Find the binomial probability for the given problem, and then compare the normal and the Poisson approximations. Find the probability of \(x\) successes in 100 Bernoulli trials with probability \(p=1 / 5\) of success (a) if \(x=25 ;\) (b) if \(x=21.\)

Bernoulli trials are named after the Swiss mathematician Jacob Bernoulli. Each trial is a random experiment with exactly two possible outcomes: success or failure. A common example is flipping a coin, where success might be landing on heads, and failure might be landing on tails.
In a Bernoulli trial, the probability of success is denoted by \( p \), and the probability of failure is \( 1 - p \). These trials are the foundation of the binomial distribution. When we have a fixed number of independent Bernoulli trials, the number of successes follows a binomial distribution. For example, if we flip a coin 100 times and are interested in the number of heads, each flip is a Bernoulli trial. The main formula related to Bernoulli trials is:

• \( P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \)

The normal approximation to the binomial distribution is useful when the number of trials \( n \) is large and the probability of success \( p \) is not too close to 0 or 1. The central limit theorem tells us that as the number of trials increases, the binomial distribution approaches a normal distribution with mean \( \mu = np \) and standard deviation \( \sigma = \sqrt{np(1-p)} \).
For the binomial distribution with \( n = 100 \) and \( p = \frac{1}{5} \), the mean \( \mu \) and the standard deviation \( \sigma \) are:

• \( \mu = 100 \times \frac{1}{5} = 20 \)
• \( \sigma = \sqrt{100 \times \frac{1}{5} \times \frac{4}{5}} = 4 \)

To use the normal approximation, convert the number of successes \( x \) to a z-score:

• \( z = \frac{x - \mu}{\sigma} \)

Lookup the z-score in the standard normal table to find the corresponding probability. This method simplifies the calculation when \( n \) is large.

The Poisson approximation is another method to approximate the binomial distribution. It is particularly useful when the number of trials \( n \) is large, and the probability of success \( p \) is small, making \( np \) moderate. In such cases, the binomial distribution can be approximated using a Poisson distribution with parameter \( \lambda = np \).
In our example, for \( n = 100 \) and \( p = \frac{1}{5} \), we have:

• \( \lambda = 100 \times \frac{1}{5} = 20 \)

The Poisson probability of getting \( x \) successes is given by:

• \( P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \)

This approximation is handy for calculations that would be otherwise complex using the exact binomial formula, especially when \( x \) is much smaller than \( n \).

The probability of success, denoted as \( p \), is a fundamental concept in probability and statistics. It represents the likelihood of a success occurring in a single Bernoulli trial. In the context of binomial, normal, and Poisson approximations, \( p \) helps determine the distribution parameters. For example, in our exercise, the probability of success is \( \frac{1}{5} \), or 0.2.

• In binomial distribution: \( p \) directly influences the shape of the distribution and the calculations involved.
• In normal approximation: \( p \) helps calculate the mean \( \mu = np \) and standard deviation \( \sigma = \sqrt{np(1-p)} \).
• In Poisson approximation: \( p \) combined with \( n \) determines the parameter \( \lambda = np \).

The probability of success is crucial in deciding the likelihood of different outcomes and in making statistical inferences about the data.