Problem 38
Question
SPORTS MEDICINE Suppose that the number of injuries a team suffers during a typical football game follows a Poisson distribution with an average of \(2.5\) injuries. a. Find the probability that during a randomly chosen game, the team suffers exactly two injuries. b. Find the probability that during a randomly chosen game, the team suffers no injuries. c. Find the probability that during a randomly chosen game, the team suffers at least one injury.
Step-by-Step Solution
Verified Answer
a. Roughly 0.257, b. Roughly 0.082, c. Roughly 0.918
1Step 1: Identify the Poisson distribution parameters
For a Poisson distribution, the average number of events (\textbf{injuries} in this case) denoted as \(\textbf{λ}\) is given. Here, \(\textbf{λ} = 2.5\).
2Step 2: Poisson probability formula
The probability of observing \textbf{k} events in a Poisson distribution is given by the formula:\[P(X = k) = \frac{e^{-λ}λ^k}{k!}\]
3Step 3: Calculate the probability of exactly 2 injuries (Part a)
For exactly two injuries, use \(\textbf{k} = 2\).Substitute \(λ = 2.5\) and \(k = 2\) into the Poisson formula:\[P(X = 2) = \frac{e^{-2.5} (2.5)^2}{2!} = \frac{e^{-2.5} \times 6.25}{2} ≈ 0.2565\]
4Step 4: Calculate the probability of no injuries (Part b)
For no injuries, use \(\textbf{k} = 0\).Substitute \(λ = 2.5\) and \(k = 0\) into the Poisson formula:\[P(X = 0) = \frac{e^{-2.5} (2.5)^0}{0!} = e^{-2.5} ≈ 0.0821\]
5Step 5: Calculate the probability of at least one injury (Part c)
Use the complement rule. The probability of at least one injury is \(1\) minus the probability of no injuries.\[P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0821 = 0.9179\]
Key Concepts
Poisson distributionProbability calculationComplement rule
Poisson distribution
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:
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.
Probability calculation
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:
\[P(X = k) = \frac{e^{-λ}λ^k}{k!}\]
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.
Complement rule
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.
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.
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