The number of goals scored by a sports team in the first half of any match has the distribution X~ Po(3.1). The number of goals scored by the same team in the second half of any match has the distribution Y ~ Po(2.4). You may assume that the distributions of X and Y are independent.

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

1. [M1] e⁻³·¹(1 + 3.1 + 3.1²/2! + 3.1³/3!) or e⁻³·¹(1 + 3.1 + 4.805 + 4.965) or 0.0450 + 0.1397 + 0.2165 + 0.22368 Condone one end error. Any λ. Accept fully correct Σ notation. Expression must be seen.
2. [A1] = 0.625 (3sf) Correct answer with no working scores SC B1.
3. [B1] [λ=] 5.5 SOI
4. [M1] 1 - e⁻⁵·⁵(1 + 5.5 + 5.5²/2! + 5.5³/3! + 5.5⁴/4!) or 1 - e⁻⁵·⁵(1 + 5.5 + 15.125 + 27.7292 + 38.1276) or 1 - e⁻⁵·⁵(0.004087 + 0.0224772 + 0.061812 + 0.113323 + 0.155819) Condone one end error. Any λ. Accept fully correct Σ notation. Expression must be seen.
5. [A1] = 0.642 or 0.643 (3sf) Correct answer with no working scores SC B1 B1.
6. [M1] [P(X = 3) × P(Y = 2) = ] e⁻³·¹ * 3.1³/3! * e⁻²·⁴ * 2.4²/2! or 0.223676 × 0.261267 [= 0.05844] Find P(3 in first half AND 2 in second half). Must see expression.
7. [M1] [P(total 5) = ] e⁻⁵·⁵ * 5.5⁵/5! or 0.17140 Use of 5.5 to find P(5).
8. [M1] P(exactly 3 in 1st half given total 5) = P(exactly 3 in 1st half and total 5) / P(total 5) Attempt at conditional probability; numerator = their 0.05844 and denominator = P(total 5) Note: (3.1³/(3!)) * (2.4²/(2!)) / (5.5⁵/(5!)) scores M1 M1 M1.
9. [A1] ['0.05844' / '0.17140'] = 0.341 (3sf)