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The binomial distribution is a fundamental concept in probability theory. It arises when we conduct a fixed number of independent trials, each with two possible outcomes: success or failure. Think of flipping a coin or in this case, determining whether each mouse contracts the disease or not. The key elements in a binomial distribution are:

• Number of trials, denoted by \( n \). For this problem, it is 30 because there are 30 mice.
• Probability of success in each trial, denoted by \( p \). Here, it is 0.2, meaning each mouse has a 20% chance of contracting the disease.
• The number of successes we're interested in, often denoted by \( k \).

With the binomial distribution, the probability of exactly \( k \) successes out of \( n \) trials can be calculated using the formula:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( \binom{n}{k} \) is the binomial coefficient. However, when dealing with large \( n \), computations can be cumbersome, which is where approximations like the Poisson distribution come into play.

The Poisson distribution is an approximation tool often used to simplify calculations for events that occur infrequently over a large number of trials. It's particularly helpful when you encounter a binomial scenario where \( n \) is large and \( p \) is small, such that \( np \) (the expected number of successes) is relatively small too.For the disorder problem at hand, Poisson distribution was deemed suitable because:

• \( n = 30 \), which is large enough, and \( p = 0.2 \), making \( np = 6 \) not too large.
• The parameter \( \lambda = np = 6 \) summarizes the expected count of successes.

Using the Poisson distribution, you determine the probability of \( k \) events using:\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]For example, calculating for 0 to 3 infected mice gives us probabilities for the respective counts, ultimately allowing us to add them together to find the total probability for having at most 3 infections.

Probability calculation in statistics involves determining the likelihood of a specific event or outcome. In this exercise, the goal is to compute the probability that at most 3 out of 30 mice contract the disease.First, with the Poisson approximation, individual probabilities for a certain number of mice contracting the disease (ranging from 0 to 3) are calculated:

• For 0 mice: \( P(X=0) = e^{-6} \)
• For 1 mouse: \( P(X=1) = 6e^{-6} \)
• For 2 mice: \( P(X=2) = 18e^{-6} \)
• For 3 mice: \( P(X=3) = 36e^{-6} \)

The sum of these probabilities gives the total probability of up to 3 mice being infected, namely:\[ P(X \leq 3) = e^{-6}(1 + 6 + 18 + 36) = 61e^{-6} \]By evaluating this expression, we find that the probability is approximately \(0.1512\), indicating a 15.12% chance that at most 3 mice will contract the disease under the given conditions.