Probability generating functions

6(a) 1 3 1 M1 At least 2 correct probabilities used in a polynomial. P(3R, 2R, 1R, 0R) = , , , 0 5 5 5 G ( t )= 1 t+ 3 t2 + 1 t3 A1 X 5 5 5 2 6(b) 1 6 9 M1 One correct probability and all three adding to 1 used in a P(2H, 1H, 0H) = , , polynomial. 16 16 16 G ( t )= 9 + 6 t+ 1 t2 A1 Y 16 16 16 2 6(c) Attempt to multiply results from part (a) and part (b) M1 Obtain polynomial M1 9 33 28 9 1 A1 Accept exact equivalent decimals. t+ t2 + t3 + t4 + t5 80 80 80 80 80 3 6(d) E(Z) = G ' ( 1 )= 1 ( 9+66+84+36+5 ) M1 Differentiate and put t = 1. Z 80 200 A1 = =2.5 80 G ' ' ( 1 )= 1 ( 66+168+108+20 )=4.525= 362 M1 Z 80   80   © UCLES 2021 Page 11 of 12 6(d) Var (Z) = G ′′( 1 )+G ′( 1 )− ( G' ( 1 ))2 M1 Use result. 2 A1 362 200 200 31 + −   = 0.775=   80 80  80  40 5 © UCLES 2021 Page 12 of 12

6(a) 1 3 1 M1 At least 2 correct probabilities used in a polynomial. P(3R, 2R, 1R, 0R) = , , , 0 5 5 5 G ( t )= 1 t+ 3 t2 + 1 t3 A1 X 5 5 5 2 6(b) 1 6 9 M1 One correct probability and all three adding to 1 used in a P(2H, 1H, 0H) = , , polynomial. 16 16 16 G ( t )= 9 + 6 t+ 1 t2 A1 Y 16 16 16 2 6(c) Attempt to multiply results from part (a) and part (b) M1 Obtain polynomial M1 9 33 28 9 1 A1 Accept exact equivalent decimals. t+ t2 + t3 + t4 + t5 80 80 80 80 80 3 6(d) E(Z) = G ' ( 1 )= 1 ( 9+66+84+36+5 ) M1 Differentiate and put t = 1. Z 80 200 A1 = =2.5 80 G ' ' ( 1 )= 1 ( 66+168+108+20 )=4.525= 362 M1 Z 80   80   © UCLES 2021 Page 11 of 12 6(d) Var (Z) = G ′′( 1 )+G ′( 1 )− ( G' ( 1 ))2 M1 Use result. 2 A1 362 200 200 31 + −   = 0.775=   80 80  80  40 5 © UCLES 2021 Page 12 of 12

4(a) 4t M1 kt G’(t) = Differentiate to obtain OE. ( 3−2t2 )2 ( 3−2t2 )2 so E(X) = G’(1) = 4 A1 CAO WWW G’’(t) = 12+24t2 or 4 ( 3−2t2 )−2 +32t2 ( 3−2t2 )−3 M1 OE. Differentiate, allow only numerical or sign slips. ( 3−2t2 )3 ( ( ))2 M1 Substitute their values into correct formula, dependent on Var(X) = G' '(1) + G'(1) – G' 1 = 36 + 4 – 16 attempt at G' '(t). [Var(X)] = 24 A1 CAO WWW 5 4(b) G ( t )= 1 = ( 3−2t2 )−1 M1 Expand given expression or give expression for the term in t4 . X 3−2t2 1 2 4  = 1+ t2 + t4 +…. 3 3 9  P(X = 4) = their coefficient of t4 M1 Found from legitimate method. 4 A1 WWW [P(X = 4) =] 27 3 © UCLES 2021 Page 9 of 13

5(a) 3 2 1 6 B1 At least 2 probabilities correct. P(3, 6, 9) = × × = 9 8 7 504 3 2 6 108 P(Two of 3, 6, 9) = × × ×3= 9 8 7 504 3 6 5 270 P(one of 3, 6, 9) = × × ×3= 9 8 7 504 6 5 4 120 P(none of 3, 6, 9) = × × = 9 8 7 504 G ( t )= 20 + 45 t+ 18 t2 + 1 t3 M1 Attempt with at least 3 probabilities in a polynomial, X 84 84 84 84 A1 CAO. 3 5(b) 12 M1 P(both even ) = 72 40 P(one even) = 72 20 P(no even) = 72 G ( t )= 5 + 10 t+ 3 t2 A1 CAO Y 18 18 18 2 © UCLES 2021 Page 11 of 14 5(c) 1 20 45 18 1   5 10 3  M1 Method and attempt to multiply. + t+ t2 + t3 + t+ t2     151284 84 84 84  18 18 18  M1 Multiplication to obtain single quintic polynomial. 1 A1 CAO (100+425t+600t2 +320t3 +64t4 +3t5) 1512 3 5(d) G'(1) = 1 ( 425+1200+960+256+15 ) M1 1512 17 A1 CWO 9 2 © UCLES 2021 Page 12 of 14

5(a) ( 1+4+9+16 ) k =1 so k = 1 B1 30 1 5(b) G ( t )= 1 t+ 4 t2 + 9 t3 + 16 t4 M1A1 Using their k in a polynomial, at least two terms correct for their k. X 30 30 30 30 2 5(c)  1 4 9 16 1 1 1  M1 Method and attempt to multiply.  t+ t2 + t3+ t4  + t+ t2  30 30 30 30 4 2 4  1 ( ) M1A1 Multiplication to obtain single polynomial of order 6. t+6t2 +18t3 +38t4 +41t5 +16t6 120 3 5(d) Given: G'(1) = 13/3 M1 Differentiate twice. 1 G''(t) = (12+108t+456t2 +820t3 +480t4) 120 2 M1 Use correct formula. 13 13 − Var(X) = G''(1) +   3  3  1876 13 169 107 A1 CAO + − = or 1.19 120 3 9 90 3 © UCLES 2021 Page 11 of 13

5(a) 3 2 1 6 B1 At least 2 probabilities correct. P(3, 6, 9) = × × = 9 8 7 504 3 2 6 108 P(Two of 3, 6, 9) = × × ×3= 9 8 7 504 3 6 5 270 P(one of 3, 6, 9) = × × ×3= 9 8 7 504 6 5 4 120 P(none of 3, 6, 9) = × × = 9 8 7 504 G ( t )= 20 + 45 t+ 18 t2 + 1 t3 M1 Attempt with at least 3 probabilities in a polynomial, X 84 84 84 84 A1 CAO. 3 5(b) 12 M1 P(both even ) = 72 40 P(one even) = 72 20 P(no even) = 72 G ( t )= 5 + 10 t+ 3 t2 A1 CAO Y 18 18 18 2 © UCLES 2021 Page 11 of 14 5(c) 1 20 45 18 1   5 10 3  M1 Method and attempt to multiply. + t+ t2 + t3 + t+ t2     151284 84 84 84  18 18 18  M1 Multiplication to obtain single quintic polynomial. 1 A1 CAO (100+425t+600t2 +320t3 +64t4 +3t5) 1512 3 5(d) G'(1) = 1 ( 425+1200+960+256+15 ) M1 1512 17 A1 CWO 9 2 © UCLES 2021 Page 12 of 14

2(a) G ' (t) = 0.20.74t0.9t2 0.36t3 M1 PGF differentiated twice. Y G "(t) = 0.741.8t1.08t2 Y G '(1) = 2.2 A1 Y G "(1) = 3.62 Y Var(Y) = G "(1) + G '(1) – (G '(1))2 M1 Correct formula used and attempt at Var(Y). Y Y Y 3.62 + 2.2 – 2.22 0.98 A1 4 2(b) G (t) =  abtct22 M1* Correct method. Y So a2 0.04, a0.2 depM1 Attempt to find all coefficients. c2 0.09, c0.3 0.1 2ab0.2, b 0.5 0.2 G X (t) = 0.20.5t0.3t2 A1 Note: G = 1 1t223t2 M1, accept without 1 . Y 100 100 1 1 So G = 1t23t dep M1, must have . X 10 10 Final answer A1. 3 © UCLES 2022 Page 7 of 13

2(a) G ' (t) = 0.20.74t0.9t2 0.36t3 M1 PGF differentiated twice. Y G "(t) = 0.741.8t1.08t2 Y G '(1) = 2.2 A1 Y G "(1) = 3.62 Y Var(Y) = G "(1) + G '(1) – (G '(1))2 M1 Correct formula used and attempt at Var(Y). Y Y Y 3.62 + 2.2 – 2.22 0.98 A1 4 2(b) G (t) =  abtct22 M1* Correct method. Y So a2 0.04, a0.2 depM1 Attempt to find all coefficients. c2 0.09, c0.3 0.1 2ab0.2, b 0.5 0.2 G X (t) = 0.20.5t0.3t2 A1 Note: G = 1 1t223t2 M1, accept without 1 . Y 100 100 1 1 So G = 1t23t dep M1, must have . X 10 10 Final answer A1. 3 © UCLES 2022 Page 7 of 13

3(a) 6 5 1 M1 Obtain two correct equations and attempt to solve. G (t) =  t t2 X 12 12 12 6 P(0 heads) = 1a1b 12 5 P(1 head) = a1b1ab 12 1 P(2 heads) = ab 12  7 1  ab , ab   12 12 Solve to give A1 Correct value for a. 1  1 a b  3  4 2 3(b) 2 M1 Square their G (t).  6 5 1  X G Y (t) =   t t2  12 12 12  1  3660t37t2 10t3t4 M1 Obtain quartic polynomial from 3-term G X (t). 144 A1 3 © UCLES 2022 Page 8 of 13 3(c) G '(t) = 1  6074t30t2 4t3 M1 Differentiate their G Y (t) twice. Y 144 G "(t) = 1  7460t12t2 Y 144 Var(Y) = G " (1) + G ' (1) – (G ' (1))2 M1 Use correct formula and attempt at Var(Y). Y Y Y 1 7 49 59 146   0.819 144 6 36 72 0.819 A1 3 Question Answer Marks Guidance

4(a) 1 1 1 1 2 1 1 6 M1 P(0 H) =   = P(1 H) =   3= At least 2 probabilities correct. 3 3 3 27 3 3 3 27 1 2 2 12 2 2 2 8 P(2 H) =   3= P(3 H) =   = 3 3 3 27 3 3 3 27 1 6 12 8 M1A1 Correct form, ft their probabilities. G (t) = + t+ t2 + t3 X 27 27 27 27 1 2 M1 OR: PGF for one coin is + t 3 3 3 M1A1 1 2  For 3 coins, PGF is  + t  3 3  3 4(b) 1 M1 Correct method. G (t) = ( 1+2t+t2)( 1+6t+12t2 +8t3) Z 108 1 M1 Obtain quintic polynomial. ( 1+8t+25t2 +38t3+28t4 +8t5) 108 1 2 25 19 7 2 A1 + t+ t2 + t3+ t4 + t5 108 27 108 54 27 27 3 © UCLES 2022 Page 9 of 12 4(c) 1 M1 G ’ (t) = ( 8+50t+114t2 +112t3+40t4) Z Differentiate and evaluate at t = 1 108 1 E(Z) = (8+50+114+112+40) 108 324 A1 =3 108 Alternative method for questions 4(c) 2 2 2 1 1 M1 Summing expected values for each coin + + + + 3 3 3 2 2 3 A1 2 Question Answer Marks Guidance

5(a) 9 6 1 M1 A1 2 probabilities correct, in a polynomial G (t) = + t+ t2 X 16 16 16 2 © UCLES 2022 Page 10 of 13 5(b)  9 6 1 25 10 1  M1 Second PGF correct and multiplied by part (a). G Z (t) =  + t+ t2  + t+ t2  16 16 16 36 36 36  1 ( 225+240t+94t2 +16t3+t4) M1 Obtains a quartic polynomial. 576 A1 1 Or 25 + 5 t+ 47 t2 + 1 t3+ 1 t4 Note: (t+3)2(t+5)2 scores M1M1A0. 576 64 12 288 36 576 3 5(c) G ’ (t) = 1 ( 240+188t+48t2 +4t3) M1 Differentiate twice. Z 576 G ’’ (t) = 1 ( 188+96t+12t3) Z 576 1 5 25 M1 Use correct formula. Var (Z) = (188+96+12)+ − 576 6 36 47 A1 72 1 3 1 5 M1 One term correct. OR: Var(Z) = 2  +2  M1 Two terms present and added. 4 4 6 6 47 A1 72 3 5(d) 1 B1 FT FT their power with greatest coefficient in part (b). 1 © UCLES 2022 Page 11 of 13

4(a) 1 1 1 1 2 1 1 6 M1 P(0 H) =   = P(1 H) =   3= At least 2 probabilities correct. 3 3 3 27 3 3 3 27 1 2 2 12 2 2 2 8 P(2 H) =   3= P(3 H) =   = 3 3 3 27 3 3 3 27 1 6 12 8 M1A1 Correct form, ft their probabilities. G (t) = + t+ t2 + t3 X 27 27 27 27 1 2 M1 OR: PGF for one coin is + t 3 3 3 M1A1 1 2  For 3 coins, PGF is  + t  3 3  3 4(b) 1 M1 Correct method. G (t) = ( 1+2t+t2)( 1+6t+12t2 +8t3) Z 108 1 M1 Obtain quintic polynomial. ( 1+8t+25t2 +38t3+28t4 +8t5) 108 1 2 25 19 7 2 A1 + t+ t2 + t3+ t4 + t5 108 27 108 54 27 27 3 © UCLES 2022 Page 9 of 12 4(c) 1 M1 G ’ (t) = ( 8+50t+114t2 +112t3+40t4) Z Differentiate and evaluate at t = 1 108 1 E(Z) = (8+50+114+112+40) 108 324 A1 =3 108 Alternative method for questions 4(c) 2 2 2 1 1 M1 Summing expected values for each coin + + + + 3 3 3 2 2 3 A1 2 Question Answer Marks Guidance

5(a) 1 9 26 24 B1 Two probabilities correct, seen anywhere. P(3H) = P(2H) = P(1H) = P(0H) = 60 60 60 60 24 26 9 1 M1 Cubic polynomial with 4 probabilities as G (t) =  t t2  t3 X coefficients with the correct powers of t from their 60 60 60 60 working. Equivalent forms are acceptable. A1 Correct, AEF. 3 5(b) 1 1 1 B1 G (t) =  t t2 Y 4 2 4 24 26 9 1  1 1 1  M1 Attempt to multiply their two PGFs. G Z (t) =   t t2  t3   t t2  60 60 60 60  4 2 4  1  2474t85t2 45t311t4 t5 M1 Obtain quintic expression and collect terms 240 A1 Correct, AEF. 4 5(c) Differentiate: G' (t) = 1  74170t135t2 44t3 5t4 M1 Differentiate their G. Z 240 428 107 A1 Any correct form. G' (1) =  1.78 Z 240 60 2 © UCLES 2023 Page 10 of 12

5(a) 1 9 26 24 B1 Two probabilities correct, seen anywhere. P(3H) = P(2H) = P(1H) = P(0H) = 60 60 60 60 24 26 9 1 M1 Cubic polynomial with 4 probabilities as G (t) =  t t2  t3 X coefficients with the correct powers of t from their 60 60 60 60 working. Equivalent forms are acceptable. A1 Correct, AEF. 3 5(b) 1 1 1 B1 G (t) =  t t2 Y 4 2 4 24 26 9 1  1 1 1  M1 Attempt to multiply their two PGFs. G Z (t) =   t t2  t3   t t2  60 60 60 60  4 2 4  1  2474t85t2 45t311t4 t5 M1 Obtain quintic expression and collect terms 240 A1 Correct, AEF. 4 5(c) Differentiate: G' (t) = 1  74170t135t2 44t3 5t4 M1 Differentiate their G. Z 240 428 107 A1 Any correct form. G' (1) =  1.78 Z 240 60 2 © UCLES 2023 Page 10 of 12

5(a) 1 B1 Working required. G (t)k  13t4t2 k1341, k   X  8 G' (t) = k38t, M1 1 3 4 11 X Or  px 0 1 2    8 8 8  8  11 A1 AG E(X) = G' (1) = X Evidence of using t 1 or  px required. 8 CWO 3 5(b) 1 1 M1 Multiply the two PGFs to obtain a single polynomial of degree 5. G (t) =  13t4t2  t212t Z 8 3 1  t2 5t310t4 8t5 A1 May have t2 as a factor. 24 2 © UCLES 2023 Page 10 of 13 5(c) 1 M1 Differentiate twice, allow one slip. G' (t) =  2t15t2 40t340t4 Z 24 1 G'' (t) =  230t120t2 160t3 Z 24 2 M1 Use correct formula. 1 97 97 Var(X) = G'' Z (1) + G' Z (1) – (G' Z (1))2 = 312    24 24 24 0.707 A1 407 576 3 5(d) Z  4 B1 FT FT their final polynomial in part (b) 1 © UCLES 2023 Page 11 of 13

5(a) P(X = r) = p(1− p)r−1 = pqr−1 B1 Implied by G(t) = pqr−1 tr . G(t) = pt(qt)r−1 M1 (Infinite summation) with common ratio (qt) and first term or common factor pt indicated or GP identified.  1  pt A1 AG G(t) = pt = Convincingly obtained. 1−qt 1−qt 3 5(b) G'(t) = pqt(1−qt)−2 + p(1−qt)−1 [= p(1−qt)−2 ] M1 Attempt to differentiate as a product/quotient. G''(t) = 2pq2t(1−qt)−3 +2pq(1−qt)−2 [=2qp(1−qt)−3 ] M1 Attempt to differentiate their G'(t). 1 M1 Substitute t=1 into their G'(t) and G''(t) and use formula G'(1) = p(1−q)−2 = , G''(1) = 2𝑞 p 𝑃2 for Var(X) (at some point). 2𝑞 1 1 2 M1 (1−q) replaced by p in denominator throughout, and Var(X) = + −( ) 𝑃2 𝑝 𝑝 some cancellation seen (at some point). q A1 AG p2 Convincingly obtained. 5 © UCLES 2023 Page 11 of 13 5(c) 1t  t  B1 Implied by correct final answer. G (t) = 6 = X 1−5t   6−5t   6 2 B1 2  t   1t  G (t) = (G (t))2 = Z X   Allow  6  . 6−5t 1−5t   6 2 © UCLES 2023 Page 12 of 13

3(a) 120 270 108 6 B1 2 correct probabilities. P(3R) = P(2R) = P(1R) = P(0R) = 504 504 504 504 G (t) = 1 ( 6+108t+270t2 +120t3) = 1 ( 1+18t+45t2 +20t3) M1 Cubic polynomial with their probabilities. X 504 84 1 3 15 5 A1 Correct. + t+ t2 + t3 84 14 28 21 3 3(b) G (t) = 1 ( 1+18t+45t2 +20t3)  1 ( 1+6t+5t2) M1 Attempt to multiply out the brackets. Z 84 12 1 ( 1+24t+158t2 +380t3+345t4 +100t5) M1 Obtain quintic polynomial (may not be simplified). 1008 1 1 79 95 115 25 A1 Correct. + t+ t2 + t3+ t4 + t5 1008 42 504 252 336 252 3 © UCLES 2023 Page 8 of 13 3(c) G' = 1 ( 24+316t+1140t2 +1380t3+500t4) M1 Differentiate twice. May not see derivatives in terms of t. 1008 G'' = 1 ( 316+2280t+4140t2 +2000t3) 1008  3360 10 B1 Seen or implied, NFWW (e.g. sampling with  E(Z)=  replacement).  1008 3 8736 10 10 2  26 10 100 M1 Use formula with their G(1) and G'(1). Var(Z) = + − = − −     1008 3  3   3 3 9  8 A1 Or 0.889 9 4 © UCLES 2023 Page 9 of 13

5(a) P(X = r) = p(1− p)r−1 = pqr−1 B1 Implied by G(t) = pqr−1 tr . G(t) = pt(qt)r−1 M1 (Infinite summation) with common ratio (qt) and first term or common factor pt indicated or GP identified.  1  pt A1 AG G(t) = pt = Convincingly obtained. 1−qt 1−qt 3 5(b) G'(t) = pqt(1−qt)−2 + p(1−qt)−1 [= p(1−qt)−2 ] M1 Attempt to differentiate as a product/quotient. G''(t) = 2pq2t(1−qt)−3 +2pq(1−qt)−2 [=2qp(1−qt)−3 ] M1 Attempt to differentiate their G'(t). 1 M1 Substitute t=1 into their G'(t) and G''(t) and use formula G'(1) = p(1−q)−2 = , G''(1) = 2𝑞 p 𝑃2 for Var(X) (at some point). 2𝑞 1 1 2 M1 (1−q) replaced by p in denominator throughout, and Var(X) = + −( ) 𝑃2 𝑝 𝑝 some cancellation seen (at some point). q A1 AG p2 Convincingly obtained. 5 © UCLES 2023 Page 11 of 13 5(c) 1t  t  B1 Implied by correct final answer. G (t) = 6 = X 1−5t   6−5t   6 2 B1 2  t   1t  G (t) = (G (t))2 = Z X   Allow  6  . 6−5t 1−5t   6 2 © UCLES 2023 Page 12 of 13

4(a) t2 43t42t12t243t3 M1 ‘Two’ terms obtained, correct denominator(s). G (t) = , G’ (t) = or Y Y 43t4 43t8 2t 12t2  43t4 43t5 E(Y) = G’ (1) = 14 M1 A1 t = 1 in their expression. Y CWO 3 4(b) t B1 G (t) = X 43t2 1 © Cambridge University Press & Assessment 2024 Page 9 of 15 4(c) G X (t) = t43t2  t  1 3 t   2  t  1 3 t3 9 t2 4 27 t3    M1 Expansion as far as t3 OR term in t4 calculated. 16 4  16 2 16 64  Use P(X = 4) = their coefficient of t4 M1 Find the numerical value of their coefficient of t4. 27 A1 Accept 0.105. 256 3 © Cambridge University Press & Assessment 2024 Page 10 of 15

4(a) t2 43t42t12t243t3 M1 ‘Two’ terms obtained, correct denominator(s). G (t) = , G’ (t) = or Y Y 43t4 43t8 2t 12t2  43t4 43t5 E(Y) = G’ (1) = 14 M1 A1 t = 1 in their expression. Y CWO 3 4(b) t B1 G (t) = X 43t2 1 © Cambridge University Press & Assessment 2024 Page 9 of 15 4(c) G X (t) = t43t2  t  1 3 t   2  t  1 3 t3 9 t2 4 27 t3    M1 Expansion as far as t3 OR term in t4 calculated. 16 4  16 2 16 64  Use P(X = 4) = their coefficient of t4 M1 Find the numerical value of their coefficient of t4. 27 A1 Accept 0.105. 256 3 © Cambridge University Press & Assessment 2024 Page 10 of 15

4(a) 1 B1 Accept 0.03125 or 0.0313. c  32 1 4(b) G (t) = ct1t5 , G’ (t) = M1 Must use product rule OR differentiate their expansion of G X X Allow one error in their differentiation. c  5t1t4 1t5 Condone missing c. E(X) = G’ (1) = c80323.5 A1 Correct work only. X 2 © Cambridge University Press & Assessment 2024 Page 10 of 15 4(c) G (t) = c2t21t10 *M1 Or Y c2 t2 10t3 45t4 120t5 210t6 252t7 210t8 120t9 45t10 10t11t12 G' (t) = c2  10t21t9 2t1t10 DM1 Y Differentiate their G (t) twice, allow one error. Y G'' (t) = Y c2  90t21t8 40t1t9 21t10 ) G' (1) = 7, G’’ (1) = 44.5 M1 A1 Use correct formula and substitute values for t = 1 (may be in terms of c). Y Y Var(Y) = G’’ (1) + G’ (1)   G’12 = 2.5 A1 for correct work only. Y Y Alternative solution for question 4(c) Var (Y) = 2Var(X) *M1 Differentiate G (t) twice DM1 Note that G' (t) has already been found in part (b). X X Var(X)G'' (1)G' (1)  G' 12 M1 Use correct formula and substitute values for t = 1. X X X 1.25  2 = 2.5 A1 4 © Cambridge University Press & Assessment 2024 Page 11 of 15 4(d) P(Y = 5) = coefficient of t5 in expansion of their M1 Identify required term. c2t21t10 Required term is c2120 Expansion = c2t2 110t45t2 120t3   15 A1 0.117, 120 or any equivalent fraction. 128 1024 2 Question Answer Marks Guidance

5(a) 8 14 7 1 B1 All correct. P(0 heads) = , P(1 head) = , P(2 heads) = , P(3 heads) = 30 30 30 30 8 14 7 1 M1 Cubic polynomial. G (t) = + t+ t2 + t3 X 30 30 30 30 A1 FT FT their probabilities that sum to one. 3 © Cambridge University Press & Assessment 2024 Page 9 of 11 5(b) 25 10 1 B1 G (t) = + t+ t2 Y 36 36 36 G Z (t) = 1   8 + 14 t+ 7 t2 + 1 t3   ( 25 +1 0t +t2) M1 With attempt to multiply. 303630 30 30 30  1 ( 200+430t+323t2 +109t3+17t4 +t5) M1 A1 Obtain quintic polynomial. 1080 4 5(c) 1 M1 Differentiate and substitute t = 1. G ’ (1) = (430+646+327+68+5) Z 1080 1476 41 A1 = =1.37 1080 30 2 © Cambridge University Press & Assessment 2024 Page 10 of 11

2(a) 1 B1 [G (t) = + pt+qt2, G’ (t) = p+2qt, ] p+2q=1.1 X X 5 1 4 M1 + p+q=1, p+q= and solve 5 5  1  3 A1 p= , q=    2  10 3 A1 G’’ (t) = 2q = , Var(X) = 0.6+1.1−1.12 =0.49 X 5 4 © Cambridge University Press & Assessment 2024 Page 6 of 13 2(b) 1 1 3  2  1  M1 Wrong p and q (or swapped) or missing t, M1A0. G Z (t) =  + t+ t2  t1+ t2  p and q must have their numerical values from 5 2 10  3  2  part (a). 1 ( 4t+10t2 +8t3+5t4 +3t5) A1 Accept answer in any equivalent form. 30 2 2(c) 16 8 B1 FT FT their fully expanded form of PGF. = oe 30 15 1 2(d) (Z =) 2 B1 Correct work only, must see correct fully expanded polynomial form of PGF. 1 Question Answer Marks Guidance

5(a) 8 14 7 1 B1 All correct. P(0 heads) = , P(1 head) = , P(2 heads) = , P(3 heads) = 30 30 30 30 8 14 7 1 M1 Cubic polynomial. G (t) = + t+ t2 + t3 X 30 30 30 30 A1 FT FT their probabilities that sum to one. 3 © Cambridge University Press & Assessment 2024 Page 9 of 11 5(b) 25 10 1 B1 G (t) = + t+ t2 Y 36 36 36 G Z (t) = 1   8 + 14 t+ 7 t2 + 1 t3   ( 25 +1 0t +t2) M1 With attempt to multiply. 303630 30 30 30  1 ( 200+430t+323t2 +109t3+17t4 +t5) M1 A1 Obtain quintic polynomial. 1080 4 5(c) 1 M1 Differentiate and substitute t = 1. G ’ (1) = (430+646+327+68+5) Z 1080 1476 41 A1 = =1.37 1080 30 2 © Cambridge University Press & Assessment 2024 Page 10 of 11

6(a) k=1−a B1 1 6(b) ( 1−at2)−1 =1+at2(+) *M1 Attempt at series for ( 1−at2)−1 with 1+at2 seen or implied. P(Y 2)=1−k(1+a) DM1 Correct un-simplified expression in terms of their k and/or a. P(Y 2)=1−(1−a)(1+a)=a2 A1 AG, shown convincingly. Alternative method for question 6(b) P(Y =0)=G (0)=k P(Y =1)=G ' (0)=0 *M1 Attempt at G '' (0) with their k . Y  Y  Y P(Y =2)= 1G '' (0)= 1(2ak). 2 Y 2 P(Y 2)=1−k−ak DM1 Correct un-simplified expression in terms of their k and/or a. P(Y 2)=1−(1−a)−a(1−a)=a2 A1 AG, shown convincingly. 3 6(c) G ' (t)= 0.32t M1 Attempt to differentiate G Y (t) with their k , Y ( 1−0.2t2)2 may be in terms of a and/or k . E(Y)=G ' (1)= 0.32 =0.5 A1 CWO Y (1−0.2)2 2 © Cambridge University Press & Assessment 2025 Page 15 of 16

6(a) k=1−a B1 1 6(b) ( 1−at2)−1 =1+at2(+) *M1 Attempt at series for ( 1−at2)−1 with 1+at2 seen or implied. P(Y 2)=1−k(1+a) DM1 Correct un-simplified expression in terms of their k and/or a. P(Y 2)=1−(1−a)(1+a)=a2 A1 AG, shown convincingly. Alternative method for question 6(b) P(Y =0)=G (0)=k P(Y =1)=G ' (0)=0 *M1 Attempt at G '' (0) with their k . Y  Y  Y P(Y =2)= 1G '' (0)= 1(2ak). 2 Y 2 P(Y 2)=1−k−ak DM1 Correct un-simplified expression in terms of their k and/or a. P(Y 2)=1−(1−a)−a(1−a)=a2 A1 AG, shown convincingly. 3 6(c) G ' (t)= 0.32t M1 Attempt to differentiate G Y (t) with their k , Y ( 1−0.2t2)2 may be in terms of a and/or k . E(Y)=G ' (1)= 0.32 =0.5 A1 CWO Y (1−0.2)2 2 © Cambridge University Press & Assessment 2025 Page 15 of 15

6(a) Probabilities: 1 , 7 , 7 for 0, 1, 2 heads. B1 15 15 15 G (t)= 1 + 7 t+ 7 t2 B1 FT FT their probabilities provided p=1. X 15 15 15 Probabilities: 8 , 7 for 1, 2 colours. B1 SC B1 for 8 , 7 for 0,1 colours or 0,2 colours. 15 15 15 15 G (t)= 8 t+ 7 t2 B1 FT FT their probabilities provided p=1. Y 15 15 SC B1 for G (t)= 8 + 7 t from 0,1 colours or Y 15 15 G (t)= 8 + 7 t2 from 0,2 colours. Y 15 15 4 6(b) X and Y are not independent. B1 1 © Cambridge University Press & Assessment 2025 Page 15 of 16 6(c) B1 Correct 3 cases identified correctly where x 0 1 2 probabilities are not zero from 1,2 colours. y 1 2 1 (x,y)=(0,1), (1,2) or (2,1). Award SC B1 for either of the following: z=x+y 1 3 3 (x,y)=(0,0), (1,1) or (2,0) from 0,1 colours. P(Z =z) 1 7 7 (x,y)=(0,0), (1,2) or (2,0) from 0,2 colours. 15 15 15 M1 At least one non-zero probability correct. Allow for 1 , 7 , or 14. 15 15 15 A1 All probabilities correct. G (t)= 1 t+14t3 A1 FT FT their probabilities provided p=1. Z 15 15 Allow A1 FT for either of the following: G (t)= 1 +14t2 from 0,1 colours. Z 15 15 G (t)= 1 + 7 t2 + 7 t3 from 0, 2 colours. Z 15 15 15 6(d) G ' (t)= 1 +42t2 M1 Attempt to differentiate G (t) and find G ' (1). Z 15 15 Z Z G ' (1)= 1 +42 FT their G (t). Z 15 15 Z 43 =2.87 A1 Allow A1 FT for either of the following: 15 28 =1.87 from 0,1 colours. 15 35 =2.33 from 0,2 colours. 15 2 © Cambridge University Press & Assessment 2025 Page 16 of 16

5(a) Probabilities: 8 , 12, 6 , 1 for 0, 1, 2, 3 heads B1 27 27 27 27 8 +12t+ 6 t2 + 1 t3 B1 FT FT their probabilities to form a 4-term cubic. Must have p=1. 27 27 27 27 (2 + 1t )3 scores B2. 3 3 2 5(b) ( 8 +12t+ 6 t2 + 1 t3)( 25+10t+ 1 t2) M1 Attempt to multiply their G (t) and their G (t). 27 27 27 27 36 36 36 X Y 1 ( 200+380t+278t2 +97t3+16t4 +t5) M1 Expand to form polynomial of degree 5 (need not be simplified). 972 A1 CWO 3 5(c) G' (t)= 1 ( 380+556t+291t2 +64t3+5t4) M1 Attempt at differentiating their G (t) . Z 972 Z E(Z)=G' (1)=1296 = 4 A1 FT FT their G (t). Z 972 3 Z G'' (t)= 1 ( 556+582t+192t2 +20t3) M1 Attempt at differentiating their G' (t). Z 972 Z Var (Z) =1 9 3 7 5 2 0+ 3 4− ( 3 4)2  = 1 2 8 5+ 3 4− ( 3 4)2  M1 Use correct formula using their G' Z (1) and their G' Z ' (1). 17 =0.944 A1 CWO 18 5 © Cambridge University Press & Assessment 2025 Page 13 of 15

7(a) G (t)=a+2at+bt2 B1 X G ' (t)=2a+2bt B1 X E(X)=G ' (1)=2a+2b X 2 7(b) G ''(t)=2b B1 X Var(X)=2b+(2a+2b)−(2a+2b)2 M1 Use correct formula with their expressions. 2b+(2a+2b)−(2a+2b)2 =2b+(2a+2b)(1−(2a+2b)) A1 AG, shown convincingly. =2b+2(a+b)(1−2a−2b) Alternative method for question 7(b) E ( X2) =02a+122a+22b=2a+4b B1 Var(X)=(2a+4b)−(2a+2b)2 M1 Use correct formula with their expressions. (2a+4b)−(2a+2b)2 =2b+2a+2b−(2a+2b)2 A1 AG, shown convincingly. =2b+(2a+2b)(1−(2a+2b)) =2b+2(a+b)(1−2a−2b) 3 © Cambridge University Press & Assessment 2025 Page 15 of 16 7(c) G (t)= ( a+2at+bt2)10 B1 FT FT their G X (t). Y G '(t)=20(a+bt)( a+2at+bt2)9 M1 Attempt to differentiate their G Y (t) and evaluate at t =1. Y E(Y)=20(a+b)(3a+b)9 3a + b =1 , hence E(Y)=20(a+b)=10E(X). A1 3a+b=1 since total probability is 1, OE. 3 7(d) Binomial *B1 Uses PGF to identify binomial.  2 DB1 Identifies binomial with correct parameters. B10,   3 2 © Cambridge University Press & Assessment 2025 Page 16 of 16

6(a) G ' (t)= (3−2t)2 +4(3−2t)t   = 3+2t   M1 Attempt to differentiate G X (t). X (3−2t)4  (3−2t)3  May be implied by expressions for G ' (1).   X E(X)=G ' (1)=5 A1 X 2(3−2t)3 +6(3+2t)(3−2t)2  24+8t  M1 Attempt to differentiate G ' (t). G ''(t)= =  X X (3−2t)6  (3−2t)4  May be implied by expressions for G ''(1).   X Var(X)=G ''(1)+G ' (1)− ( G ' (1) )2 =32+5−25 M1 Use correct formula using their values. X X X Var(X)=12 A1 CWO   5 © Cambridge University Press & Assessment 2025 Page 14 of 15 6(b) G (t)= t3 = 1 t3( 1+(−4)( −2t ) + ) M1 Multiply and attempt at binomial expansion. Z (3−2t)4 81 3 P(Z =4) = their coefficient of t4. M1 Attempt at finding their coefficient of t3 or t4. Either seen earns M1. OR P(Z =3) = their coefficient of t3. P(Z =4)= 8 and P(Z =3)= 1 A1 Both values seen.     243 81 P(Z 4)=1−' 8 '−' 1 ' M1 Attempt to find P(Z 4) using their values. 243 81 P(Z 4)= 232 A1 AWRT 0.955. 243 Note: considering the expansions of the probability generating functions for X and Y separately, then finding P(Z 4)=1−P(Z≤4) with P(Z≤4)=P(X =1,Y =2)+P(X =1,Y =3)+P(X =2,Y =2) earns full credit. 5 © Cambridge University Press & Assessment 2025 Page 15 of 15

7(a) G (t)=a+2at+bt2 B1 X G ' (t)=2a+2bt B1 X E(X)=G ' (1)=2a+2b X 2 7(b) G ''(t)=2b B1 X Var(X)=2b+(2a+2b)−(2a+2b)2 M1 Use correct formula with their expressions. 2b+(2a+2b)−(2a+2b)2 =2b+(2a+2b)(1−(2a+2b)) A1 AG, shown convincingly. =2b+2(a+b)(1−2a−2b) Alternative method for question 7(b) E ( X2) =02a+122a+22b=2a+4b B1 Var(X)=(2a+4b)−(2a+2b)2 M1 Use correct formula with their expressions. (2a+4b)−(2a+2b)2 =2b+2a+2b−(2a+2b)2 A1 AG, shown convincingly. =2b+(2a+2b)(1−(2a+2b)) =2b+2(a+b)(1−2a−2b) 3 © Cambridge University Press & Assessment 2025 Page 15 of 16 7(c) G (t)= ( a+2at+bt2)10 B1 FT FT their G X (t). Y G '(t)=20(a+bt)( a+2at+bt2)9 M1 Attempt to differentiate their G Y (t) and evaluate at t =1. Y E(Y)=20(a+b)(3a+b)9 3a + b =1 , hence E(Y)=20(a+b)=10E(X). A1 3a+b=1 since total probability is 1, OE. 3 7(d) Binomial *B1 Uses PGF to identify binomial.  2 DB1 Identifies binomial with correct parameters. B10,   3 2 © Cambridge University Press & Assessment 2025 Page 16 of 16

4(a) G (t)= 2t+3t2 B1 CAO X 5 5 1 4(b) G (t)= 1 t4(2+3t)4  =t4(2+3t )4 M1 Attempt at 4th power of their G X (t). Y 625  5 5  A1 2 4(c) Attempt at coefficient of t6intheirG (t). M1 Identify required term, coefficient must a non-zero probability. Y May be seen in part 4(b). P(Y =6)= 216 A1 AWRT 0.346. 625 2 © Cambridge University Press & Assessment 2025 Page 14 of 17 4(d) G '(t)= 4 t3(2+3t)4 + 12 t4(2+3t)3 M1 Attempt at differentiating their G (t) from part 4(b). Y 625 625 Y OR G '(t)= 1 ( 64t3+480t4 +1296t5 +1512t6 +648t7) Y 625 G ''(t)= 12 t2(2+3t)4 + 96 t3(2+3t)3 +108t4(2+3t)2 M1 Attempt at differentiating their G '(t). Y 625 625 625 Y OR G ''(t)= 1 ( 192t2 +1920t3+6480t4 +9072t5+4536t6) Y 625 G '(1)=32 =6.4, G ''(1)=888=35.52 B1 WWW Y 5 Y 25 Var(Y)=888+32− (32)2 M1 Use correct formula with their values. 25 5 5 Var(Y)= 24 =0.96 A1 CWO 25 Alternative method for question 4(d) E(X)=2+32=8or useG ' (t)= 2+6t. M1 5 5 5 X 5 5 E ( X2) = 2+34=14 or useG ''(t)= 6 M1 5 5 5 X 5 Var(X)=14− (8)2 or Var(X)= 6+8− (8)2 M1 5 5 5 5 5 Var(X)= 6 A1 25 Var(Y)=4Var(X)= 24 A1 CWO 25 5 © Cambridge University Press & Assessment 2025 Page 15 of 17

6(a) k(a+1)2 =1 *B1 © Cambridge University Press & Assessment 2026 Page 16 of 17 k a 2 + 2 a k + k = 1 k a 2 − k = 13 *M1 Uses G  ( 1 ) = 13 to form equation in k and a. − k ( a + 1 ) 2 + 2 k a ( a + 1 ) = 13 ( a a 2 + − 1 1 ) 2 = 1 3 DM1 Uses their equations to eliminate k or a. a = 2 A1 AG, convincingly obtained. k = 19 B1 5 6(b) EITHER: G ''X ( t ) = 9 2 t 3 , G ''X ( 1 ) = 2 9  = 2 k  (B1FT CWO with their k. Differentiates their G 'X ( t ) and evaluates at t = 1 . May be in terms of k. Var(X)= 2+1−(1)2 = 4 B1FT) Uses their values correctly in correct formula to 9 3 3 9 obtain positive value. Trace ID: PC-6B4604D1D5 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) OR: © Cambridge University Press & Assessment 2026 Page 17 of 17 E ( X 2 ) = 59  = 5 k  (B1FT x –1 0 1 P(X =x) k 4k 4k Var(X)=5−(1)2 = 4 B1FT) Uses their values correctly in correct formula to 9 3 9 obtain positive value. Available marks 2 6(c) EITHER: (B1FT SOI 1  k3  FT their k. May be in terms of k. G (t)= (2t+1)6 = (2t+1)6  Y 729t3  t3  Coefficient of t 0 term: 7 1 2 9  6 3  ( 2 ) 3   = k 3  6 3  ( 2 ) 3  M1 Identifies appropriate term and attempt to calculate. May be in terms of k. P ( Y = 0 ) = 17 62 09 A1) AWRT 0.219 OR: Relevant values for ( X 1 , X 2 , X 3 ) (B1 All seven cases considered, SOI. are (–1, 0, 1), 6 possible orderings, and (0,0,0). 6144+(4)3 =6k(4k)2 +(4k)3 M1 Must be correct form using their k, may be in 9 9 9 9   terms of k. 17 62 09 A1) Available marks 3 Trace ID: PC-6B4604D1D5 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

6(a) k(a+1)2 =1 *B1 © Cambridge University Press & Assessment 2026 Page 16 of 17 k a 2 + 2 a k + k = 1 k a 2 − k = 13 *M1 Uses G  ( 1 ) = 13 to form equation in k and a. − k ( a + 1 ) 2 + 2 k a ( a + 1 ) = 13 ( a a 2 + − 1 1 ) 2 = 1 3 DM1 Uses their equations to eliminate k or a. a = 2 A1 AG, convincingly obtained. k = 19 B1 5 6(b) EITHER: (B1FT CWO with their k. G '' (t)= 2 , G '' (1)= 2 =2k Differentiates their X 9t3 X 9 G 'X ( t ) and evaluates at t = 1 . May be in terms of k. Var(X)= 2+1− (1)2 = 4 B1FT) Uses their values correctly in correct formula to 9 3 3 9 obtain positive value. Trace ID: PC-E48FEEF9A8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) OR: (B1FT x –1 0 1 E ( X2) = 5 =5k P(X =x) k 4k 4k 9 Var(X)=5− (1)2 = 4 B1FT) Uses their values correctly in correct formula to 9 3 9 obtain positive value. Available marks 2 6(c) EITHER: (B1FT SOI 1  k3  FT their k. May be in terms of k. G (t)= (2t+1)6 = (2t+1)6  Y 729t3  t3  Coefficient of © Cambridge University Press & Assessment 2026 Page 17 of 17 t 0 term: 7 1 2 9  6 3  ( 2 ) 3   = k 3  6 3  ( 2 ) 3  M1 Identifies appropriate term and attempt to calculate. May be in terms of k. P ( Y = 0 ) = 17 62 09 A1) AWRT 0.219 OR: Relevant values for ( X 1 , X 2 , X 3 ) (B1 All seven cases considered, SOI. are (–1, 0, 1), 6 possible orderings, and (0,0,0). 6  19  49  49 + ( 49 ) 3   = 6  k  ( 4 k ) 2 + ( 4 k ) 3  M1 Must be correct form using their k, may be in terms of k. 17 62 09 A1) Available marks 3 Trace ID: PC-E48FEEF9A8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

3(a) © Cambridge University Press & Assessment 2026 Page 11 of 15 G 'X ( t ) = 0 .3 + 0 .8 t + 0 .6 t 2 M1 Attempt at differentiating G X ( t ) . May be implied by expression for G 'X ( 1 ) seen. E ( X ) = 1 .7 A1 CWO G ''X ( t ) = 0 .8 + 1 .2 t M1 Attempt at differentiating G 'X ( t ) . May be implied by expression for G ''X ( 1 ) seen. V a r ( X ) = 2 + 1 .7 − 1 .7 2 M1 Use correct formula using their G ' (1) and X G ''X ( 1 ) . 0.81 A1 CWO 5 3(b) G Y ( t ) = 0 .1 t 2 + 0 .3 t 5 + 0 .4 t 8 + 0 .2 t 1 1 B1 Correct coefficients in a 4-term polynomial. B1 Correct powers of t. 2 Question Answer Marks Guidance

2(a) G ' (t)= 1 ( t2 +2 ) + 2t(t+1)= 1 ( 3t2 +2t+2 )  X 6 6 6  © Cambridge University Press & Assessment 2026 Page 10 of 18  E ( X ) =  76 B1 AWRT 1.17  G ''X ( t ) = t + 13   G ''X ( 1 ) = 43 M1 Attempt to differentiate their G 'X ( t ) and evaluate at t = 1 . May be implied by an expression for G '' (1) or 4 seen. X 3 V a r ( X ) = 43 + 76 − ( 76 ) 2 M1 Use correct formula with their values. 43 16  = 1 .1 4  A1 AWRT 1.14, NFWW (for example, an incorrect G ''X ( t ) , w i t h G ''X ( 1 ) = 43 scores max B1 M1 M1 A0 XP). 4 Trace ID: PC-1430F78DC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2(b) EITHER: (*M1 SOI, must be cubed. G Y (t)=    1 6 (t+1)( t2 +2 )   3    =    1 6 ( t3+t2 +2t+2 )   3   © Cambridge University Press & Assessment 2026 Page 11 of 18 2 11 6 ( 8 + 2 4 t + 3 6 t 2 +  )  = 12 7 + 19 t + 16 t 2 +   DM1 Attempt to expand to form a polynomial up to at least t2. May be unsimplified. May be implied by a sum of coefficients or AWRT 0.315. ISW with powers of t that are 3 or above. P(Y 2)= 8 + 24 + 36 = 1 + 1+ 1 =17 =0.315 A1) AWRT 0.315 216 216 216 27 9 6 54 OR: Alternative method for X 1 , X 2 , X 3 : P(all zero)= (1)3 , P( two 0s, one 1)=3 (1)2  (1) 3 3 3 P ( t w o 0 s , o n e 2 ) = 3  ( 13 ) 2  ( 16 ) , P ( o n e 0 , t w o 1 s ) = 3  13  ( 13 ) 2 (M1 Consider all 4 cases. M1 Correct probability calculations for at least 2 cases. Only 2 cases considered can score a maximum of M0 M1 A0. 15 74 A1) AWRT 0.315, CWO Available marks 3 Question Answer Marks Guidance