A-Level Mathematics May/June 2024 Q5(c): Given that the team scores a total of 5 goals in a randomly chosen match, find the prob…

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  Let $X$ be the number of goals scored in the first half and $Y$ be the number of goals scored in the second half. From the question context, we have: $X \sim \text{Po}(3.1)$ $Y \sim \text{Po}(2.4)$

  Let $T$ be the total number of goals scored in the match. Then $T = X + Y$. Since $X$ and $Y$ are independent Poisson variables, their sum is also a Poisson variable with a parameter that is the sum of their individual parameters. $T \sim \text{Po}(3.1 + 2.4) \implies T \sim \text{Po}(5.5)$

  We are asked to find the probability that the team scores exactly 3 goals in the first half, given that they score a total of 5 goals. This is a conditional probability problem: $P(X=3 | T=5)$.

  Using the formula for conditional probability, $P(A|B) = \frac{P(A \cap B)}{P(B)}$: $P(X=3 | T=5) = \frac{P(X=3 \text{ and } T=5)}{P(T=5)}$

  The event '$X=3$ and $T=5$' means the team scored 3 goals in the first half and 5 in total, which implies they must have scored $5-3=2$ goals in the second half. So, this is the event '$X=3$ and $Y=2$'.

  Numerator: $P(X=3 \text{ and } Y=2)$ Since $X$ and $Y$ are independent, $P(X=3 \text{ and } Y=2) = P(X=3) \times P(Y=2)$. $P(X=3) = e^{-3.1} \frac{3.1^3}{3!} = 0.223676...$ $P(Y=2) = e^{-2.4} \frac{2.4^2}{2!} = 0.261267...$ Numerator = $(e^{-3.1} \frac{3.1^3}{3!}) \times (e^{-2.4} \frac{2.4^2}{2!}) = 0.223676... \times 0.261267... = 0.058440...$

  Denominator: $P(T=5)$ Using the distribution $T \sim \text{Po}(5.5)$: $P(T=5) = e^{-5.5} \frac{5.5^5}{5!} = 0.171400...$

  Finally, we calculate the conditional probability: $P(X=3 | T=5) = \frac{0.058440...}{0.171400...} = 0.340956...$

  Rounding to 3 significant figures: $P(X=3 | T=5) = 0.341$

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  How the marks are awarded

  • M1 — Finding the probability of the intersection event '3 goals in the first half AND 2 in the second half' by correctly multiplying the two independent Poisson probabilities: $P(X=3) \times P(Y=2)$.
  • M1 — Finding the probability of the condition 'a total of 5 goals' by first identifying the combined distribution $T \sim \text{Po}(5.5)$ and then calculating $P(T=5)$.
  • M1 — Correctly applying the conditional probability formula by setting up the fraction with the intersection probability (the first M1 mark) as the numerator and the total probability (the second M1 mark) as the denominator.
  • A1 — Obtaining the correct final answer of 0.341, correctly rounded to 3 significant figures.

  Common mistakes

  • Calculating $P(X=3) / P(T=5)$ instead of the correct $P(X=3 \cap Y=2) / P(T=5)$, which fails to identify the correct numerator for the conditional probability.
  • Incorrectly calculating the probability of the total, for example by finding $P(X=5)$ or $P(Y=5)$ instead of using the sum of the Poisson distributions $T \sim \text{Po}(5.5)$.
  • Misinterpreting the event '3 goals in the first half and 5 in total' and calculating the numerator as $P(X=3) \times P(T=5)$ instead of $P(X=3) \times P(Y=2)$.
  • Adding probabilities $P(X=3) + P(Y=2)$ instead of multiplying them for an 'AND' event, showing a misunderstanding of basic probability rules for independent events.

  Examiner tip: For conditional probability questions ('given that'), always identify the intersection event for the numerator and the condition event for the denominator.

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