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The Poisson approximation is a useful tool when dealing with binomial distribution problems, especially where the number of trials, denoted as \( n \), is large, and the probability of success \( p \) is small. The central idea is simple: when you have a lot of trials, and each trial has a tiny probability of success, it's often difficult and cumbersome to directly use the binomial formula. Instead, using the Poisson distribution provides a workable approximation that is much simpler to compute.

In our specific case, we have \( n = 100 \) and \( p = 0.01 \), which means the expected number of successes, \( \lambda = np = 1 \). The Poisson probability function is given by:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
For our problem, finding the probability of zero successes means setting \( k = 0 \):
\[ P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} = e^{-1} \approx 0.368 \]

• The Poisson approximation is particularly useful as it provides a close approximation to the binomial probability without immense computation.
• This method is best applied under conditions where \( n \) is large, and \( p \) is very small, as it essentially treats individual successes as rare events in a large number of trials.

The normal approximation is another method to simplify the computation of binomial probabilities, particularly for large \( n \). When both the number of trials \( n \) is large and neither \( p \) nor \( 1-p \) are very small, the binomial distribution begins to resemble a normal distribution. This is where the normal approximation comes in handy.

The parameters for the normal distribution approximation are:
- The mean \( \mu = np \)
- The standard deviation \( \sigma = \sqrt{np(1-p)} \)

In our example, \( n = 100 \) and \( p = 0.01 \), making \( \mu = 1 \) and \( \sigma \approx 0.995 \).
To use the normal approximation correctly, especially when calculating probabilities like \( P(X = 0) \), we need to apply a continuity correction by considering \( X < 0.5 \). Convert this to a z-score:
\[ z = \frac{0.5 - 1}{0.995} \approx -0.502 \]
Referencing a standard normal distribution (z) table, the probability \( P(Z < -0.502) \) is approximately 0.309.

This approximation provides a quick way to estimate binomials but should be used with care when \( n \) is small or \( p \) is close to 0 or 1.

When dealing with binomial distributions, the concept of probability calculation is fundamental. It involves finding the likelihood of a certain number of successes across a fixed number of trials with a specific success probability.

The exact calculation for a binomial distribution is done using the formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Here \( \binom{n}{k} \) represents the number of combinations. This is also known as the binomial coefficient, which determines how many ways \( k \) successes can occur in \( n \) trials.

In our problem, we sought the probability of zero successes out of 100 trials with a success probability of 0.01.
The exact computation simplifies due to \( k=0 \):
\[ P(S_n = 0) = \binom{100}{0} (0.01)^0 (0.99)^{100} = 1 \times (0.99)^{100} \approx 0.366 \]
This methodology, while exact, can be computationally intensive for larger values of \( n \). Hence, the Poisson and normal approximations offer shortcuts that yield reasonably close results.