91Ó°ÊÓ

The Poisson distribution is a powerful tool in probability and statistics. It helps predict the likelihood of a certain number of events happening within a fixed interval. This interval can be time, distance, area, etc. The main parameter of a Poisson distribution is the average rate of occurrence, denoted as \(λ\).
For example, in the given exercise, the average number of injuries per game is 2.5. This is our \(λ\). Importantly, the Poisson distribution assumes the following:

• Events occur independently of each other.
• The average rate (\(λ\)) remains constant over the interval.
• Two events cannot occur at the exact same instant.

Using these properties, you can estimate the probability of having a specific number of events (like injuries) in a given interval.

Calculating probabilities with the Poisson distribution involves a specific formula. The formula for finding the probability of exactly \(k\) events occurring is:

\[P(X = k) = \frac{e^{-λ}λ^k}{k!}\]

This might look complicated, but let's break it down:

• \(e\) is the base of the natural logarithm, approximately equal to 2.71828.
• \(λ\) is the average number of events in the interval.
• \(k\) is the exact number of events you're interested in.
• \(k!\) is the factorial of \(k\).

Let's apply this to our exercise:

For part (a), to find the probability of exactly 2 injuries:
\[P(X = 2) = \frac{e^{-2.5} (2.5)^2}{2!} = \frac{e^{-2.5} \times 6.25}{2} ≈ 0.2565\]
You substitute \(λ = 2.5\) and \(k = 2\).

For part (b), to find the probability of no injuries:
\[P(X = 0) = \frac{e^{-2.5} (2.5)^0}{0!} = e^{-2.5} ≈ 0.0821\]
Here, \(λ = 2.5\) and \(k = 0\). This gives you the desired probability.

The complement rule is a handy trick in probability, especially when calculating with the Poisson distribution. It states that the probability of an event not occurring is 1 minus the probability that it does occur.
This can save time when calculating complex probabilities. In our exercise, part (c) asks for the probability of at least one injury. Instead of summing up all probabilities from one to however many injuries can happen, we use the complement rule:

\[P(X \geq 1) = 1 - P(X = 0)\]

In simpler terms, if you know the probability of zero injuries, you can find the probability of one or more injuries by subtracting from 1.
From part (b), we already have \(P(X = 0) = 0.0821\). So:

\[P(X \geq 1) = 1 - 0.0821 = 0.9179\]

This complements our understanding and makes calculating probabilities easier and faster.