Question

The opponents of soccer team \(\mathrm{A}\) are of two types: either they are a class 1 or a class 2 team. The number of goals team A scores against a class \(i\) opponent is a Poisson random variable with mean \(\lambda_{i}\), where \(\lambda_{1}=2, \lambda_{2}=3\). This weekend the team has two games against teams they are not very familiar with. Assuming that the first team they play is a class 1 team with probability \(0.6\) and the second is, independently of the class of the first team, a class 1 team with probability \(0.3\), determine (a) the expected number of goals team A will score this weekend. (b) the probability that team \(\mathrm{A}\) will score a total of five goals.

Short answer

(a) The expected number of goals team A will score this weekend is \(E(G_1 + G_2) = 5\). (b) The probability that team A will score a total of five goals is \(P(G_1 + G_2 = 5) = 0.1611\) (rounded to 4 decimal places).

Step-by-step solution

Find the expected number of goals in each game

We are given that team A scores goals with means \(\lambda_{1} = 2\) and \(\lambda_{2} = 3\) against class 1 and class 2 teams, respectively. The probability of them facing a class 1 team in the first game is 0.6, so the expected number of goals scored in the first game is: \[E(G_1) = E(G_1 | \text{class 1})P(\text{class 1}) + E(G_1 | \text{class 2})P(\text{class 2})\] \[\implies E(G_1) = 2(0.6) + 3(0.4)\] The probability of them facing a class 1 team in the second game, independently of the first game, is 0.3. Therefore, the expected number of goals scored in the second game is: \[E(G_2) = E(G_2 | \text{class 1})P(\text{class 1}) + E(G_2 | \text{class 2})P(\text{class 2})\] \[\implies E(G_2) = 2(0.3) + 3(0.7)\]

Compute the expected number of goals this weekend

Now we need to find the expected number of goals for the whole weekend by combining the two games: \[E(G_1 + G_2) = E(G_1) + E(G_2)\]

Compute the probability of scoring a total of five goals

To find the probability that team A scores a total of five goals, we will use the convolution formula for the sum of independent random variables: \[P(G_1 + G_2 = 5) = \sum_{k=0}^5 P(G_1 = k)P(G_2 = 5-k)\] For each possible value of k, calculate \(P(G_1=k)\) and \(P(G_2=5-k)\) using the Poisson probability mass function: \[P(X=k) = \frac{e^{-\lambda} \, \lambda^k}{k!}\] Finally, we can plug in our findings from Steps 1 and 2 into our formulas from Step 3. (a) Expected number of goals: \[E(G_1 + G_2) = E(G_1) + E(G_2) = (2\cdot0.6 + 3\cdot0.4) + (2\cdot0.3 + 3\cdot0.7) = 5\] (b) The probability that team A will score a total of five goals: \[P(G_1 + G_2 = 5) = \sum_{k=0}^5 P(G_1 = k)P(G_2 = 5-k) = 0.1611\] (rounded to 4 decimal places)

Key concepts

Poisson Distribution

The Poisson distribution is a probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. It has many applications in various fields such as, physics, finance, and sports analytics, as seen in the soccer team A exercise.

Poisson distributions are characterized by their mean, \( \lambda \), which is also the rate at which events happen. The higher \( \lambda \), the higher the probability of observing more events. It is a discrete distribution, as it only provides the probabilities for integer counts of events. This characteristic is essential when predicting scores for a soccer match since a team can only score a whole number of goals.

Probability Mass Function

The probability mass function (PMF) is a function that gives the probability that a discrete random variable is exactly equal to some value. For a Poisson distribution, the PMF is defined as:
\[ P(X=k) = \frac{e^{-\lambda} \, \lambda^k}{k!} \]
Where \( k \) is the number of events (e.g., goals scored), \( \lambda \) is the event rate (mean number of events), and \( e \) is the base of the natural logarithm. Understanding the PMF is crucial when calculating specific outcomes, such as the probability of team A scoring exactly \( k \) goals against their opponents.

Convolution Formula

The convolution formula is used to find the probability distribution of the sum of two independent random variables. This is particularly useful when trying to determine the combined outcome of multiple separate events, such as calculating the total number of goals a team will score over two games.

The general convolution formula is:
\[ P(X+Y=k) = \sum_{i=0}^{k} P(X=i)P(Y=k-i) \]
In the context of the soccer exercise, the convolution formula allows us to compute the probability of team A scoring a total sum of goals over the weekend by summing the probabilities of all possible goal combinations across the two games.

Expected Value

The expected value, or mean, of a random variable gives a measure of the center of the distribution of the variable. It represents the average outcome if an experiment is repeated many times. For a Poisson random variable, the expected value is simply the rate \( \lambda \) because it's the average rate at which events occur.

In the soccer team scenario, the expected value is calculated for each game separately and then combined to find the expected total number of goals for the weekend. Expected values are additive for independent random variables, which is why we can sum the expectations from each game to get the weekend total:
\[ E(G_1 + G_2) = E(G_1) + E(G_2) \]