and whether they preferred soccer or hockey. The numbers of stu…

Accepted Answer

2(a) 56 14 or or 0.112 500 125 B1 1 2(b) ( ) ( ) ( ) P D S 120 P D|S P S 280 ∩ = = M1 120 3 or 280 7 A1 2 Question Answer Marks 2(c) P(hockey) = 220 0.44 500 = P(Amos or Benn) = 242 0.484 500 = P(hockey ∩ A or B) = 104 0.208 500 = P(H) × P(A U B) = P(H ∩ (A U B)) if independent M1 220 242 1331 500 500 6250 × = so not independent A1 2

Question 6

On any given day, the probability that Moena messages her friend Pasha is 0.72. (a) Find the probability that for a random sample of 12 day…

Accepted Answer

7(a) = 1 – [12C100.72100.282 + 12C11 0.72110.281+0.7212] 1 – (0.19372 + 0.09057 + 0.01941) A1 0.696 A1 3 7(b) 0.283 × 0.72 = 0.0158 B1 1 Question Answer Marks 7(c) Mean = 100 × 0.72 = 72 Var = 100 × 0.72 × 0.28 = 20.16 M1 P(less than 64) = P 63.5 72 20.16 z −   <     (M1 for substituting their µ and σ into ±standardisation formula with a numerical value for ‘63.5’) M1 Using either 63.5 or 64.5 within a ±standardisation formula M1 Appropriate area Φ, from standardisation formula P(z<…) in final solution = P(z < –1.893) M1 0.0292 A1 5

Question 7

Juan goes to college each day by any one of car or bus or walking. The probability that he goes by car is 0.2, the probability that he goes…

Accepted Answer

1(a) Fully correct labelled tree for method of transport with correct probabilities. B1 Fully correct labelled branches with correct probabilities for lateness with either 1 branch after W or 2 branches with the probability 0. B1 2 1(b) P(C|E) = ( ) ( ) P C E 0.2 0.6 P E 0.2 0.6 0.45 0.1 0.35 1 ∩ × = × + × + × M1 Summing three appropriate 2-factor probabilities M1 0.12 0.515 A1 0.233 or 12 515 A1 4

Question 8

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 9

(a) Find the number of diﬀerent possible arrangements of the 9 letters in the word CELESTIAL. [1] .........................................…

Accepted Answer

7(a) 9! 2!2!= 90 720 B1 1 7(b) 6! 2! M1 360 A1 2 Question Answer Marks 7(c) 2 Es together = ( ) 8! 20160 2! = M1 Es not together = 90720 – 20160 = 70560 M1 Probability = 70560 90720 M1 7 or 0.778 9 A1 Alternative method for question 7(c) _ ^ _ ^ _ ^ _ ^ _ ^ _ ^ _ ^ _ 7! 8 7 2! 2 × × = 70560 7! × k in numerator, k integer ⩾ 1, denominator ⩾ 1 M1 Multiplying by 8C2 OE M1 Probability = 70560 90720 M1 7 or 0.778 9 A1 4 Question Answer Marks 7(d) Scenarios are: E L _ _ _ 5C3 10 E E L _ _ 5C2 10 E _ _ _ _ 5C4 5 E E _ _ _ 5C3 10 M1 Summing the number of ways for 3 or 4 correct scenarios M1 Total = 35 A1 3

Question 10

The probability that a student at a large music college plays in the band is 0.6. For a student who plays in the band, the probability that…

Accepted Answer

2(a) 0·6 × 0·7 + 0·4(1 – x) = 0·58 ≡ 0·42 + 0·4(1 – x) = 0·58 M1 Equation of form 0·6 × a + 0·4 × b = 0·58; a = 0·3, 0·7, b = x, (1 – x) B1 Single correct product seen, condone 0·42, in an equation of appropriate form x = 0·6 A1 Alternative method for question 2(a) 0·6 × 0·3 + 0·4x = 0·42 ≡ 0·18 + 0·4x = 0·42 M1 Equation of form 0·6 x a + 0·4 x b = 0·42; a = 0·3, 0·7, b = x, (1 – x) B1 Single correct product seen, condone 0·18, in an equation of appropriate form x = 0·6 A1 3 2(b) ( ) 2 0.6 0.3 × M1 (a × b)2, a = 0·6, 0·4 and b = 0·7, 0·3, x, (1–x) or 0·182, alone. 0.0324 A1 2 M1 pn, n = 6, 7 0 < p < 1

Question 11

Kayla is competing in a throwing event. A throw is counted as a success if the distance achieved is greater than 30 metres. The probability…

Accepted Answer

3(a) P(X > 6) = 6 0.75 0.178, 729 4096 A1 0·17797… 2 Question Answer Marks Guidance 3(b) 1 – P(0, 1, 2) = 1 – ( 10 0.75 + 10C1 1 9 0.25 0.75 + 10C2 2 8 0.25 0.75 ) M1 Binomial term of form 10Cx ( ) 10 1 , x x p p − − 0 < p < 1, any p, x ≠ 0, 10 1 – (0·0563135 + 0·1877117 + 0·2815676) A1 Correct unsimplified expression 0·474 A1 0·474 ⩽ p ⩽ 0·4744 3

Question 12

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 13

(a) Find the number of diﬀerent ways in which the 10 letters of the word SHOPKEEPER can be arranged so that all 3 Es are together. [2] ....…

Accepted Answer

7(a) 8! 2! M1 ( ) 8! 7! 8 ! ,where , , where 2 ! a k a k k × ≡ ∈ ∈   20160 A1 2 Question Answer Marks Guidance 7(b) Total number of ways: ( ) 10! 302400 2!3! = (A) B1 Accept unsimplified With Ps together: 9! ( 3! = 60 480) (B) B1 Accept unsimplified With Ps not together: 302 400 – 60 480 M1 10! 9! m n − , m, n integers or (A) – (B) if clearly identified 241 920 A1 Alternative method for question 7(b) 8! 3! B1 k × 8! in numerator, k a positive integer, no ± B1 m × 3! in denominator, m a positive integer, no ± 9 8 2 × × M1 Their 8! 3!multiplied by 9C2 or 9P2 no additional terms 241 920 A1 Exact value, WWW 4 Question Answer Marks Guidance 7(c) Probability = Number of ways Es at beginning and end Total number of ways Probability = 8! 2! 10! 2! 3! × = 20160 302400 M1 8! ! 10! ! ! k k l      1 ⩽ k, l ∈ ℕ ⩽ 3, FT denominator from 7(b) or correct 1 15 , 0·0667 A1 Alternative method for question 7(c) Probability = 3 2 10 9 × M1 1 3,2 10 9 a a a − × = 1 15 , 0·0667 A1 Alternative method for question 7(c) Probability = 1 1 3! 10 9 × × M1 1 1 !, 3,2 10 9 m m × × = 1 15 , 0.0667 A1 2 Question Answer Marks Guidance 7(d) Scenarios: P E E E 5C0 = 1 P E E _ 5C1 = 5 P E _ _ 5C2 = 10 P _ _ _ 5C3 = 10 M1 5Cx seen alone, 1 ⩽ x ⩽ 4 M1 Summing the number of ways for 3 or 4 correct scenarios (can be unsimplified), no incorrect scenarios Total = 26 A1 3

Question 14

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 15

Mr and Mrs Ahmed with their two children, and Mr and Mrs Baker with their three children, are visiting an activity centre together. They wi…

Accepted Answer

6(a) Condone 9C6 + 3C3, 9P6 × 3P3 84 A1 Accept unevaluated. 2 6(b) Number with 3 Baker children = 6C2 or 15 B1 Correct seen anywhere, not multiplied or added Total no of selections = 9C5 or 126 Probability = number of selections with 3 Baker children total number of selections M1 Seen as denominator of fraction 15 126 , 0·119 A1 OE, e.g. 5 42 Alternative method for question 6(b) 5 3 3 2 1 6 5 9 8 7 6 5    × × × × ×       C B1 5C3 (OE) or 10 seen anywhere, multiplied by fractions only, not added M1 3 2 1 6 5 9 8 7 6 5    × × × × ×       k , 1 ⩽ k, k integer 15 126 , 0·119 A1 OE, e.g. 5 42 3 Question Answer Marks Guidance 6(c) [Total no of arrangements = 9!] [Arrangements with men together = 8! × 2] Not together: 9! – M1 9! – k or 362880 – k, k an integer<362 880 8! × 2 B1 8! × 2(!) or 80 640 seen anywhere 282 240 A1 Exact value Alternative method for question 6(c) 7! × 8 × 7 B1 7! × k, k positive integer > 1 M1 m × 8 × 7, m × 8P2, m × 8C2 m positive integer > 1 282 240 A1 Exact value 3 6(d) 7! × 2 × 7 M1 7! × k, k positive integer > 1 If 7! not seen, condone 7 × 6 × 5 × 4 × 3 × 2 × (1) × k or 7 × 6! × k only M1 m × 2 × 7, m positive integer > 1 70 560 A1 3

Question 16

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 17

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 18

The 8 letters in the word RESERVED are arranged in a random order. (a) Find the probability that the arrangement has V as the ﬁrst letter a…

Accepted Answer

5(a) Total number of ways = 8! 3!2! (= 3360) B1 Correct unsimplified expression for total number of ways Number of ways with V and E in correct positions = 6! 2! 2! × (= 180) B1 6! 2! 2! × alone or as numerator in an attempt to find the number of ways with V and E in correct positions. No , × ± Probability = 180 3360 3 56   =     or 0.0536 B1 FT Final answer from their 6! 2! 2! × divided by their total number of ways Alternative method for question 5(a) 1 3 8 7 × M1 8 7 × a b seen, no other terms (correct denominators) M1 1 3 × c d seen, no other terms (correct numerators) 3 56 or 0.0536 A1 3 Question Answer Marks Guidance 5(b) Rs together and Es together: 5! (120) B1 Alone or as numerator of probability to represent the number of ways with Rs and Es together, no ×, +, – Es together: ( ) 6! 360 2! = B1 Alone or as denominator of probability to represent the number of ways with Es together, no ×, + or – Probability = 5! 6! 2! M1 5! 6! 2! their their seen 1 3 A1 OE Alternative method for question 5(b) P(Rs together and Es together): 5! 1 total number of ways 28 their   =     B1 P(Es together): 6! 3 2! 28 total number of ways their   =     B1 Alone or as numerator of probability to represent the P(Rs and Es together), no ×, +, – Probability = 1 28 3 28 M1 Alone or as denominator of probability to represent the P(Es together), no ×, + or – 1 3 A1 OE, 1 28 3 28 their their seen 4

Question 19

A fair spinner with 5 sides numbered 1, 2, 3, 4, 5 is spun repeatedly. The score on each spin is the number on the side on which the spinne…

Accepted Answer

1(a) 7 4 1 5 5     =           16384 390625 or 0∙0419[43…] 1 1(b) 5 4 1 5  −  or 2 3 4 1 4 1 4 1 4 1 4 1 5 5 5 5 5 5 5 5 5       + × × + × + ×             M1 1 – pn n = 5,6 or p + pq + pq2+pq3+ pq4 (+ pq5) 0 < p < 1, p + q = 1, Sum of a geometric series may be used. 2101 3125 or 0∙672[32] A1 Final answer. Alternative method for question 1(b) [P(at least 1 three scored in 5 throws) =] 5 4 3 2 2 3 4 5 5 5 5 4 3 2 4 1 1 4 1 4 1 4 1 4 C C C C 5 5 5 5 5 5 5 5 5              + + + +                           M1 5 5 4 5 3 2 5 2 3 5 4 4 3 2 1 ( ) C ( ) ( ) C ( ) ( ) C ( ) ( ) C ( )( ) p p q p q p q p q + + + + or 6 6 5 6 4 2 6 3 3 5 4 3 6 2 4 6 5 2 1 , 0 1, 1 ( ) C ( ) ( ) C ( ) ( ) C ( ) ( ) C ( ) ( ) C ( )( ) p p p q p q p q p q p q p q + + < < + + + = + At least first, last and one intermediate term is required to show pattern of terms if not all terms stated. 2101 3125 or 0∙672[32] A1 Final answer. 2

Question 20

Georgie has a red scarf, a blue scarf and a yellow scarf. Each day she wears exactly one of these scarves. The probabilities for the three …

Accepted Answer

2(a) 0.2 1 0.45 0.4 0.35 0.3     × + × + × M1 0∙2 [× 1] + 0∙45 × b + 0∙35 × c, b = 0∙4, 0∙6 c = 0∙3, 0∙7 0∙485 or 97 200 A1 2 2(b) ( ) ( ) ( ) 0.35 0.7 0.245 | 1 0.515 ∩ × = = = − P Y H P Y H their P H (a) B1 0∙35 × 0∙7 or 0∙245 seen as numerator or denominator of fraction. M1 0∙515 or 1 – their (a) or [0∙3 × 0 +] 0∙45 × d + 0∙35 × e, where d = their b′, e = their c′ seen as denominator of fraction. 0∙476 or 49 103 A1 0∙4757 ⩽ p ⩽ 0∙476 3

Question 21

There are 400 students at a school in a certain country. Each student was asked whether they preferred swimming, cycling or running and the…

Accepted Answer

7(a)(i) 104 31 135 27 , 400 400 80 +   =     , 0∙3375 B1 Evaluated, exact value. 1 7(a)(ii) Method 1 ( ) 180 P 400 = M , 0∙45 ( ) 135 P 400 = S , 0∙3375 ( ) 31 P 400 ∩ = M S , 0∙0775 180 135 243 31 ,0.151875 400 400 1600 400 × = ≠ so NOT independent M1 Their P(M) × their P(S) seen, accept unsimplified. A1 P(M), P(S) and P(M ∩ S) notation seen, numerical comparison and correct conclusion, WWW. Method 2 ( ) 31 P 400 ∩ = M S ( ) 135 P 400 = S ( ) 180 P 400 = M ( ) 31 31 400 P | , 0.2296 135 135 400 = = … M S 180 400 ≠ so NOT independent M1 ( ) ( ) ( ) P [P | ] P ∩ = their M S M S their S (oe) seen, accept unsimplified. A1 P(M), P(S) and P(M ∩ S) notation seen, numerical comparison and correct conclusion, WWW. 2 Question Answer Marks Guidance 7(b)(i) Method 1 [1 – P(0,1,2)] = 1 – (10C0 0∙30 0∙710 + 10C1 0∙31 0∙79 + 10C2 0∙32 0∙78) M1 10Cx px (1 – p)10-x for 0 < × < 10, 0 < p < 1, any p. = 1 – (0∙028248 + 0∙121061 + 0∙233474) A1 Correct expression, accept unsimplified, condone omission of final bracket, condone recovery from poor notation. = 0∙617 A1 Accept 0∙61715 ⩽ p ⩽ 0∙61722, WWW. Method 2 [P(3,4,5,6,7,8,9,10) =] 10C3 0∙33 0∙77 + 10C4 0∙34 0∙76 + 10C5 0∙35 0∙75 + 10C6 0∙36 0∙74 + 10C7 0∙37 0∙73 + 10C8 0∙38 0∙72 + 10C9 0∙39 0∙71 + 10C10 0∙310 0∙70 M1 10Cx px (1 – p)10–x for 0 < × < 10, 0 < p < 1, any p. A1 Correct unsimplified expression. = 0∙617 A1 Accept 0∙61715 ⩽ p ⩽ 0∙61722, WWW. 3 Question Answer Marks Guidance 7(b)(ii) [p = 0∙3] Mean = 0∙3 × 90 = 27; variance = 0∙3 × 90 × 0∙7 = 18∙9 B1 Correct mean and variance, allow unsimplified. Condone σ = 4∙347 evaluated. P(X < 32) = 31.5 27 18.9 −   <     P z M1 Substituting their μ and σ (not σ2, √σ) into the ±standardising formula with a numerical value for ‘31∙5’. M1 Using either 31∙5 or 32∙5 within a ±standardising formula with numerical values for their μ and σ (condone σ2, √σ). ( ) 1.035 =Φ M1 Appropriate area Φ, from standardisation formula P(z<…) in final solution, must be probability. = 0∙850 A1 Allow 0∙8495 < p ⩽ 0∙85(0), final answer WWW. 5

Question 22

(a) How many diﬀerent arrangements are there of the 8 letters in the word RELEASED? [1] ...................................................…

Accepted Answer

3(a) 8! 3!       = 6720 1 3(b) _ _ _ L E D _ _ : With LED together: 6! 2! M1 6! k or 5! 6 x k k ⩾ 1 and no other terms M1 2! m , m an integer, m ⩾ 5 360 A1 CAO 3 3(c) Method using _ _ _ A _ D _ _ : Arrange the 6 letters RELESE = 6! 3! [= 120] *M1 6! 3!×k seen, k an integer > 0 Multiply by number of ways of placing AD in non-adjacent places = their 120 × 7P2 [= 5040] *M1 ( ) 1 × − m n n or 2 × n m C or 2 × n m P , n = 6, 7 or 8, m an integer > 0 [Probability =] 5040 6720 their their DM1 Denominator = their (a) or correct, dependent on at least one M mark already gained. 5040 3 or or 0.75 6720 4 A1 Alternative method for Question 3(c) Method using ‘Total arrangements – Arrangements with A and D together’: Their 6720 – 7! 2 3! × [= 5040] *M1 Their 6720 – k, k a positive integer *M1 ( ) 7! , 1,2 3! × − = k m k Question Answer Marks Guidance [Probability =] 5040 6720 their their DM1 With denominator = their (a) or correct, dependent on at least one M mark already gained. 5040 3 or or 0.75 6720 4 A1 Alternative method for Question 3(c) Method using ‘1 – Probability of arrangements with A and D together’: 7! 2 3! × [= 1680] *M1 7! , 1,2 3! × = k k [Probability =] 1 680 6720 their their *M1 With denominator = their (a) or correct 1 – 1 680 6720 their their DM1 1 – m, 0 < m < 1 , dependent on at least one M mark already gained 5040 3 or or 0.75 6720 4 A1 4

Question 23

To gain a place at a science college, students ﬁrst have to pass a written test and then a practical test. Each student is allowed a maximu…

Accepted Answer

4(a) branches clearly identifying written and practical, pass and fail for each intersection (no additional branches) B1 ‘One written test’ branch all probabilities (or %) correct B1 ‘Two written tests’ branch all probabilities (or %) correct, condone additional branches after W2F with probabilities 1 for PF and 0 for PP 3 4(b) [P(W1P) × P(PP) + P(W1F) × P(W2P) × P(PP)] 0.8 0.3 0.2 0.6 0.3 × + × × M1 Consistent with their tree diagram or correct 0.276 or 69 250 A1 2 4(c) ( ) ( ) ( ) ( ) 1 Practical 0.8 0.3 1 getting place ∩ × = = P W P W P P their b 0.24 0.276   =     M1 Correct expression or FT their (b) 20 23 or 0.87[0] A1 2

Question 24

In Questa, 60% of the adults travel to work by car. (a) A random sample of 12 adults from Questa is taken. Find the probability that the nu…

Accepted Answer

6(a) [= 1 – (0.063852 + 0.017414 + 0.0021768)] A1 Correct unsimplified expression, or better. [1 – 0.083443] = 0.917 A1 AWRT Alternative method for Question 6(a) P (0,1,2,3,4,5,6,7,8,9) = 12C00.60 0.412 + 12C1 0.61 0.411+ ………….12C9 0.69 0.43 [= 0.000016777 + 0.00030199 + 0.0024914 + 0.012457 + 0.042043 + 0.10090 + 0.17658 + 0.22703 + 0.21284 + 0.14189] M1 One term: 12Cx px (1 – p)12-x for 0 < x < 12, any p allowed. A1 Correct unsimplified expression with at least the first two and last terms 0.917 A1 WWW, AWRT 3 Question Answer Marks Guidance 6(b) [Mean =] 0.6 × 150 [= 90]; [Variance =] 0.6 × 150 × 0.4 [= 36] B1 Correct mean and variance. Accept evaluated or unsimplified ( ) 80.5 90 81 6 −   < = <     P X P Z M1 Substituting their mean and variance into ±standardisation formula (with a numerical value for 80.5), allow σ2, √ σ, but not µ ± 0.5 M1 Using continuity correction 80.5 or 81.5 ( ) 1.5833 1 0.9433 Φ − = − M1 Appropriate area Φ, from final process, must be probability 0.0567 A1 AWRT 5 6(c) np = 90, nq = 60 both greater than 5 B1 At least nq evaluated and statement >5 required 1

Question 25

An ordinary fair die is thrown repeatedly until a 5 is obtained. The number of throws taken is denoted by the random variable X. (a) Write …

Accepted Answer

1(a) 6 B1 WWW 1 1(b) 3 4 5 6 5 1 5 1 5 1 5 1 6 6 6 6 6 6 6 6         + + +                 M1 p3(1 – p) + p4(1 – p) + p5(1 – p) + p6(1 – p), 0 < p < 1 0.300 (0.2996…) A1 At least 3s.f. Award at most accurate value. Alternative method for Question 1(b) 3 7 5 5 6 6     −         M1 p3 – p7, 0 < p < 1 0.300 (0.2996…) A1 At least 3s.f. Award at most accurate value. 2 1(c) 9 5 1 6   −    M1 1 – pn, 0 < p < 1, n = 9, 10 0.806 A1 Alternative method for Question 1(c) 2 8 1 1 5 1 5 1 5 6 6 6 6 6 6 6       + + + +              M1 p + p(1 – p) + p(1 – p)2 + p(1 – p)3 + p(1 – p)4 + p(1 – p)5 + p(1 – p)6 + p(1 – p)7 + p(1 – p)8 (+ p(1 – p)9), 0 < p < 1 As per answer for minimum terms shown 0.806 A1 2

Question 26

On each day that Alexa goes to work, the probabilities that she travels by bus, by train or by car are 0.4, 0.35 and 0.25 respectively. Whe…

Accepted Answer

3(a) P(not late) = 0.4 × 0.45 + 0.35 × 0.3 + 0.25 × (1 – x) or P(late) = 0.4 × 0.55 + 0.35 × 0.7 + 0.25x M1 0.4 0.35 0.25 × + × + × p q r , p = 0.45, 0.55, q = 0.3, 0.7 and r = (1 – x), x 0.18 + 0.105 + 0.25 (1 – x) = 0.48 or 0.22 + 0.245 + 0.25x = 0.52 A1 Linear equation formed using sum of 3 probabilities and 0.48 or 0.52 as appropriate. Accept unsimplified. x = 0.22 A1 Final answer 3 3(b) ( ) ( ) ( )   ∩ =       P train late P train late P late 0.35 0.7 1 0.48 × = − or 0.35 0.7 0.4 0.55 0.35 0.7 0.25 0.22 × × + × + ×their B1 0.35 0.7 × or 0.245 seen as numerator of fraction M1 P(late) seen as a denominator with their probability as numerator (Accept 0.52 0.22 0.245 0.25 0.22 + + × their p their p or their ) = 0.471 or 49 104 A1 3

Question 27

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 28

(a) Find the total number of diﬀerent arrangements of the 8 letters in the word TOMORROW. [2] .............................................…

Accepted Answer

6(a) 8! 2!3! M1 8! ! ! × k m k = 1 or 2, m = 1 or 3, not k = m = 1 no additional terms 3360 A1 2 Question Answer Marks Guidance 6(b) Method 1 Arrangements Rs at ends – Arrangements Rs at ends and Os together [Os not together = ] 6! 3! – 4! M1 6! ! k – m, 1 ⩽ k ⩽ 3, m an integer, condone 6! 2 !   × −     m k . M1 w – 4! or w – 24, w an integer Condone w – 2 × 4! 96 A1 Method 2 identified scenarios R _ _ _ R, Arrangement No Os together + 2Os and a single O 4C3 × 3! + 4C2 × 2 × 3! M1 4C3 × 3! + r or 4× 3! + r or 4P3 × 3! + r, r an integer. Condone 2 × 4C3 × 3! + r. 2 × 4× 3! + r or 2 × 4P3 × 3! + r. M1 q + 4C2 × 3! × k or q + 4P2 × 3! × k, k = 1,2, q an integer [24 + 72 =] 96 A1 3 6(c) Method 1 Identified scenarios OORR 3 2 3 2 2 0 C C C 3 1 3   × × = × =   ORR_ 3 2 3 1 2 1 C C C 3 1 3 9 × × = × × = OOR_ 3 2 3 2 1 1 C C C 3 2 3 18 × × = × × = OR_ _ 3 2 3 1 1 2 C C C 3 2 3 18 × × = × × = OOOR 3 2 3 3 1 0 C C C 1 2 2   × × = × =   B1 Outcomes for 2 identifiable scenarios correct, accept unsimplified. M1 Add 4 or 5 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. Total 50 A1 All correct and added Probability = 8 4 50 C M1 8 4 '50' their C , accept numerator unevaluated Question Answer Marks Guidance 6(c) cont’d 50 or 0.714 70 A1 Method 2 Identified outcomes ORTM 3 2 1 1 C C × = 6 ORTW 3 2 1 1 C C × = 6 ORMW 3 2 1 1 C C × = 6 ORRM 3 2 1 2 C C × = 3 ORRW 3 2 1 2 C C × = 3 ORRT 3 2 1 2 C C × = 3 OROR 3 2 2 2 C C × = 3 OROT 3 2 2 1 C C × = 6 OROM 3 2 2 1 C C × = 6 OROW 3 2 2 1 C C × = 6 OROO 3 2 3 1 C C × = 2 B1 Outcomes for 5 identifiable scenarios correct, accept unsimplified. M1 Add 9, 10 or 11 identified correct scenarios only values, no additional incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. Total 50 A1 All correct and added Probability = 8 4 50 C M1 8 4 '50' their C , accept numerator unevaluated. 50 or 0.714 70 A1 5

Question 29

Three fair six-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown at the same time, repeatedly. For a single throw of the thre…

Accepted Answer

4(a) [Possible cases: 1 1 2, 1 2 1, 2 1 1] Probability = 3 1 3 6  ×     3 1 6  ×     k , where k is an integer. M1 Multiply a probability by 3, not +, – or ÷ 1 72 A1 Accept 3 216 or 0.0138 or 0.0139 3 4(b) P(18) 3 1 1 6 216    = =       B1 P(18 on 5th throw) 4 215 1 216 216   = ×     M1 (1 – p)4p, 0 < their p < 1 0.00454 A1 3

Question 30

In the region of Arka, the total number of households in the three villages Reeta, Shan and Teber is 800. Each of the households was asked …

Accepted Answer

7(a)(i) 40 1 or or 0.05 800 20 B1 1 7(a)(ii) 177 223 177 40 + + M1 Their 223 + 177 + 40 seen as denominator of fraction in the final answer, accept unsimplified 177 440 or 0.402 A1 CAO Alternative method for Question 7(a)(ii) ( ) ( ) ( ) 177 177 800 800 G | S 223 177 40 440 800 800 ∩ = = = + + P G S P P S = 177 800 11 or 0.55 20 M1 Their P(S) seen as denominator of fraction in the final answer, accept unsimplified 177 440 or 0.402 A1 CAO 2 7(b)(i) P(0, 1, 2) = 10C0 ( ) 0 0.35 ( ) 10 0.65 + 10C1 ( ) 1 0.35 ( ) 9 0.65 + 10C2 ( ) 2 0.35 ( ) 8 0.65 M1 One term:10Cx px (1 – p)10–x for 0 < x < 10, any 032) = P( Z > 32.5 42) 27.3 − = P(Z > - 1.818) M1 Substituting their mean and variance into ±standardisation formula (any number), condone σ2 or √σ M1 Using continuity correction 31.5 or 32.5 ( ) Φ 1.818 M1 Appropriate area Φ, from final process, must be probability 0.966 A1 0.965 ⩽ p ⩽ 0.966 5

Question 31

Two fair coins are thrown at the same time. The random variable X is the number of throws of the two coins required to obtain two tails at …

Accepted Answer

1(a) 6 3 1 4 4       ) 6 1 , 0 1 − < < p p p 0.0445, 729 16384 A1 2 1(b) 9 3 4       M1 3 , 0 1, 8, 9,1 0 4   < < =     n n or p p n 0.0751, 19683 262144 A1 2

Question 32

For her bedtime drink, Suki has either chocolate, tea or milk with probabilities 0.45, 0.35 and 0.2 respectively. When she has chocolate, t…

Accepted Answer

3 ( ) ( ) ( ) P P | P T B T B B   ∩ =      ′  ′ ′ ( ) P 0.45 0.7 0.35 0.4 0.2 1 B × ′ = × + + × 131 0.655, 200   =     M1 [ ] 0.45 0.35 0.2 1 , 0.7,0.3 0.4,0.6 × + × + × = = a b a b , seen anywhere. A1 Correct, accept unsimplified. ( ) P ' 0.35 0.4 T B ∩ = × [= 0.14, 7 50 ] M1 Seen as numerator or denominator of a fraction. ( ) 0.14 P | ' 0.655 their T B their = M1 Values substituted into conditional probability formula correctly. Accept unsimplified. Denominator sum of 3 two-factor probabilities (condone omission of 1 from final factor). If clearly identified, condone from incomplete denominator. 0.214, 28 131 A1 If 0 marks awarded, SC B1 0.214 WWW. 5

Question 33

Raman and Sanjay are members of a quiz team which has 9 members in total. Two photographs of the quiz team are to be taken. For the ﬁrst ph…

Accepted Answer

5(a) [8! =] 40 320 B1 Evaluated, exact value only. 1 5(b) Method 1 [^ ^ ^ R ^ ^ S ^ ^] 7! × 8C2 × 2 M1 7! × k seen, k an integer > 1. M1 ( ) 1 × − m n n or 2 × n m C or 2 × n m P , n = 7, 8 or 9, m an integer > 1. 282 240 A1 Exact value only. SC B1 for final answer 282 240 WWW. Method 2 [Total number of arrangements – Arrangements with R & S together] 9! – 8! × 2 M1 9! – k, k an integer < 362 880 . M1 m – 8! × n, m an integer > 40 320, n = 1,2. 282 240 A1 Exact value only. SC B1 for final answer 282 240 WWW. 3 5(c) 9C5 [× 4C4] M1 9Cx [× 9–xC9–x,] x = 4, 5. Condone × 1 for 9–xC9–x. Condone use of P. 126 A1 WWW 2 Question Answer Marks Guidance 5(d) [Number of ways with Raman and Sanjay together on back row =] 7C3 [Number of ways with Raman and Sanjay together on front row =] 7C2 M1 7Cx seen, x = 3 or 2. [Total =] 35 + 21 M1 Summing two correct scenarios. 56 A1 Evaluated – may be seen used in probability. If M0 scored, SC B1 for 56 WWW. Probability = ( ) 56 56 4 , 126 9 = their their c , 0.444 B1 FT FT their 56 from adding 2 or more scenarios in numerator and their (c) or correct as denominator. 4

Question 34

The times, in minutes, that Karli spends each day on social media are normally distributed with mean 125 and standard deviation 24. (a) (i)…

Accepted Answer

7(a)(i) P(X > 142) = P 142 125 24 Z −   >     M1 Substitution of correct values into the ±Standardisation formula, allow continuity correction, not σ2, √σ. [= P( 0.7083) ]1 0.7604 > = − Z M1 Appropriate numerical area Φ, from final process, must be probability, expect p < 0.5. 0.2396 A1 0.239 ⩽ p ⩽ 0.240 to at least 3sf. Their 0.2396 × 365 [= 87.454] M1 FT their 4sf (or better) probability. 87 or 88 A1 FT Final answer must be positive integer, no indication of approximation/rounding, only dependent on previous M mark. SC B1 FT for their 3sf probability × 365 = integer value, condone 0.24 used. 5 7(a)(ii) P(0, 1) = 0.760410 + 10C1 × 0.23961 × 0.76049 [= 0.064628 + 0.20364] M1 One term: 10Cx px (1 – p)10–x for 0 < x < 10, any p. A1 FT Correct unsimplified expression using their probability to at least 3sf from (a)(i) or correct. 0.268 A1 AWRT, WWW. 3 7(b) 1.282 = ± z B1 Correct value only, critical value. 125 1.282 24 − = − t M1 Use of ± Standardisation formula with correct values substituted, allow continuity correction, σ2, √ σ, to form an equation with a z-value and not probability. 94.2 = t A1 AWRT, condone AWRT 94.3. Not dependent on B mark. 3

Question 35

Each of the 180 students at a college plays exactly one of the piano, the guitar and the drums. The numbers of male and female students who…

Accepted Answer

1(a) 82 41 , 180 90 , 0.456 1 1(b) ( ) ( ) ( ) P M|D P M D P D   ∩ =       = 11 0.6011 180 or 20 11 0.1722 180 180 + M1 Their identified ( ) ( ) P M D P D ∩ 11 from data table 20 11 or + , accept unsimplified, condone × 180. 11 31, 0.355 A1 Final answer. 2 Question Answer Marks Guidance 1(c) P(F) = 100 5 , , 0.5556 180 9 OE P(G) = 82 41 , 0.4556 180 90 OE P(F∩G) = 38 19 , , 0.2111 180 90 OE ( ) ( ) 100 82 41 38 P F P G , 0.2531 OE 180 180 162 180   × = × = ≠     Not independent M1 Their identified P(F) × their identified P(G) or correct seen, can be unsimplified. A1 ( ) ( ) ( ) 41 38 , ,P F G and P F P G 162 180 ∩ × seen with correct conclusion, WWW. Values and labels must be seen. Alternative method for question 1(c) P(F∩G) = 38 19 , ,0.2111 180 90 OE P(G) = 82 41 , ,0.4556 180 90 OE ( ) 38 19 180 P F|G , 0.4634 82 41 180 = = OE ( ) 100 5 P F , 180 9 ≠ = ,0.5556 OE Not independent M1 P(F|G) (OE) unsimplified with their identified probs or correct A1 ( ) ( ) 19 100 , , P F G and P F|G 41 180 ∩ seen with correct conclusion WWW. Values and labels must be seen. 2

Question 36

In a certain region, the probability that any given day in October is wet is 0.16, independently of other days. (a) Find the probability th…

Accepted Answer

5(a) [= 0.17490 + 0.333145 + 0.28555] A1 Correct unsimplified expression, or better. 0.794 A1 0.7935 < p ⩽ 0.794, mark at most accurate. If M0 scored, SC B1 for final answer 0.794. 3 5(b) ( ) 7 0.84 0.16 M1 (1 – p)7p, 0 < p < 1 0.0472 A1 0.0472144 to at least 3sf. 2 Question Answer Marks Guidance 5(c) 4 × 0.0472 × (1 – 0.0472)3 M1 4 × q(1 – q)3, q = their (b) or correct. 0.163 A1 0.163 ⩽ p ⩽ 0.1634, mark at most accurate from their probability to at least 3sf. 2

Question 37

A security code consists of 2 letters followed by a 4-digit number. The letters are chosen from {A, B, C, D, E} and the digits are chosen f…

Accepted Answer

5(a) M1 16 800 A1 2 Question Answer Marks Guidance 5(b) Method 1 [Identify scenarios] With A and no 5: 8 × 6P4 or (1 × 4 × 6 ×5 × 4 × 3) ×2 or 4C1 × 2! × 6P4 = 2880 With 5 and no A: 4P2 × 4 × 6P3 or (4 × 3 × 1 × 6 × 5 × 4) × 4 or 4P2 × 6C3 × 4! = 5760 With A and 5: 8× 4 × 6P3 or (4 × 1 × 1 × 6 × 5 × 4) × 8 or 4C1 × 2! × 6C3 × 4! = 3840 M1 One number of ways correct, accept unsimplified. M1 Add 2 or 3 identified correct scenarios only, accept unsimplified. [Total =] 12 480 A1 CAO Method 2 [total number of codes – number of codes with no A or 5] No A or 5 : (4 × 3 ) ×( 6 × 5 × 4 × 3) = 4320 M1 4P2 × 6P4 or 4C2 × 6C4 seen, accept unsimplified. Required number = their (a) – their 4320 M1 Their 5(a) (or correct) – their (No A or 5) value. 12 480 A1 Method 3 [subtracting double counting] With A 4P1 × 7P4 × 2 or 4C1 × 2 × 7C4 × 4! = 6720 With 5 5P2 × 6P3 × 4 or 5C2 × 2 × 6C3 × 4! = 9600 With A and 5 = 4P1 × 6P3 × 8or 4C1 × 2! × 6C3 × 4! × 8 = 3840 M1 One outcome correct, accept unsimplified. Required number = 6720 + 9600 – 3840 M1 Adding ‘with a’ to ‘with 5’ and subtracting ‘A and 5’. 12 480 A1 CAO 3 Question Answer Marks Guidance 5(c) Method 1 – number of successful codes divided by total (1 ×) 3 × 5P2 M1 3 × 5Pn, n = 2, 3. Condone 3 × 5C2, no + or –. Probability = 3 5 2 1 6 800 their P their × M1 Probability = 60 1 6 800 their their . 1 280 , 0.00357 A1 Method 2 – product of probabilities of each part of code 1 1 1 3 5 4 5 4 7 6 5 4   × × × × ×     or 1 1 3 5 2 5 4 7 4 P P × × × M1 1 1 5 4 k × × where 0 1 k < < for considering letters. M1 t 1 3 7 6 × × or 3 5 2 7 4 P t P × × where 0 1 t < < . 1 280 A1 CAO 3

Question 38

Box A contains 6 red balls and 4 blue balls. Box B contains x red balls and 9 blue balls. A ball is chosen at random from box A and placed …

Accepted Answer

7(a) Probabilities: 1 10 x x + + , 9 10 x + , 10 x x + , 10 10 x + B1 One probability correct in correct position. B1 Another probability correct in correct position. B1 Other two probabilities correct in correct positions. 3 7(b) 4 10 10 10 their x × + M1 Method consistent with their tree diagram. 4 10 x + A1 AG 2 Question Answer Marks Guidance 7(c) 4 1 10 6 x = + 10 24, 14 x x + = = B1 Find value of x. Can be implied by correct probabilities in calculation. P(ARed|BRed) = P(ARed ∩ BRed) ÷ P(BRed) 6 1 10 10 6 1 4 10 10 10 10 x their x x x their their x x + × + + × + × + + = 6 15 10 24 6 15 4 14 10 24 10 24 × × + × = 3 8 73 120 B1 FT 6 1 10 10 x their x + × + as numerator or denominator of fraction. M1 6 1 4 10 10 10 10 x x their their x x + × + × + + seen anywhere. A1 FT Seen as denominator of fraction. 45 73 , 0.616[4…] A1 If B0 M0: SC B1 for 3 8 73 120 or 0.375 0.6083 SC B1 45 73 or 0.616. 5

Question 39

In a certain country, the probability of more than 10cm of rain on any particular day is 0.18, independently of the weather on any other da…

Accepted Answer

2(a) [P(>2) = 1 – P(0,1,2) =] 1 – (7C0 0 7 0.18 0.82 + 7C1 1 6 0.18 0.82 + 7C2 2 5 0.18 0.82 ) M1 One term 7Cx ( ) 7 1 , 0 1, 0 7 − − < < < < x x p p p x = 1 – (0.249285 + 0.383048 + 0.252251) = 1 – 0.88458 A1 Correct unsimplified expression or better Condone omission of brackets if recovered 0.115 B1 WWW. 0∙115 ⩽ p < 0∙1155 not from wrong working 3 2(b) [P(at least 1 day of rain) = 1 – P(0) = ( ) 7 1 0.82 ] 0.7507 − = B1 AWRT 0.751 seen [P(exactly 2 periods) =] ( ) 2 0.7507 1 0.7507 3 × − × M1 FT their 7 1−p or their 0.7507 if identified, not 0.18, 0.82 Accept ×3Cr, r=1,2 or ×3P1 for ×3 Condone ×2 0.421 A1 Accept 0.421 ⩽ p ⩽ 0.4215 SC B1 if 0/3 scored for final answer only 0.421 ⩽ p ⩽ 0.4215 3

Question 40

The weights of male leopards in a particular region are normally distributed with mean 55kg and standard deviation 6kg. (a) Find the probab…

Accepted Answer

4(a) P( 46 55 62 55 46 62) P 6 6 X Z − −   < < = < <     M1 46 or 62, 55 and 6 substituted into ±standardisation formula once. Condone 62 and continuity correction ±0.5 7 P 1.5 6   = − < <     Z B1 Both standardisation values correct, accept unsimplified ( ) ( ) 7 =Φ 1 Φ 1.5 6    − −         = ( ) 0.8784 0.9332 1 + − M1 Calculating the appropriate area from stated Φs of z-values, must be probabilities. 0.812 A1 0∙8115 < p ⩽ 0∙812 4 4(b) z = ±0.674 B1 CAO, critical z-value 36 42 0.674 σ − = − M1 36 and 42 substituted in ±standardisation formula, no continuity correction, not σ2, √ σ, equated to a z-value [ ] 8.9 0 σ = A1 WWW. Only dependent on M. 3 Question Answer Marks Guidance 4(c) P(male < 46) = 1−their 0.9332 = 0.0668 M1 FT value from part (a) or Correct: 46 55 1 Φ 6 −   −     ,condone continuity correction, σ2, √ σ, and probability found. Condone unsupported correct value stated. P(female < 46) = P( ( ) 46 42 ) Φ 0.449 8.90 − < =    Z their 0.6732 = M1 46, 42 and their 4(b) σ (or correct σ) substituted in ±standardisation formula, condone continuity correction, σ2, √ σ, and probability found Condone 4 8.90 their . P(both) = 0.0668 ×0.6732 M1 Product of their 2 probabilities (0 < both < 1) Not 0.25 or their final answer to 4(a) used. 0.0450 or 0.0449 A1 0∙0449 ⩽ p ⩽ 0∙0450 4

Question 41

A factory produces chocolates in three ﬂavours: lemon, orange and strawberry in the ratio 3 : 5 : 7 respectively. Nell checks the chocolate…

Accepted Answer

6(a) [Probability of lemon = 3 1] 15 5 = 6 4 1 4096 , 0.0524 5 5 78125    × =           B1 0.0524288 rounded to more than 3SF if final answer 1 6(b) 6 1 1 5   −     M1 or 6 4 5       . FT their 1 5 or correct. From final answer Condone 5 5 6 4 1 4 4 or 5 5 5 5         × +                 4096 , 15625 0.262 A1 0.262144 rounded to more than 3SF Alternative method for question 6(b) [1 – P(1,2,3,4,5,[6]) =] 1 – 2 3 4 5 1 4 1 4 1 4 1 4 1 4 1 5 5 5 5 5 5 5 5 5 5 5           + × + × + × + × + ×                       M1 From final answer Condone omission of 5 4 1 5 5  ×     4096 , 15625 0.262 A1 0.262144 rounded to more than 3SF 2 Question Answer Marks Guidance 6(c) 10 9 8 15 14 13 × × M1 1 2 15 14 13 − − × × a a a , no additional terms 24 91 , 0∙264 A1 0.263736 rounded to more than 3SF Alternative method for question 6(c) 3 2 1 3 2 7 3 15 14 13 15 14 13 × × + × × × 3 7 6 7 6 5 3 15 14 13 15 14 13 + × × × + × × M1 [3Ls + 2Ls1S + 1L2Ss + 3Ss] Condone one numerator error. Condone no multiplications seen if tree diagram complete with probabilities on each branch, scenarios listed and attempt at evaluation 24 91 , 0∙264 A1 0.263736 rounded to more than 3SF Alternative method for question 6(c) 5 4 3 5 4 10 5 10 9 1 3 3 15 14 13 15 14 13 15 14 13   − × × + × × × + × × ×     M1 1 ‒ P(3,2,1 oranges) Condone one numerator error. 24 91 , 0∙264 A1 0.263736 rounded to more than 3SF Alternative method for question 6(c) 10 3 15 3 C c M1 24 91 , 0∙264 A1 0.263736 rounded to more than 3SF 2 Question Answer Marks Guidance 6(d) 7 5 3 3! 15 14 13 × × × M1 All probabilities of the form: 7 5 3 × × a b c , 13 ⩽ a,b,c ⩽ 15 M1 3! × × × e g i f h j e,f,g,h,i.j positive integers forming probabilities or 6 identical probability calculations or values added, no additional terms 3 13 , 0∙231 A1 0∙230769 rounded (not truncated) to more than 3SF Alternative method for question 6(d) 3 5 7 1 1 1 15 3 C C C C × × M1 3 5 7 1 1 1 C C C × × k , k integer > 1 Condone use of permutations M1 3 5 7 15 3 C C C C × × a b c , 0<a<3, 0<b<5, 0<c<7, Condone use of permutations 3 13 , 0∙231 A1 0∙230769 rounded (not truncated) to more than 3SF 3 Question Answer Marks Guidance 6(e) ( ) 7 6 5 3 7 6 3 15 14 13 15 14 13 × × + × × × their c 14 24 65 91   = ÷     B1 3 7 6 3 15 14 13 × × × seen (SSL, SLS, LSS) SC B1 3 126 3, 3 65 2730 × × seen B1 7 6 5 15 14 13 × × seen in numerator (SSS) SCB1 210 1 , 2730 13 seen in numerator M1 Fraction with their (c) or correct in denominator 720 24 , , 0.263736 2730 91       = 49 60 , 0∙817 A1 Accept 0.816 Alternative method for question 6(e) 7 3 7 2 1 3 10 3 C C C C × + B1 7 3 2 1 C C × seen (SSL, SLS, LSS) SCB1 21 × 3 seen or use of permutations B1 7 3 C seen in numerator (SSS) SCB1 35 seen in numerator or use of permutations M1 Fraction with 10 3 C or consistent with their numerator of 6(c) in denominator = 49 60 , 0∙817 A1 Accept 0.816 4

Question 42

Janice is playing a computer game. She has to complete level 1 and level 2 to ﬁnish the game. She is allowed at most two attempts at any le…

Accepted Answer

6(a) 0.6 0.4 0.3 0.72 or 1 – 0.4 x 0.7 = 0.72 1 6(b)   0.72 0.4 0.6 0.2    M1 0.72  u, 0 < u < 1 M1 v  (0.4 + 0.6  0.2), or v  (1 – 0.6  0.8) 0 < v ⩽ 1 no additional terms SC B1 for 0.72  (0.4 + 0.12) or 0.72  (1 – 0.48) 0.3744 A1 WWW. Condone 0.374. SC B1 for 0.3744 only 3 Alternative method for question 6(b) [p(P1P2) + p(F1P1P2) + p(P1F2P2) + p(F1P1F2P2)] = 0.6  0.4 + 0.4  0.3  0.4 + 0.6  0.6  0.2 + 0.4  0.3  0.6  0.2 M1 Any two terms unsimplified and correct M1 Summing 4 appropriate scenarios by listing or on a tree diagram SC B1 for 0.24 + 0.048 + 0.072 + 0.0144 0.3744 A1 WWW. Condone 0.374. SC B1 for 0.3744 only 3 Question Answer Marks Guidance 6(c) (fails first or second level finishes game) (fails first or second level | finishes game)  P P their(b) Numerator = P(S SF) + P(FS S) = 0.6  0.6  0.2 + 0.4  0.3  0.4 = 0.072 + 0.048 = 0.12 Required probability = 0.12 their(b) M1 Either 0.6  0.6  0.2 or 0.4  0.3  0.4 seen Condone 0.072 or 0.048 if seen in (b) A1 Both correct accept unsimplified expression. No additional terms M1 sum of two 3-term probabilities as numerator or correct Their their (b) 0.321 or 25 78 A1 0.3205 < p ⩽ 0.321 4

Question 43

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 44

Hanna buys 12 hollow chocolate eggs that each contain a sweet. The eggs look identical but Hanna knows that 3 contain a red sweet, 4 contai…

Accepted Answer

7(a) YYY: 5 4 3 60 1 , 12 11 10 1320 22    OOO: 4 3 2 24 1 , 12 11 10 1320 55    RRR: 3 2 1 6 1 , 12 11 10 1320 220    or a  (a –1)  ( a – 2), a = 5, 4, 3 in numerator seen in at least one expression. A1 One expression 1 2 12 11 10 a a a     , a = 5, 4, 3 (consistent in expression). Correct order of values in the numerator is essential. M1 5 4 3 4 3 2 3 2 1 12 12 12 d e d e d e         , either d = 11, e = 10 or d = 12, e = 12. Condone 1 1 1 OE 22 55 220   [Total =] 90 3 , , 0.0682 1320 44 A1 0.06818. Dependent only upon the second M mark. Question Answer Marks Guidance 7(a) Alternative method for question 7(a) YYY: 5 3 12 3 C 10 1 , 220 22 C  OOO: 4 3 12 3 C 4 1 , 220 55 C  RRR: 3 3 12 3 C 1 220 C  M1 Either 12 3 C in denominator or 3 C a in numerator seen in at least one expression. A1 One expression 3 12 3 C C a a = 5, 4, 3 M1 5 4 3 3 3 3 12 12 12 3 3 3 C C C C C C   Condone 1 1 1 OE 22 55 220   [Total =] 90 3 , , 0.0682 1320 44 A1 0.06818. Dependent only upon the second M mark. 4 7(b) [P(YYY | all same colour) =] 60 90 1320 1320  M1    60 1 P YYY or or 1320 22 90 3 7 a or or 1320 44 their their 2 3 , 0.667 A1 OE 2 Question Answer Marks Guidance 7(c) In each method, the M mark requires the scenarios to be identifiable. This may be implied by a list of scenarios and then the calculations which will be assumed to be in the same order. A correct value/expression will be condoned as identifying the connected scenario. Method 1 [1 – no orange = ]1−8 7 6 12 11 10   or 8 3 12 3 C 1 C  = 14 1 55  B1 8 7 6 12 11 10   or 8 3 12 3 C C seen, condone 336 56 or 1320 220 only, not OE. M1 1 12 f g h d e    Either d = 11, e = 10 or d = 12, e = 12 or 1 - 8 3 12 3 C C . Condone 14 1 55  OE (not 41) 55 . 41 55 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. Question Answer Marks Guidance 7(c) Method 2 P(1 O)= 4 3 2 4 5 4 672 12 11 10 12 11 10 3 4 5 3 1320 2 12 11 10                        P(2O) = 4 3 8 288 3 12 11 10 1320    P(3O) = 24 1320 B1 P(1 O)or P(2 O) correct, accept unsimplified. M1 3 correct scenarios added, with at least one 3-term product of form 12 f g h d e   seen, either d = 11, e = 10 or d = 12, e = 12. [Total =] 984 41 1320 55  , 0.745 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. Method 3 O Y R = 4 5 3 1 1 1 C C C   = 60 O R R = 4 3 1 2 C C  = 12 O Y Y = 4 5 1 2 C C  = 40 O O Y = 4 5 2 1 C C  = 30 O O R = 4 3 2 1 C C  = 18 O O O = 4 3 C = 4 Total = 164 Prob = 12 3 164 C B1 Number of ways either 1 or 2 orange sweets obtained correctly (112 or 48). Accept unsimplified Note 4C1  8C2 = 112 or 4C2  8C1 = 48 are correct alternatives. M1 3 correct scenarios (1, 2 or 3 orange sweets) added on numerator, denominator 12 3 C 984 41 1320 55  , 0.745 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. Question Answer Marks Guidance 7(c) Method 4 P(R R O) = 3 2 4 1 12 11 10 55    P(R O ) = 3 4 1 12 11 11   P(R Y O) = 3 5 4 1 12 11 10 22    P(O ) = 4 1 12 3  P(Y R O) = 5 3 4 1 12 11 10 22    P(Y O ) = 5 4 5 12 11 33   P(Y Y O) = 5 4 4 2 12 11 10 33    B1 P(R ^ ^) = 17 110 or P(Y ^ ^) = 17 66 . Accept unsimplified. M1 3 correct scenarios added, with at least one 3-term product of form 12 f g h d e   seen, either d = 11, e = 10 or d = 12, e = 12. 984 41 1320 55  , 0.745 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. Question Answer Marks Guidance 7(c) Method 5 P(O ) = 4 1 12 3  P(^ O ) = 8 4 8 12 11 33   P(^ ^ O) = 8 7 4 28 12 11 10 165    B1 P(^ O ) = 8 33 or P(^ ^ O) = 28 165 . Accept unsimplified. M1 3 correct scenarios added, with at least one 3-term product of form 12 f g h d e   seen, either d = 11, e = 10 or d = 12, e = 12 with correct numerator. 984 41 1320 55  , 0.745 A1 0.745 ⩽ p ⩽ 0.74545 If M0 scored SC B1 0.745 ⩽ p ⩽ 0.74545. 3

Question 45

Ramesh throws an ordinary fair 6-sided die. (a) Find the probability that he obtains a 4 for the ﬁrst time on his 8th throw. [1] ..........…

Accepted Answer

4(a) 7 5 1 78125 0.0465, 6 6 1679616                1 Question Answer Marks Guidance 4(b) P(X < 6) = 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                                         M1 1 – pn, 0 < p < 1, n = 4, 5, 6 or sum of 4, 5 or 6 terms   1   n p p for  0,1 ,2,3,4 5  n . 0.598, 4651 7776 A1 2 4(c) [Probability of total less than 4 is] 3 1 or 36 12 B1 SOI [1 – P(0, 1, 2)] =1 (  10C0 0 10 1 11 12 12          + 10C1 1 9 1 11 12 12          + 10C2 2 8 1 11 12 12          ) M1 One term 10Cx   10 1   x x p p , for 0 < x < 10, 0 < p < 1. 1 – (0.418904 + 0.380822 + 0.155791) A1 FT Correct expression. Accept unsimplified. 0.0445 A1 0.04448 ⩽ p ⩽ 0.0445 4

Question 46

Sajid is practising for a long jump competition. He counts any jump that is longer than 6m as a success. On any day, the probability that h…

Accepted Answer

6(a) and outcomes identified. B1 Third jump correct with probabilities and outcomes identified. 2 6(b) SFF 0.2 0.7 0.9 0.126    FSF 0.8 0.1 0.7 0.056    FFS 0.8 0.9 0.1 0.072    M1 Two or three correct 3 factor probabilities added, correct or FT from part 6(a). Accept unsimplified. [Total = probability of 1 success =] 0.254 127 500       A1 Accept unsimplified. [Probability of at least 1 success = 1−0.8 0.9 0.9 ]0.352    44 125       B1 FT Accept unsimplified. P(exactly 1 success | at least 1 success)= 0.254 0.352 their their M1 Accept unsimplified. 0.722, 127 176 A1 0.7215 < p ⩽ 0.722 5 S S F S F F S F S S S F F F 0.2 0.8 0.1 0.9 0.1 0.9 0.3 0.7 0.3 0.7 0.3 0.7 0.1 0.9 1st 2nd 3rd Question Answer Marks Guidance 6(c) 0.8 0.9 0.9 0.1 0.3 0.3 0.005832       [FFFSSS] 0.2 0.3 0.3 0.7 0.9 0.9 0.010206       [SSSFFF] M1 a × b × c × d × e × f FT from their tree diagram. Either a, b and c all = 0.8 or 0.9 (at least one of each) and d, e and f all = 0.1 or 0.3 (at least one of each). Or a, b, c = 0.2 or 0.3 (at least one of each) and d, e, f = 0.7 or 0.9 (at least one of each). A1 Either correct. Accept unsimplified. [Total =] 0.0160[38] A1 3

Question 47

The residents of Persham were surveyed about the reliability of their internet service. 12% rated the service as ‘poor’, 36% rated it as ‘s…

Accepted Answer

2(a) [P(3, 4, …7) = 1 – P(0, 1, 2, 8)] M1 x 8 − x One term 8Cx p (1 − p ) , for 0 < x < 8, 0 < p < 1 = 1 − ( 8C0 0.480 0.528 + 8C1 0.481 0.52 7 + 8C2 0.482 0.526 + 8C8 0.488 0.52 0 ) = 1 – (0.00534597 + 0.039478 + 0.127544 + 0.0028179) A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.825 B1 Mark the final answer at the most accurate value. 0.8248 < p ⩽ 0.825 WWW. Alternative method for Question 2(a) [P(3, 4, 5, 6, 7) =] M1 x 8 − x One term 8Cx p (1 − p ) , for 0<x<8, 0<p<1 8C3 0.483 0.525 + 8C4 0.484 0.52 4 + 8C5 0.485 0.523 + 8C6 0.486 0.52 2 + 8C7 0.487 0.521 A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.825 B1 Final answer 0.8248 < p ⩽ 0.825 WWW. 3 2(b) [Mean = 0.52  125 =]65, B1 65 and 31.2 seen, allow unsimplified. May be seen in  var = 0.52  0.48  125 =  31.2 standardisation formula. (5.585 < σ ⩽ 5.586 imply correct variance). 72.5 − M1 Substituting their 65 and their 31.2 into ±standardisation [P(X > 72) = ]P( Z  65) [= P( Z  1.343 )] 31.2 formula (any number for 72∙5), not their 31.2, √ their 5.586 . M1 Using continuity correction 72∙5 or 71∙5 in their standardisation formula . 7.5 7.5 Note or seen gains M2 BOD 31.2 5.586 = 1 – 0.9104 M1 Appropriate area Φ, from final process, must be probability. 0.0896 A1 0.0896 ⩽ p ⩽ 0.0897 WWW. 5

Question 48

On any day, Kino travels to school by bus, by car or on foot with probabilities 0.2, 0.1 and 0.7 respectively. The probability that he is l…

Accepted Answer

Question Answer Marks Guidance 1(a) 0.2 +x 0.1 2 x + 0.7  0.25 = 0.235 M1 0.2 +x 0.1 2 x + 0.7  0.25 or 0.2x + 0.2x + 0.175 seen. M1 Equating their 3 term expression (2 terms involving x) to 0.235 x = 0.15 A1 3 1(b)  P ( car and notlate )  M1 0.1  (1 − 2 their x ) or 0.1  0.7 as numerator  P (car |notlate ) =   P ( notlate )  and 0.2  (1 – their x) + 0.1  (1 – 2 × their x ) + 0.7  0.75 with values 0.1  (1 − 0.3 ) substituted or 1 – 0.235 or 0.765 as denominator of fraction. 1 − 0.235 Condone 0.2  (1 – their x) + 0.1  (1 –  their x) + 0.7  0.75 as denominator consistent with 1(a).  0.07  70 14 A1 0.091503267 to at least 3SF. = 0.0915, ,   If M0 scored SC B1 for 0.091503267 to at least 3SF.  0.765  765 153 2

Question 49

Three fair 6-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown at the same time repeatedly. The score on each throw is the su…

Accepted Answer

3(a) 4 1 B1 May be seen used in calculation. [P(17 or 18) =] = ,0.0185(185…) 216 54 5 M1 p(1 – p)5, 0 < p < 1  53  1 P(X = 6) =   .  54  54 0.0169 A1 0.01686 < p ⩽ 0.0169 If A0 scored SC B1 for 0.01686 < p ⩽ 0.0169 3 3(b) 7 M1 r  53    53   [P(X < 8) =] 1 −  1 − their  or 0.98148  or correct  ,  54    54   r = 7,8 0 < their p < 1 0.123 A1 0.1225 ⩽ p ⩽ 0.123 Alternative method for Question 3(b) [P(X < 8) =] M1 q + pq + p 2 q + p 3 q + p 4 q + p 5 q  + p 6 q  , p + q = 1, 0  p , q  1, q 2 3 4    1   53  1   53  1   53  1   53  1    +    +    +    +    + 53 = their  54   54  54   54  54   54  54   54  54  54  53 5 1   53 6 1  +        54  54   54  54  0.123 A1 0.1225 ⩽ p ⩽ 0.123 2

Question 50

At a company’s call centre, 90% of callers are connected immediately to a representative. A random sample of 12 callers is chosen. (a) Find…

Accepted Answer

6(a) [1 – P(10, 11, 12) =] M1 x 12 − x One term 12Cx p (1 − p ) , for 0 < x < 12, 0 < p < 1 1 − ( 12C10 0.910 0.12 + 12C11 0.911 0.11 + 12C12 0.912 0.10 ) = 1 – (0.230128 + 0.376573 + 0.282430) A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. 0.111 B1 Mark the final answer at the most accurate value, 0.1108 < p ⩽ 0.111 WWW. Alternative method for Question 6(a) [P(0,1,2,3,4,5,6,7,8,9) =] M1 x 12 − x One term 12Cx p (1 − p ) , for 0 < x < 12, 0 < p < 1 12C0 0.9 0 0.112 +12C1 0.91 0.111 +12C2 0.9 2 0.110 +12C3 0.93 0.19 +12C4 0.9 4 0.18 +12C5 0.95 0.17 +12C6 0.96 0.16 +12C7 0.97 0.15 +12C8 A1 Correct expression, accept unsimplified, no terms omitted, leading 0.98 0.14 +12C9 0.99 0.13 ) to final answer. If answer correct condone omission of any 7 of the 8 middle terms. 0.111 B1 Final answer 0.1108 < p ⩽ 0.111 WWW. 3 6(b) [Mean = 80  0.9 =] 72, B1 72 and 7.2 seen, allow unsimplified.  Variance = 80  0.9  0.1 = 7.2 May be seen in standardisation formula. (2.683 ⩽ σ < 2.684 imply correct variance). 69.5 − M1 Substituting their mean and their variance into ±standardisation P(X > 69) = P( Z  72) 7.2 formula (any number for 69∙5), not their 7.2, not √their 2.683 M1 Using continuity correction 69∙5 or 68∙5 in their standardisation formula. [= P( Z −0.9317) =] M1 Appropriate area Φ, from final process, must be probability. Φ ( 0.9317 ) 0.824 A1 0.8239 ⩽ p ⩽ 0.8243 WWW. 5 6(c) np = 72, nq = 8 Both greater than 5, [so approximation is valid] B1 np, nq evaluated accurately. both np & nq referenced correctly. > 5 or greater than 5 seen. 1

Question 51

(a) Find the number of diﬀerent arrangements of the 9 letters in the word ALLIGATOR in which the two As are together and the two Ls are tog…

Accepted Answer

7(a) 7! M1 7! b,c = 1,2 b ! c ! 2! 2! 7!   oe, no further terms present. 2! 2! 5040 A1 2 7(b) Method 1 for first 3 marks: Arrangements of 6 letters including Ls between As 5!  5  2 M1 5!  d, d integer > 1 M1 e!  f  g, e = 5, 6, 7; f = 1, 5; g = 1, 2; f ≠ g, 1 can be implicit. 1200 A1 Method 2 for first 3 marks: Number of arrangements of LL^^^^^ – number of arrangements with the Ls split by an A 6!  2 – 5!  2 M1 6!  2 – h h an integer 1 < h < 1440 M1 k – 5!  2 k an integer k > 240 1200 A1 Method 3 for first 3 marks: Alternative approaches to Method 1 ^A ^ ^ ^ ^ ^ A 5P1  1P1  5P5  1P1 = 600 M1 LL treated as a single unit. M1 1200 A1 7(b) Final 2 marks of Question 7(b)  9!  B1 Accept unsimplified. [Total number of arrangements =] = 90720   May be seen as denominator of probability.  2!2!  1200 5 B1 FT their 1200 Probability = , , 0.0132 unsimplified B1 FT if their 1200 and their 90 720 90720 378 their 90720 supported by work in this part. 5 7(c) Method 1: Scenarios identified Both As and Ls removed A _ _ _ _ 5C4 = 5 B1 1 correct, identified outcome/value AA _ _ _ 5C3 = 10 for A, AL or AAL scenario, accept unsimplified AL _ _ _ 5C3 = 10 5C5–x cannot be used in place of 5Cx AAL _ _ 5C2 = 10 M1 Add 4 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. [Total =] 35 A1 Value stated WWW. Method 2: 1 A fixed, 1 L removed No other scenarios can be present anywhere in solution A ^ ^ ^ ^ 7C4 M1 7Ch, 3 ⩽ h ⩽ 5 B1 7C4 oe, no other terms, scenario identified. [Total =] 35 A1 Value stated. Method 3: 1 A fixed, both Ls removed A ^ ^ ^ ^ = 6C4 = 15 B1 Correct outcome/value for 1 identified scenario, accept A L ^ ^ ^ = 6C3 = 20 unsimplified. WWW M1 Add 2 values of appropriate scenarios, no incorrect scenarios, no repeated scenarios, accept unsimplified, condone use of permutations. [Total =] 35 A1 Value stated. 3

Question 52

In a large college, 32% of the students have blue eyes. A random sample of 80 students is chosen. Use an approximation to ﬁnd the probabili…

Accepted Answer

2 Mean = 80  0.32 = 25.6, B1 25.6 and 17.4[08] seen, allow unsimplified. var = 80  0.32  0.68 = 17.408 4.172… implies correct variance. 19.5 − M1 Substituting their 25.6 and 17.408 into ±standardisation P(X < 20) = P( Z  25.6) = P( Z −1.462) formula (any number for 19∙5), not σ2, √ σ. 17.408 M1 Using continuity correction 19∙5 or 20∙5 in their standardisation formula. = [1 − Φ (1.462)] = 1 – 0.9282 M1 Appropriate area Φ, from final process, must be probability. (Expect final ans < 0∙5 ). Note: the correct final answer may imply M1 from use of calculator. 0.0718 A1 0.0718 ⩽ p ⩽0.0719 5

Question 53

Company A produces bags of sugar. An inspector ﬁnds that on average 10% of the bags are underweight. 10 of the bags are chosen at random. (…

Accepted Answer

5(a) [P(0, 1, 2) =] 10C0 0.10 0.910 + 10C1 0.11 0.99 + 10C2 0.12 0.98 M1 One term 10Cx p x (1 − p )10 − x , 0  p  1, x  0 = 0.348678+0.38742+0.19371 A1 Correct expression, accept unsimplified. 0.930 B1 0.9298 ⩽ p ⩽ 0.9303 Alternative method for Question 5(a) [1 – P(3, 4, 5, 6, 7, 8, 9, 10) = 1 – (10C3 0.97 0.13 +10C4 0.96 0.14 +10C5 M1 One term 10Cx p x (1 − p )10 − x , 0.95 0.15 +10C6 0.9 4 0.16 +10C7 0.93 0.17 +10C8 0.9 2 0.18 +10C9 0.91 0.19 0  p  1, x  0 +10C10 0.9 0 0.110 ) A1 Correct expression, accept unsimplified. 0.930 B1 0.9298 ⩽ p ⩽ 0.9303 3 5(b) 1.11 − M1 1.11, 1.04 and 0.06 substituted into ±Standardisation [P(X > 1.11) = ]P( Z  1.04) = P( Z  1.167) formula, no continuity correction not 0.062 or √0.06 0.06 = 1 – 0.8784 M1 1 – their 0.8784 as final answer, must be probability. (Expect final ans < 0∙5 ). 0.122 A1 0.1216 ⩽ p ⩽ 0.122 SC M0 M1 B1 for 0.122 with no standardisation formula. 3 5(c) w − B1 0.8775 < z ⩽ 0.878 or −0.878 ⩽ z < −0.8775 seen. [P(X < w) = P( Z  1.04) = 0.81] 0.06 M1 1.04 and 0.06 substituted in ±standardisation formula, no w − 1.04 = 0.878 continuity correction, not σ2, √ σ, equated to a z-value. 0.06 w = 1.09 A1 1.09 ⩽ w ⩽ 1.093 3

Question 54

(a) Find the number of diﬀerent arrangements of the 9 letters in the word ACTIVATED. [2] ..................................................…

Accepted Answer

6(a) 9! M1 h ! , h = 7, 8, 9; j = 1, 2 2!2! 2! j ! 90720 A1 2 6(b) Arrangements with 5 letters between As + Arrangements with 6 letters between As + Arrangements with 7 letters between As 7! M1 7! With gap of 5: 2! 3 [= 7560] 2!k , k positive integer 1< k < 7 7! With gap of 6: 2 [= 5040] M1 Add their no of ways for 3 identified correct scenarios, no 2! additional incorrect scenarios, accept unsimplified. 7! With gap of 7: 2! 1 [= 2520] 7! A1 [Total no = 2!=]6 15120 3 6(c) Method 1: Summing number of ways AT _ _ _ 2×2× 5C3 40 B1 Correct no of ways for 4 correctly identified scenarios, A _ _ _ _ 2×5C4 10 accept unsimplified. AATT _ 5C1 5 AAT _ _ 2×5C2 20 M1 Add no of ways for 5 or 6 identified correct scenarios, no AA _ _ _ 5C3 10 additional incorrect scenarios, no repeated scenarios, accept _ _ _ _ _ 5C5 1 unsimplified. [Total no of ways not containing more Ts than As = ] A1 All correct and added = 40+10+5+20+10+1 [=86] 86 M1 their 86 Probability = 9 accept numerator unevaluated C 5 9C 5 ortheiridentified total 86 43 A1 , , 0.683 126 63 Method 2: Subtracting no of ways with more Ts from total T _ _ _ _ 2×5C4 10 B1 Correct no of ways for 2 correctly identified scenarios, no TTA _ _ 2×5C2 20 additional incorrect scenarios, no repeated scenarios, accept TT _ _ _ 5C3 10 unsimplified, condone use of permutations M1 Add no of ways for 2 or 3 correct scenarios and subtract from their total no of ways All correct and subtracted Total no of ways with more Ts than As =40 A1 9C5 − 40 = 86 86 M1 their 86 Probability = 9 accept numerator unevaluated C 5 9C 5 ortheiridentified total 6(c) 43 A1 , 0.683 63 5

Question 55

Sam and Tom are playing a game which involves a bag containing 5 white discs and 3 red discs. They take turns to remove one disc from the b…

Accepted Answer

7(a) 3 2 5 3 M1 3 2 5 3 [P(SR TR) + P(SW TR) =]  +   + k or l +  0 < k,l < 1 8 7 8 7 8 7 8 7 21 3 A1 3 = , , 0.375 SC B1 for with no explanation. 56 8 8 2 7(b) [RRWR, WRRR, WRWR] M1 m n o q    1 ⩽ m,n,o,q ⩽ 5, m ≠ n ≠ o ≠ q 3 2 5 1 5 3 2 1 5 3 4 2    +    +    8 7 6 5 8 7 6 5 8 7 6 5 8 7 6 5 1 1 1 A1 Probability for one scenario correct, accept unsimplified. [= + + ] 56 56 14 M1 Adding probabilities for 3 correct scenarios and no incorrect. 180 3 A1 Or 0.1071428… to 4SF or better. = , , 0.107 SC B1 for 3/28 with inadequate explanation. 1680 28 4 7(c) 30 1 M1 3 2 5 1 their P ( RRWR ) or    1680 56 8 7 6 5 [P(S first disc R |T2 ) =] = 3 3 3 their 7 ( b ) − must bea probor 28 28 28 1 A1 , 0.167 6 2

Question 56

80% of the residents of Kinwawa are in favour of a leisure centre being built in the town. 20 residents of Kinwawa are chosen at random and…

Accepted Answer

3(b) 4 256 B1 8192  ( 0.8 ) ( 0.2 ) =  0.08192, Accept OE.   3125 100000 1 3(c) ( 0.8 )5 ( 0.2 ) 2  6 M1 ( 0.8 )5 ( 0.2 ) 2  k or ( 0.8 )5 ( 0.2 ) k 0.2 , 2 ⩽ k ⩽ 7. 8144 A1 786432 = 0.0786, 0.0786 ⩽ p < 0.07865, . 78125 10000000 If A0 awarded, SC B1 for correct answer WWW. 2

Question 57

The probability that it will rain on any given day is x. If it is raining, the probability that Aran wears a hat is 0.8 and if it is not ra…

Accepted Answer

4 (1 – x) × 0.7 × 0.9 = 0.36 M1 (1 − x ) =a b 0.36, a = 0.7 or 0.3, b = 0.9 or 0.1 B1 (1-x)×0.7×0.9=0.36, (1-x)×0.63=0.36, 0.36 0.63 – 0.63x = 0.36 or 1 − x = seen. 0.63 Condone recovery from omission of brackets. 3 A1 Accept 0.428571 to at least 3 sf. x = Condone 0.4285 rounding to 0.429 . 7 3 If M0 awarded, SC B1 for x = or 0.428571 to at least 3 sf. 7 3

Question 58

Marco has four boxes labelled K, L, M and N. He places them in a straight line in the order K, L, M, N with K on the left. Marco also has f…

Accepted Answer

5 1 8 B1 Both stated, accept unsimplified. P(A) = , P(B) = = 1, 2 24 3 1 M1 Evidence that independence properties not used. P(A  B ) = 6 1 1 1 A1 Evaluated and conclusion stated. P(A) × P(B) =  = P(A) × P(B) and P(A  B ) seen. 2 3 6 so events are independent 3

Question 59

(a) Find the number of diﬀerent arrangements of the 9 letters in the word DELIVERED in which the three Es are together and the two Ds are n…

Accepted Answer

7(c) Scenarios B1 1 correct unsimplified outcome/value for one identified D E _ _ _ 4C3 4 scenario excluding DDEEE. D E E _ _ 4C2 6 Note: 4C1 cannot be used for 4C3 . D E E E _ 4C1 4 D D E _ _ 4C2 6 M1 Add values of 6 appropriate scenarios, no additional, incorrect D D E E _ 4C1 4 or repeated scenarios. Accept unsimplified. D D E E E [4C0] 1 [Total =] 25 A1 3

Question 60

(a) Find the number of diﬀerent arrangements of the 8 letters in the word COCOONED. [1] ...................................................…

Accepted Answer

3(a) 8! 2!3!        3360 1 3(b) 6! 2!2! M1 6! 2! ! f ; f = 1, 2, 3. 180 A1 2 3(c)       | P OOO CC P OOO CC P CC            5! 7! 3! M1 5! g g a positive integer, g 3360, 1. Condone numerator of 5! 3360g . M1 7! 3! h or 8! 3! h , where h is a positive integer. Condone division by 3360 in denominator. = 120 1 , , 0.143 840 7 A1 0.1428571… to at least 3SF. If M0 scored SC B1 for 1 7 WWW. 3

Question 61

A children’s wildlife magazine is published every Monday. For the next 12 weeks it will include a model animal as a free gift. There are ﬁv…

Accepted Answer

7(a) Method 1 [P(X < 6) = P(X ⩽ 5) =] 5 1 0.8  M1 1 – 0.8r, r = 5, 6. = 0.672 A1 Method 2 [P(X < 6) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) =] 1 4 1 5 5 5   + 4 5       2 1 5  + 4 5       3 1 5  + 4 5       4 1 5  M1 Condone an extra term ( 5 4 1 ) 5 5  . First, last and one of the 3 middle terms implies M1. = 0.672 A1 2 Question Answer Marks Guidance 7(b) Method 1 [1 − P(0, 1, 2)] = 1 – (12C0 (0.8)12 + 12C1 (0.2)(0.8)11 + 12C2 (0.2)2 (0.8)10) [= 1 – (0.06872 + 0.20615 + 0.28347)] M1 One term 12Cx   12 1 x x p p   , 0 < p < 1, 0 x , 1, 2. A1 Correct expression, accept unsimplified, no terms omitted, leading to final answer. Correct unsimplified expression or better. = 0.442 B1 0.411 < p ⩽ 0.442 WWW. Method 2 [P(3,4,5,6,7,8,9,10,11,12) = ] 12C3 (0.2)3 (0.8)9 + 12C4 (0.2)4 (0.8)8 + … + 12C11 (0.2)11 (0.8)1 + 12C12 (0.2)12 [= 0.23622 + 0.13288 + … + 1.966  10–7 + 4.096  10–9] M1 One term 12Cx   12 1 x x p p   , 0 < p < 1, 0 x , 1, 2. A1 Correct expression, accept unsimplified, leading to final answer. Accept first, last and 8 of the middle terms. =0.442 B1 0.411 < p ⩽ 0.442 . 3 Question Answer Marks Guidance 7(c)   5 0.2  5! M1   5 0.2  s, s a positive integer. 1 may be implied. M1 t  5! where 0 < t < 1. = 0.0384, 24 625 A1 Alternative Method for Question 7(c) 5 4 3 2 1 1 1 1 1 1 5 5 1 [ ] ( ) C C C C C C     M1 5 5 1 ( ) C or 55 as denominator. M1 5 4 3 2 1 1 1 1 1 1 [ ] C C C C C     or 5! as numerator. = 0.0384, 24 625 A1 3

Question 62

A sports event is taking place for 4 days, beginning on Sunday. The probability that it will rain on Sunday is 0.4. On any subsequent day, …

Accepted Answer

2(a) [P(no rain) =   3 0.6 0.8  =] 0.3072, 192 625 1 Question Answer Marks Guidance 2(b) 0.6 0.8 0.2   M1 a  b  c where a, b = 0.6, 0.8, c = 0.2, 0.4, 0.7 . Condone including Wednesday with both 0.3 and 0.7 used. = 0.096[0], 12 125 A1 2 2(c)   P RDDD 0.4 0.3 0.8 0.8     = 0.0768, 48 625   P DRDD 0.6 0.2 0.3 0.8     = 0.0288, 18 625   P DDRD 0.6 0.8 0.2 0.3     = 0.0288, 18 625   P DDDR 0.6 0.8 0.8 0.2     = 0.0768, 48 625 B1 Correct probability for one clearly identified outcome evaluated accept unsimplified. A correct unsimplified expression is not sufficient. M1 Add 4 probability values, 0 < p < 1, for appropriate identified scenarios. Accept unsimplified. Ways of identifying scenarios for this mark: Stating the days. All the unsimplified probability calculations exactly as stated in the mark scheme. Identifying the correct branches on a tree diagram and linking with the values. No repeated scenarios. No incorrect scenarios. 0.2112, 132 625 A1 Accept 0.211 If 0/3 scored SC B1 for 0.2112, 132 625. 3

Question 63

Anil is a candidate in an election. He received 40% of the votes. A random sample of 120 voters is chosen. Use an approximation to ﬁnd the …

Accepted Answer

2 Mean =120 0.4 48 Var =120 0.4 0.6 28.8    B1 48 and 28 4 5 , 28.8 seen, allow unsimplified. (5.366 ⩽ σ ⩽ 5.367 or 12 5 5 implies correct variance). P(36 54 X   ) = P( 35.5 48 54.5 48) 28.8 28.8 Z     M1 Substituting their µ and σ into one ±standardisation formula (any number for 35.5 or 54.5), condone σ2 and √σ. M1 Using continuity correction 35.5, 36.5 or 53.5, 54.5 once in their standardisation formula. Note: 12.5 28.8  or 6.5 28.8  seen gains M2 BOD. [= P( 2.3292 1.211) Z    =] 0.8871 + 0.9900 – 1 M1 Appropriate area Φ, from final process. Must be a probability. Expect final answer > 0.5 . Note: correct final answer implies this M1. = 0.877 A1 0.877 ≤ p < 0.8772 . 5

Question 64

The mass of grapes sold per day by a large shop can be modelled by a normal distribution with mean 28kg. On 10% of days less than 16kg of g…

Accepted Answer

6(a)   16 28 P 16 P 0.1 X Z                   16 28 1.282    M1 Use of the ±standardisation formula with 16, 28, σ and a z-value (not 0.1, 0.9, 0.282, 0.5398, 0.8159) equated to a z-value. Condone continuity correct ±0.5, not 2,  . Condone 12 1.282    . 9.36  A1 3 6(b) [1 − P(0, 1, 2) =] 1 – (12C0(0.1)0 (0.9)12 + 12C1 (0.1)1 (0.9)11 + 12C2 (0.1)2 (0.9)10 ) [1 – (0.2824 + 0.3766 + 0.2301)] M1 One term 12Cx   12 1 x x p p   , 0 1 p  . 0,1,2 x  . A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. 0.111 B1 0.1108699… rounded to at least 3SF. Alternative Method for Question 6(b) P(3,4,5,6,7,8,9,10,11,12) = 12C3 (0.1)3 (0.9)9 + 12C4 (0.1)4 (0.9)8+ … + 12C11 (0.1)11 (0.9)1 + 12C12 (0.1)12 (0.9)0 [0.08523 + 0.02131 + … + 1.08×10-10 + 1×10-12] M1 One term 12Cx   12 1 x x p p   , 0 1 p  . 0,1,2 x  . A1 Correct expression, accept unsimplified, no terms omitted leading to final answer. 0.111 B1 0.1108699… rounded to at least 3SF. 3 Question Answer Marks Guidance 6(c) [P( 1.3 1.3 Z    ) = 2 Φ(1.3) – 1 ] = 2 × 0.9032 – 1 B1 Identifying at least one of −1.3 or 1.3 as the appropriate z-values. M1 Calculating the appropriate probability area from 2 symmetrical z-values (leading to their final answer, expect > 0.5). = 0.806, 504 625 A1 0.8064, 0.806 ⩽ p < 0.8065 . [In 365 days 0.8064 365  ] = 294 or 295 B1 FT Strict FT their at least 4-figure probability (not z-value). Final answer must be positive integer, no approximation or rounding stated. 4 Question Answer Marks Guidance 7(a) Method 1: Total number of arrangements – number of arrangements with Cs together 10! 9! 2!4! 4!  [75600-15120] M1 10! , ! ! c a b  a ≠ b, a = 1, 2, b = 1, 4, with c being a positive integer. M1 ! 4! e d  , e = 8, 9, 10, with d being a positive integer. = 60480 A1 Exact value only. SC B1 for final answer 60480 www. Method 2: Arrangements ^ ^ C ^ C ^ ^ ^ ^ ^ 8! 9 8 4! 2   M1 8! 4! f  seen, with f being a positive integer. M1 9 8 g h   , with g being a positive integer, h = 1, 2. g × 9C2 and g × 9P2 are acceptable. = 60480 A1 Exact value only. SC B1 for final answer 60480 www. 3

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106	see sheet	6	9709/55 May/June 2025
107	see sheet	8	9709/55 May/June 2025
108	see sheet	9	9709/55 May/June 2025
109	see sheet	10	9709/55 May/June 2025
110	see sheet	6	9709/51 Oct/Nov 2025
111	see sheet	4	9709/51 Oct/Nov 2025
112	see sheet	5	9709/51 Oct/Nov 2025
113	see sheet	8	9709/51 Oct/Nov 2025
114	see sheet	10	9709/51 Oct/Nov 2025
115	see sheet	3	9709/52 Oct/Nov 2025
116	see sheet	7	9709/52 Oct/Nov 2025
117	see sheet	7	9709/52 Oct/Nov 2025
118	see sheet	9	9709/52 Oct/Nov 2025
119	see sheet	10	9709/52 Oct/Nov 2025
120	see sheet	4	9709/53 Oct/Nov 2025
121	see sheet	8	9709/53 Oct/Nov 2025
122	see sheet	8	9709/53 Oct/Nov 2025
123	see sheet	7	9709/55 Oct/Nov 2025
124	see sheet	8	9709/55 Oct/Nov 2025

Questions190 pages

Mark scheme124 answers

Another paper, or another topic

Questions as text

Q1 · In Greenton, 70% of the adults own a car 9709/52 Feb/March 2020

Q2 · Box A contains 7 red balls and 1 blue ball 9709/52 Feb/March 2020

Q3 · A company produces small boxes of sweets that contain 5 jellies and 3 chocolates 9709/51 May/June 2020

Q4 · On Mondays, Rani cooks her evening meal 9709/51 May/June 2020

Q5 · A total of 500 students were asked which one of four colleges they attended and whether… 9709/52 May/June 2020

Q6 · On any given day, the probability that Moena messages her friend Pasha is 0.72 9709/52 May/June 2020

Q7 · Juan goes to college each day by any one of car or bus or walking 9709/53 May/June 2020

Q8 · In a certain large college, 22% of students own a car 9709/53 May/June 2020

Q9 · Find the number of diﬀerent possible arrangements of the 9 letters in the word CELESTIAL 9709/53 May/June 2020

Q10 · The probability that a student at a large music college plays in the band is 0.6 9709/51 Oct/Nov 2020

Q11 · Kayla is competing in a throwing event 9709/51 Oct/Nov 2020

Q12 · The random variable X takes each of the values 1, 2, 3, 4 with probability 1 Two… 9709/51 Oct/Nov 2020

Q13 · Find the number of diﬀerent ways in which the 10 letters of the word SHOPKEEPER can be… 9709/51 Oct/Nov 2020

Q14 · 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

Q15 · Mr and Mrs Ahmed with their two children, and Mr and Mrs Baker with their three children… 9709/52 Oct/Nov 2020

Q16 · An ordinary fair die is thrown until a 6 is obtained 9709/53 Oct/Nov 2020

Q17 · The 13 00 train from Jahor to Keman runs every day 9709/53 Oct/Nov 2020

Q18 · The 8 letters in the word RESERVED are arranged in a random order 9709/53 Oct/Nov 2020

Q19 · A fair spinner with 5 sides numbered 1, 2, 3, 4, 5 is spun repeatedly 9709/52 Feb/March 2021

Q20 · Georgie has a red scarf, a blue scarf and a yellow scarf 9709/52 Feb/March 2021

Q21 · There are 400 students at a school in a certain country 9709/52 Feb/March 2021

Q22 · How many diﬀerent arrangements are there of the 8 letters in the word RELEASED? 9709/51 May/June 2021

Q23 · To gain a place at a science college, students ﬁrst have to pass a written test and then… 9709/51 May/June 2021

Q24 · In Questa, 60% of the adults travel to work by car 9709/51 May/June 2021

Q25 · An ordinary fair die is thrown repeatedly until a 5 is obtained 9709/52 May/June 2021

Q26 · On each day that Alexa goes to work, the probabilities that she travels by bus, by train… 9709/52 May/June 2021

Q27 · Every day Richard takes a ﬂight between Astan and Bejin 9709/52 May/June 2021

Q28 · Find the total number of diﬀerent arrangements of the 8 letters in the word TOMORROW 9709/52 May/June 2021

Q29 · Three fair six-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown at the… 9709/53 May/June 2021

Q30 · In the region of Arka, the total number of households in the three villages Reeta, Shan… 9709/53 May/June 2021

Q31 · Two fair coins are thrown at the same time 9709/51 Oct/Nov 2021

Q32 · For her bedtime drink, Suki has either chocolate, tea or milk with probabilities 0.45… 9709/51 Oct/Nov 2021

Q33 · Raman and Sanjay are members of a quiz team which has 9 members in total 9709/51 Oct/Nov 2021

Q34 · The times, in minutes, that Karli spends each day on social media are normally… 9709/51 Oct/Nov 2021

Q35 · Each of the 180 students at a college plays exactly one of the piano, the guitar and the… 9709/52 Oct/Nov 2021

Q36 · In a certain region, the probability that any given day in October is wet is 0.16… 9709/52 Oct/Nov 2021

Q37 · A security code consists of 2 letters followed by a 4-digit number 9709/53 Oct/Nov 2021

Q38 · Box A contains 6 red balls and 4 blue balls 9709/53 Oct/Nov 2021

Q39 · In a certain country, the probability of more than 10cm of rain on any particular day is… 9709/52 Feb/March 2022

Q40 · The weights of male leopards in a particular region are normally distributed with mean… 9709/52 Feb/March 2022

Q41 · A factory produces chocolates in three ﬂavours: lemon, orange and strawberry in the ratio… 9709/52 Feb/March 2022

Q42 · Janice is playing a computer game 9709/51 May/June 2022

Q43 · In a large college, 28% of the students do not play any musical instrument, 52% play… 9709/52 May/June 2022

Q44 · Hanna buys 12 hollow chocolate eggs that each contain a sweet 9709/52 May/June 2022

Q45 · Ramesh throws an ordinary fair 6-sided die 9709/53 May/June 2022

Q46 · Sajid is practising for a long jump competition 9709/53 May/June 2022

Q47 · The residents of Persham were surveyed about the reliability of their internet service 9709/51 Oct/Nov 2022

Q48 · On any day, Kino travels to school by bus, by car or on foot with probabilities 0.2, 0.1… 9709/52 Oct/Nov 2022

Q49 · Three fair 6-sided dice, each with faces marked 1, 2, 3, 4, 5, 6, are thrown at the same… 9709/52 Oct/Nov 2022

Q50 · At a company’s call centre, 90% of callers are connected immediately to a representative 9709/52 Oct/Nov 2022

Q51 · Find the number of diﬀerent arrangements of the 9 letters in the word ALLIGATOR in which… 9709/52 Oct/Nov 2022

Q52 · In a large college, 32% of the students have blue eyes 9709/53 Oct/Nov 2022

Q53 · Company A produces bags of sugar 9709/53 Oct/Nov 2022

Q54 · Find the number of diﬀerent arrangements of the 9 letters in the word ACTIVATED 9709/53 Oct/Nov 2022

Q55 · Sam and Tom are playing a game which involves a bag containing 5 white discs and 3 red… 9709/53 Oct/Nov 2022

Q56 · 80% of the residents of Kinwawa are in favour of a leisure centre being built in the town 9709/52 Feb/March 2023

Q57 · The probability that it will rain on any given day is x 9709/52 Feb/March 2023

Q58 · Marco has four boxes labelled K, L, M and N 9709/52 Feb/March 2023

Q59 · Find the number of diﬀerent arrangements of the 9 letters in the word DELIVERED in which… 9709/52 Feb/March 2023

Q60 · Find the number of diﬀerent arrangements of the 8 letters in the word COCOONED 9709/51 May/June 2023

Q61 · A children’s wildlife magazine is published every Monday 9709/51 May/June 2023

Q62 · A sports event is taking place for 4 days, beginning on Sunday 9709/52 May/June 2023

Q63 · Anil is a candidate in an election 9709/53 May/June 2023

Q64 · The mass of grapes sold per day by a large shop can be modelled by a normal distribution… 9709/53 May/June 2023

Q65 · Hazeem repeatedly throws two ordinary fair 6-sided dice at the same time 9709/51 Oct/Nov 2023

Q66 · A red spinner has four sides labelled 1, 2, 3, 4 9709/51 Oct/Nov 2023

Q67 · George has a fair 5-sided spinner with sides labelled 1, 2, 3, 4, 5 9709/52 Oct/Nov 2023

Q68 · A factory produces a certain type of electrical component 9709/52 Oct/Nov 2023

Q69 · Freddie has two bags of marbles 9709/52 Oct/Nov 2023

Q70 · Find the number of diﬀerent arrangements of the 9 letters in the word ANDROMEDA in which… 9709/52 Oct/Nov 2023

Q71 · Tim has two bags of marbles, A and B 9709/53 Oct/Nov 2023

Q72 · A bag contains 9 blue marbles and 3 red marbles 9709/52 Feb/March 2024

Q73 · Sam is a member of a soccer club 9709/52 Feb/March 2024

Q74 · Anil is taking part in a tournament 9709/52 Feb/March 2024

Q75 · A game for two players is played using a fair 4-sided dice with sides numbered 1, 2, 3… 9709/51 May/June 2024

Q76 · In a certain area in the Arctic the probability that it snows on any given day is 0.7… 9709/51 May/June 2024