Question Simulator

AS & A Level Further Mathematics (9231)

Probability generating functions Banker

9 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 42 · Q6 (a, b, c)10 marksFinding & using a PGF · Mean & variance from a PGF · PGF of a sum of independent variables

2025

  • 2025 Oct/Nov · Paper 42 · Q6 (a, b)10 marksMean & variance from a PGF · PGF of a sum of independent variables
  • 2025 May/June · Paper 42 · Q6 (a, b, c)6 marksFinding & using a PGF · Mean & variance from a PGF

2024

  • 2024 Oct/Nov · Paper 42 · Q2 (a, b, c, d)8 marksFinding & using a PGF · Mean & variance from a PGF · PGF of a sum of independent variables
  • 2024 May/June · Paper 42 · Q4 (a, b, c)7 marksFinding & using a PGF · Mean & variance from a PGF · PGF of a sum of independent variables

2023

  • 2023 Oct/Nov · Paper 42 · Q3 (a, b, c)10 marksFinding & using a PGF · Mean & variance from a PGF · PGF of a sum of independent variables
  • 2023 May/June · Paper 42 · Q5 (a, b, c)9 marksFinding & using a PGF · Mean & variance from a PGF · PGF of a sum of independent variables

2022

  • 2022 Oct/Nov · Paper 42 · Q5 (a, b, c, d)9 marksFinding & using a PGF · Mean & variance from a PGF · PGF of a sum of independent variables
  • 2022 May/June · Paper 42 · Q2 (a, b)7 marksMean & variance from a PGF · PGF of a sum of independent variables

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Mean & variance from a PGF · Core · Paper 4 style · 5 marks
  1. (a) The number of goals, X, scored by a school hockey team in a match has probability generating function G(t) = (1/8)(1 + 3t + 4t²). Write down P(X = 2). [1]
  2. (b) Use G(t) to find E(X) and Var(X). [4]
Show answer
  1. (a) 1/2
    B1: coefficient of t² = 4/8.
  2. (b) E(X) = 11/8; Var(X) = 31/64
    M1: G′(t) = (3 + 8t)/8, G′(1) = 11/8.A1: E(X) = 11/8.M1: G″(1) = 1; Var = 1 + 11/8 − 121/64.A1: = 31/64.
Q2Mean & variance from a PGF · Core · Paper 4 style · 4 marks
  1. (a) A discrete random variable Y has probability generating function G(t) = (t + 2t³ + t⁵)/4. Find E(Y) and Var(Y). [4]
Show answer
  1. (a) E(Y) = 3; Var(Y) = 2
    M1: G′(t) = (1 + 6t² + 5t⁴)/4, G′(1) = 3.A1: E(Y) = 3.M1: G″(t) = (12t + 20t³)/4, G″(1) = 8.A1: Var(Y) = 8 + 3 − 9 = 2.
Q3Mean & variance from a PGF · Core · Paper 4 style · 5 marks
  1. (a) The number of seeds that germinate in a tray, X, has probability generating function G(t) = (0.3 + 0.7t)⁶. Use differentiation of G(t) to find E(X) and Var(X). [4]
  2. (b) Name the distribution of X. [1]
Show answer
  1. (a) E(X) = 4.2; Var(X) = 1.26
    M1: G′(t) = 4.2(0.3 + 0.7t)⁵, G′(1) = 4.2.A1: E(X) = 4.2.M1: G″(t) = 14.7(0.3 + 0.7t)⁴, G″(1) = 14.7.A1: Var(X) = 14.7 + 4.2 − 17.64 = 1.26.
  2. (b) B(6, 0.7)
    B1: binomial, n = 6, p = 0.7.
Q4Mean & variance from a PGF · Core · Paper 4 style · 5 marks
  1. (a) A random variable W has probability generating function G(t) = (1/27) t(2 + t)³. Find E(W) and Var(W). [4]
  2. (b) Find P(W = 1). [1]
Show answer
  1. (a) E(W) = 2; Var(W) = 2/3
    M1: G′(t) = (1/27)[(2 + t)³ + 3t(2 + t)²], G′(1) = (27 + 27)/27 = 2.A1: E(W) = 2.M1: G″(t) = (1/27)[6(2 + t)² + 6t(2 + t)], G″(1) = (54 + 18)/27 = 8/3.A1: Var(W) = 8/3 + 2 − 4 = 2/3.
  2. (b) 8/27
    B1: coefficient of t is 2³/27 = 8/27.
Q5Mean & variance from a PGF · Standard · Paper 4 style · 7 marks
  1. (a) The number of seats, X, booked at a cinema counter in Peshawar in a 5-minute period has probability generating function G(t) = e^(3(t² − 1)). Find E(X) and Var(X). [5]
  2. (b) Explain why X must always be even, and describe X in terms of a Poisson variable. [2]
Show answer
  1. (a) E(X) = 6; Var(X) = 12
    M1: G′(t) = 6t e^(3(t²−1)), G′(1) = 6.A1: E(X) = 6.M1: G″(t) = (6 + 36t²) e^(3(t²−1)), G″(1) = 42.M1: Var = 42 + 6 − 36.A1: = 12.
  2. (b) Only even powers of t appear; X = 2Y where Y ~ Po(3).
    B1: e^(3(t²−1)) is a series in t² only.B1: X = 2Y with Y ~ Po(3) (so E = 6, Var = 4 × 3 = 12, consistent).
Q6Mean & variance from a PGF · Standard · Paper 4 style · 5 marks
  1. (a) The PGF of a random variable X is G(t) = (a + bt)², where a and b are positive constants. Given that E(X) = 1.2, find a and b. [3]
  2. (b) Find Var(X) using G(t). [2]
Show answer
  1. (a) a = 0.4, b = 0.6
    M1: G(1) = 1 ⇒ a + b = 1.M1: G′(1) = 2b(a + b) = 2b = 1.2.A1: b = 0.6, a = 0.4.
  2. (b) 0.48
    M1: G″(t) = 2b², G″(1) = 0.72.A1: Var = 0.72 + 1.2 − 1.44 = 0.48.
Q7Mean & variance from a PGF · Standard · Paper 4 style · 4 marks
  1. (a) The score, S, on a fair six-sided dice has probability generating function G(t) = (1/6)(t + t² + t³ + t⁴ + t⁵ + t⁶). Use G(t) to find E(S) and Var(S). [4]
Show answer
  1. (a) E(S) = 7/2; Var(S) = 35/12
    M1: G′(1) = (1 + 2 + … + 6)/6 = 21/6.A1: E(S) = 7/2.M1: G″(1) = (0 + 2 + 6 + 12 + 20 + 30)/6 = 70/6.A1: Var(S) = 70/6 + 7/2 − 49/4 = 35/12.
Q8Mean & variance from a PGF · Standard · Paper 4 style · 5 marks
  1. (a) A random variable X takes only the values 0, 1 and 2. Its PGF is G(t) = p + qt + rt². Given that E(X) = 1 and Var(X) = 0.5, find p, q and r. [5]
Show answer
  1. (a) p = 0.25, q = 0.5, r = 0.25
    B1: p + q + r = 1.B1: G′(1) = q + 2r = 1.M1: G″(1) = 2r; Var = 2r + 1 − 1 = 2r = 0.5.A1: r = 0.25, q = 0.5.A1: p = 0.25.
Q9Mean & variance from a PGF · Challenge · Paper 4 style · 6 marks
  1. (a) A random variable X has probability generating function G(t) = t²/(3 − 2t). Show that G(1) = 1 and find E(X). [3]
  2. (b) Find Var(X). [3]
Show answer
  1. (a) E(X) = 4
    B1: G(1) = 1/1 = 1.M1: G′(t) = [2t(3 − 2t) + 2t²]/(3 − 2t)² = (6t − 2t²)/(3 − 2t)².A1: G′(1) = 4.
  2. (b) 6
    M1: differentiate G′(t) = (6t − 2t²)(3 − 2t)⁻²: G″(t) = (6 − 4t)(3 − 2t)⁻² + 4(6t − 2t²)(3 − 2t)⁻³.A1: G″(1) = 2 + 16 = 18.A1: Var = 18 + 4 − 16 = 6.
Q10Mean & variance from a PGF · Challenge · Paper 4 style · 7 marks
  1. (a) The PGF of a random variable X is G(t) = kt/(5 − 2t). Find the value of k. [1]
  2. (b) Find E(X) and Var(X) by differentiating G(t). [5]
  3. (c) State the distribution of X, including its parameter. [1]
Show answer
  1. (a) k = 3
    B1: G(1) = k/3 = 1.
  2. (b) E(X) = 5/3; Var(X) = 10/9
    M1: G′(t) = 15/(5 − 2t)².A1: G′(1) = 15/9 = 5/3.M1: G″(t) = 60/(5 − 2t)³, G″(1) = 60/27 = 20/9.M1: Var = 20/9 + 5/3 − 25/9.A1: = 10/9.
  3. (c) Geometric, Geo(3/5)
    B1: G(t) = (3/5)t/(1 − (2/5)t).
Q11Mean & variance from a PGF · Challenge · Paper 4 style · 6 marks
  1. (a) A random variable X, taking positive integer values, has PGF G(t) = ln(1 − t/2)/ln(1/2). Find E(X), giving the exact value and the value correct to 3 significant figures. [3]
  2. (b) Find Var(X) correct to 3 significant figures. [3]
Show answer
  1. (a) E(X) = 1/ln 2 ≈ 1.44
    M1: G′(t) = [−(1/2)/(1 − t/2)]/ln(1/2).A1: G′(1) = −1/ln(1/2) = 1/ln 2.A1: = 1.44.
  2. (b) 0.804
    M1: G″(t) = [−(1/4)/(1 − t/2)²]/ln(1/2), G″(1) = 1/ln 2.M1: Var = 2/ln 2 − 1/(ln 2)².A1: = 0.804.
Q12PGF of a sum of independent variables · Core · Paper 4 style · 5 marks
  1. (a) The independent random variables X and Y have distributionsX: P(X = 0) = 0.2, P(X = 1) = 0.5, P(X = 2) = 0.3Y: P(Y = 1) = 0.6, P(Y = 2) = 0.4.Write down the PGFs of X and Y. [2]
  2. (b) Find the PGF of Z = X + Y in expanded form, and hence find P(Z = 3). [3]
Show answer
  1. (a) G_X(t) = 0.2 + 0.5t + 0.3t²; G_Y(t) = 0.6t + 0.4t²
    B1: G_X.B1: G_Y.
  2. (b) G_Z(t) = 0.12t + 0.38t² + 0.38t³ + 0.12t⁴; P(Z = 3) = 0.38
    M1: G_Z = G_X G_Y (independent).A1: 0.12t + 0.38t² + 0.38t³ + 0.12t⁴.A1: P(Z = 3) = 0.38.
Q13PGF of a sum of independent variables · Core · Paper 4 style · 5 marks
  1. (a) At a school fun fair in Karachi, a spinner A gives scores 1, 2 or 3, each with probability 1/3, and a fair coin gives a bonus score of 0 (tail) or 1 (head). The total T is the spinner score plus the coin score. Find the PGF of T as a polynomial. [3]
  2. (b) Use your PGF to find E(T). [2]
Show answer
  1. (a) G_T(t) = (t + 2t² + 2t³ + t⁴)/6
    B1: G_A(t) = (t + t² + t³)/3, G_C(t) = (1 + t)/2.M1: multiply (independent).A1: (t + 2t² + 2t³ + t⁴)/6.
  2. (b) 5/2
    M1: G′(1) = (1 + 4 + 6 + 4)/6.A1: = 15/6 = 5/2.
Q14PGF of a sum of independent variables · Core · Paper 4 style · 5 marks
  1. (a) X ~ B(4, 0.3) and Y ~ B(6, 0.3) are independent. Write down the PGFs of X and Y, and hence show that X + Y has a binomial distribution, stating its parameters. [3]
  2. (b) Find P(X + Y = 2), correct to 3 significant figures. [2]
Show answer
  1. (a) (0.7 + 0.3t)⁴ × (0.7 + 0.3t)⁶ = (0.7 + 0.3t)¹⁰, so X + Y ~ B(10, 0.3)
    B1: G_X = (0.7 + 0.3t)⁴, G_Y = (0.7 + 0.3t)⁶.M1: product of PGFs for independent variables.A1: (0.7 + 0.3t)¹⁰, B(10, 0.3).
  2. (b) 0.233
    M1: ¹⁰C₂ (0.3)²(0.7)⁸.A1: = 0.233.
Q15PGF of a sum of independent variables · Standard · Paper 4 style · 5 marks
  1. (a) The numbers of emails, X and Y, received per hour by two independent offices have Poisson distributions with means 2 and 3 respectively. Using PGFs, show that X + Y has a Poisson distribution and state its mean. [3]
  2. (b) Find P(X + Y ≤ 1) in the form ke⁻⁵, and give its value correct to 3 significant figures. [2]
Show answer
  1. (a) e^(2(t−1)) e^(3(t−1)) = e^(5(t−1)), so X + Y ~ Po(5)
    B1: G_X = e^(2(t−1)), G_Y = e^(3(t−1)).M1: multiply.A1: e^(5(t−1)) is the PGF of Po(5); mean 5.
  2. (b) 6e⁻⁵ ≈ 0.0404
    M1: P(0) + P(1) = e⁻⁵ + 5e⁻⁵.A1: 6e⁻⁵ = 0.0404.
Q16PGF of a sum of independent variables · Standard · Paper 4 style · 5 marks
  1. (a) A random variable X has PGF G(t) = (1 + 2t + t²)/4. Three independent observations X₁, X₂, X₃ are taken and S = X₁ + X₂ + X₃. Show that the PGF of S can be written as ((1 + t)/2)⁶. [2]
  2. (b) Hence find P(S = 3) and Var(S). [3]
Show answer
  1. (a) G_S = [(1 + t)²/4]³ = ((1 + t)/2)⁶
    M1: G_S = [G(t)]³.A1: 1 + 2t + t² = (1 + t)², so G_S = (1 + t)⁶/64.
  2. (b) P(S = 3) = 5/16; Var(S) = 3/2
    M1: coefficient of t³ = ⁶C₃/64 = 20/64.A1: 5/16.B1: S ~ B(6, 1/2), Var = 6 × 1/4 = 3/2.
Q17PGF of a sum of independent variables · Standard · Paper 4 style · 6 marks
  1. (a) The independent random variables X and Y have PGFs G_X(t) = 0.4 + 0.6t and G_Y(t) = t²(0.5 + 0.5t). Show that the PGF of 2Y is G_Y(t²). [2]
  2. (b) Find the PGF of Z = X + 2Y in expanded form, and hence find P(Z = 5) and E(Z). [4]
Show answer
  1. (a) E(t^(2Y)) = E((t²)^Y) = G_Y(t²)
    M1: G_(2Y)(t) = E(t^(2Y)).A1: = E((t²)^Y) = G_Y(t²).
  2. (b) G_Z = 0.2t⁴ + 0.3t⁵ + 0.2t⁶ + 0.3t⁷; P(Z = 5) = 0.3; E(Z) = 5.6
    M1: G_Z = (0.4 + 0.6t) × t⁴(0.5 + 0.5t²).A1: 0.2t⁴ + 0.3t⁵ + 0.2t⁶ + 0.3t⁷.A1: P(Z = 5) = 0.3.A1: E(Z) = 0.8 + 1.5 + 1.2 + 2.1 = 5.6.
Q18PGF of a sum of independent variables · Standard · Paper 4 style · 6 marks
  1. (a) A game uses a fair three-sided spinner labelled 0, 1, 2. The spinner is spun three times and S is the total score. Find the PGF of S and hence find P(S = 3). [4]
  2. (b) Find Var(S). [2]
Show answer
  1. (a) G_S = ((1 + t + t²)/3)³; P(S = 3) = 7/27
    B1: single spin G(t) = (1 + t + t²)/3.M1: G_S = G(t)³.M1: (1 + t + t²)³ = 1 + 3t + 6t² + 7t³ + 6t⁴ + 3t⁵ + t⁶.A1: P(S = 3) = 7/27.
  2. (b) 2
    M1: Var(single) = 5/3 − 1 = 2/3.A1: Var(S) = 3 × 2/3 = 2.
Q19PGF of a sum of independent variables · Challenge · Paper 4 style · 8 marks
  1. (a) A cricketer in a practice session hits a six on any ball with probability 1/4, independently. X₁ is the number of balls up to and including her first six, and X₂ is the number of further balls up to and including her second six. Write down the PGF of X₁ and find the PGF of N = X₁ + X₂. [3]
  2. (b) By expanding G_N(t), find P(N = 4). [3]
  3. (c) Find E(N) using G_N(t). [2]
Show answer
  1. (a) G(t) = (t/4)/(1 − 3t/4) = t/(4 − 3t); G_N(t) = t²/(4 − 3t)²
    B1: geometric PGF t/(4 − 3t).M1: X₁, X₂ independent, identically distributed: square.A1: G_N = t²/(4 − 3t)².
  2. (b) 27/256
    M1: G_N = (t²/16)(1 − 3t/4)⁻².M1: coefficient of t² in (1 − 3t/4)⁻² is 3 × (3/4)².A1: P(N = 4) = (1/16) × 27/16 = 27/256.
  3. (c) 8
    M1: G_N′(t) = 2t/(4 − 3t)² + 6t²/(4 − 3t)³.A1: G_N′(1) = 2 + 6 = 8.
Q20PGF of a sum of independent variables · Challenge · Paper 4 style · 7 marks
  1. (a) The random variable S = X + Y, where X and Y are independent, has PGF G_S(t) = (0.2 + 0.8t)³ e^(2(t−1)). Identify the distributions of X and Y. [2]
  2. (b) Find E(S) and Var(S). [3]
  3. (c) Find P(S = 0), correct to 3 significant figures. [2]
Show answer
  1. (a) X ~ B(3, 0.8), Y ~ Po(2)
    B1: binomial factor.B1: Poisson factor.
  2. (b) E(S) = 4.4; Var(S) = 2.48
    M1: E(S) = 3 × 0.8 + 2.A1: 4.4.A1: Var(S) = 3 × 0.8 × 0.2 + 2 = 2.48.
  3. (c) 0.00108
    M1: G_S(0) = 0.2³ e⁻².A1: = 0.00108.
Q21PGF of a sum of independent variables · Challenge · Paper 4 style · 7 marks
  1. (a) Each of n independent identical components in a solar panel array either works (scored 1) or fails (scored 0); the score of one component has PGF (1 + 2t)/3. S_n is the total score of the n components. Write down the PGF of S_n. [1]
  2. (b) Find the least n for which P(S_n = 0) < 0.01. [3]
  3. (c) For this value of n, find P(S_n ≥ 4). [3]
Show answer
  1. (a) ((1 + 2t)/3)ⁿ
    B1.
  2. (b) n = 5
    M1: P(S_n = 0) = G(0) = (1/3)ⁿ.M1: (1/3)ⁿ < 0.01 ⇒ n > ln 100/ln 3 = 4.19.A1: n = 5.
  3. (c) 112/243
    M1: coefficients of t⁴ and t⁵ in (1 + 2t)⁵/243: 5 × 16 + 32.M1: = 80 + 32 = 112.A1: 112/243.
Q22Finding & using a PGF · Core · Paper 4 style · 5 marks
  1. (a) The number of power cuts, X, in a town on a summer day has the distributionx: 0, 1, 2, 3P(X = x): 0.5, 0.3, 0.15, 0.05.Write down the probability generating function of X. [1]
  2. (b) Use G(t) to find E(X) and Var(X). [4]
Show answer
  1. (a) G(t) = 0.5 + 0.3t + 0.15t² + 0.05t³
    B1.
  2. (b) E(X) = 0.75; Var(X) = 0.7875
    M1: G′(1) = 0.3 + 0.3 + 0.15.A1: E(X) = 0.75.M1: G″(1) = 0.3 + 0.3 = 0.6; Var = 0.6 + 0.75 − 0.5625.A1: = 0.7875.
Q23Finding & using a PGF · Core · Paper 4 style · 5 marks
  1. (a) X ~ B(n, p). Show that the probability generating function of X is (q + pt)ⁿ, where q = 1 − p. [2]
  2. (b) A quiz in an Urdu class has 5 multiple-choice questions, each guessed correctly with probability 0.2 independently. Use the PGF of the number of correct guesses to find its mean and variance. [3]
Show answer
  1. (a) Shown (see working).
    M1: G(t) = Σ ⁿCᵣ pʳ qⁿ⁻ʳ tʳ = Σ ⁿCᵣ (pt)ʳ qⁿ⁻ʳ.A1: = (q + pt)ⁿ by the binomial theorem.
  2. (b) Mean 1; variance 0.8
    M1: G = (0.8 + 0.2t)⁵, G′(1) = 1.M1: G″(1) = 20 × 0.04 = 0.8.A1: mean 1, variance 0.8 + 1 − 1 = 0.8.
Q24Finding & using a PGF · Standard · Paper 4 style · 7 marks
  1. (a) Bilal takes his driving test repeatedly until he passes. The probability that he passes on any attempt is 0.35, independently. X is the number of attempts needed. Show that the PGF of X is G(t) = 0.35t/(1 − 0.65t), stating the range of t for which it is valid. [3]
  2. (b) Use G(t) to find E(X) and Var(X), correct to 3 significant figures. [4]
Show answer
  1. (a) G(t) = 0.35t/(1 − 0.65t), |t| < 1/0.65
    M1: G(t) = Σ (r ≥ 1) 0.35(0.65)^(r−1) tʳ = 0.35t Σ (0.65t)^(r−1).A1: sum of geometric series = 0.35t/(1 − 0.65t).B1: valid for |0.65t| < 1.
  2. (b) E(X) = 2.86; Var(X) = 5.31
    M1: G′(t) = 0.35/(1 − 0.65t)², G′(1) = 0.35/0.1225 = 2.857.A1: E(X) = 2.86.M1: G″(t) = 0.455/(1 − 0.65t)³, G″(1) = 10.61; Var = 10.61 + 2.857 − 8.163.A1: = 5.31.
Q25Finding & using a PGF · Standard · Paper 4 style · 7 marks
  1. (a) X ~ Po(λ). Show that the PGF of X is e^(λ(t − 1)). [2]
  2. (b) Hence show that E(X) = Var(X) = λ. [3]
  3. (c) The number of calls to a helpline per minute is Po(1.5). By considering G(1) + G(−1), find the probability that the number of calls in a given minute is even, correct to 3 significant figures. [2]
Show answer
  1. (a) Shown (see working).
    M1: G(t) = Σ e^(−λ) λʳ tʳ/r! = e^(−λ) Σ (λt)ʳ/r!.A1: = e^(−λ) e^(λt) = e^(λ(t−1)).
  2. (b) Shown (see working).
    M1: G′(t) = λe^(λ(t−1)), G′(1) = λ.M1: G″(1) = λ².A1: Var = λ² + λ − λ² = λ.
  3. (c) 0.525
    M1: P(even) = (G(1) + G(−1))/2 = (1 + e⁻³)/2.A1: = 0.525.
Q26Finding & using a PGF · Standard · Paper 4 style · 5 marks
  1. (a) A random variable X has P(X = r) = k/2ʳ for r = 1, 2, 3, 4, and zero otherwise. Find k and write down the PGF of X. [3]
  2. (b) Use G(t) to find E(X). [2]
Show answer
  1. (a) k = 16/15; G(t) = (8t + 4t² + 2t³ + t⁴)/15
    M1: k(1/2 + 1/4 + 1/8 + 1/16) = 1.A1: 15k/16 = 1, k = 16/15.B1: G(t) = (8t + 4t² + 2t³ + t⁴)/15.
  2. (b) 26/15
    M1: G′(1) = (8 + 8 + 6 + 4)/15.A1: = 26/15.
Q27Finding & using a PGF · Standard · Paper 4 style · 6 marks
  1. (a) A bag contains 3 red and 2 blue marbles. Marbles are taken out one at a time, at random and without replacement, until a red marble is obtained. X is the number of marbles taken out. Find the probability distribution of X and write down its PGF. [3]
  2. (b) Use G(t) to find E(X) and Var(X). [3]
Show answer
  1. (a) P(X = 1) = 3/5, P(X = 2) = 3/10, P(X = 3) = 1/10; G(t) = (6t + 3t² + t³)/10
    M1: P(X = 2) = (2/5)(3/4); P(X = 3) = (2/5)(1/4)(1).A1: 3/5, 3/10, 1/10.B1: G(t) = 0.6t + 0.3t² + 0.1t³.
  2. (b) E(X) = 3/2; Var(X) = 9/20
    M1: G′(1) = 0.6 + 0.6 + 0.3 = 1.5.M1: G″(1) = 0.6 + 0.6 = 1.2.A1: Var = 1.2 + 1.5 − 2.25 = 0.45.
Q28Finding & using a PGF · Challenge · Paper 4 style · 6 marks
  1. (a) A random variable X has PGF G(t) = ((2 + t)/3)ⁿ, where n is a positive integer. Given that E(X) = 2, find n. [2]
  2. (b) Find P(X = 2), and find Var(X) from G(t). [4]
Show answer
  1. (a) n = 6
    M1: G′(1) = n/3.A1: n = 6.
  2. (b) P(X = 2) = 80/243; Var(X) = 4/3
    M1: coefficient of t²: ⁶C₂ × 2⁴/3⁶.A1: = 240/729 = 80/243.M1: G″(1) = n(n − 1)/9 = 30/9.A1: Var = 10/3 + 2 − 4 = 4/3.
Q29Finding & using a PGF · Challenge · Paper 4 style · 8 marks
  1. (a) A random variable X takes values 0, 1, 2, … with P(X = r) = (r + 1)/2^(r+2). Show that the PGF of X is G(t) = 1/(2 − t)². [3]
  2. (b) Find E(X) and Var(X). [4]
  3. (c) Find P(X ≤ 2). [1]
Show answer
  1. (a) Shown (see working).
    M1: G(t) = (1/4) Σ (r + 1)(t/2)ʳ.M1: Σ (r + 1)xʳ = (1 − x)⁻² with x = t/2.A1: G(t) = (1/4)(1 − t/2)⁻² = 1/(2 − t)².
  2. (b) E(X) = 2; Var(X) = 4
    M1: G′(t) = 2/(2 − t)³, G′(1) = 2.A1: E(X) = 2.M1: G″(t) = 6/(2 − t)⁴, G″(1) = 6.A1: Var = 6 + 2 − 4 = 4.
  3. (c) 11/16
    B1: 1/4 + 2/8 + 3/16 = 11/16.
Q30Finding & using a PGF · Challenge · Paper 4 style · 6 marks
  1. (a) In a board game, a player must throw a six with a fair dice to start. Y is the number of throws needed. Show that the PGF of Y is G(t) = t/(6 − 5t). [2]
  2. (b) Explain why P(Y is odd) = (G(1) − G(−1))/2, and hence find this probability. [3]
  3. (c) Two players, Ayesha and Daniyal, take turns to throw the dice, Ayesha first; the first to throw a six starts the game. Use part (b) to state the probability that Ayesha starts. [1]
Show answer
  1. (a) Shown (see working).
    M1: G(t) = Σ (1/6)(5/6)^(r−1) tʳ = (t/6)/(1 − 5t/6).A1: = t/(6 − 5t).
  2. (b) 6/11
    M1: G(1) − G(−1) = Σ P(Y = r)(1 − (−1)ʳ) = 2 Σ over odd r of P(Y = r).M1: G(−1) = −1/11.A1: P(odd) = (1 + 1/11)/2 = 6/11.
  3. (c) 6/11
    B1: Ayesha starts exactly when the total number of throws is odd.

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Mean & variance from a PGF Banker9 questions · 9/9 sessions
PGF of a sum of independent variables Banker8 questions · 8/9 sessions
Finding & using a PGF 7 questions · 7/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Mean & variance from a PGF Banker299/9
PGF of a sum of independent variables Banker248/9
Finding & using a PGF 237/9

← All Further Mathematics topics