A fair 5-sided spinner has sides labelled 1, 2, 3, 4, 5. The spinner is spun repeatedly until a 2 is obtained on the side on which the spinner lands. The random variable X denotes the number of spins required. Two fair 5-sided spinners, each with sides labelled 1, 2, 3, 4, 5, are spun at the same time. If the numbers obtained are equal, the score is 0. Otherwise, the score is the higher number minus the lower number. The two spinners are spun at the same time repeatedly .

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [B1] [P(X= 4) = (0.8)³ (0.2) =] 0.1024, 64/625. Condone 0.102.
2. [M1] 1-0.8ᵈ, d = 5, 6.
3. [A1] = 0.672, 2101/3125. 0.67232 to at least 3SF. If M0 awarded, SC B1 for 2101/3125 or 0.67232 only.
4. [M1] [P(X<6)=] (1/5) + (4/5)(1/5) + (4/5)²(1/5) + (4/5)³(1/5) + (4/5)⁴(1/5). If answer correct, condone omission of 2 from 3 middle terms. Allow M1 for (1/5) + (4/5)(1/5) + (4/5)²(1/5) + (4/5)³(1/5) + (4/5)⁴(1/5)
5. [A1] = 0.672, 2101/3125. 0.67232 to at least 3SF. If M0 awarded, SC B1 for 2101/3125 or 0.67232 only.
6. [M1] [P(X > 0 ∩ X ≠ 2) = ] 14/25, 0.56[0] seen as numerator or denominator of conditional probability fraction.
7. [M1] [P(X ≠ 2) = ] 19/25, 0.76[0] seen as denominator of conditional probability fraction.
8. [A1] Final answer 14/19, 0.7368421... to at least 3SF. If A0, SC B1 for correct final answer www.
9. [M1] [Number of outcome (X > 0 ∩ X ≠ 2) =] 14 seen as numerator or denominator of conditional probability fraction.
10. [M1] [Number of outcome (X ≠ 2)=]19 seen as denominator of conditional probability fraction.
11. [A1] Final answer 14/19, 0.7368421... to at least 3SF.
12. [M1] One term ⁹Cₓ (p)ˣ (1-p)⁹⁻ˣ, 0 < p <1, 0<x<9.
13. [A1] 1- (⁹C₀ (1-p)⁹ + ⁹C₁ (p)¹ (1-p)⁸ + ⁹C₂ (p)² (1-p)⁷), 0<p<1. Correct expression from their p, accept unsimplified, no terms omitted leading to final answer. Condone omission of last bracket only.
14. [B1] 0.371 ≤ p < 0.3715 .
15. [M1] One term ⁹Cₓ (p)ˣ (1-p)⁹⁻ˣ, 0 < p <1, 0 < x <9.
16. [A1] ⁹C₃ (p)³ (1-p)⁶ + ⁹C₄ (p)⁴ (1-p)⁵ + ... + ⁹C₈ (p)⁸(1-p)¹ + ⁹C₉ (p)⁹, 0<p<1. Correct expression from their p, accept unsimplified, no terms omitted leading to final answer.
17. [B1] 0.371 ≤ p < 0.3715 .