Bernoulli and Binomial Probability Distributions

A Bernoulli trial is a random experiment with exactly two possible outcomes: success and failure.
A random variable X associated with such a trial, where X = 1 indicates success and X = 0 indicates failure, is called a Bernoulli random variable.

The probability distribution of X, known as the Bernoulli distribution, is given by:
f(x) = \begin{cases} p, & \text{if } x = 1 \ 1 - p, & \text{if } x = 0 \end{cases} $0 \le p \le 1, the probability of success.

The mathematical expectation and variance of the Bernoulli distribution are:
E(X) = p
Var(X) = p(1 - p)

Selecting a defective item from a manufacturing process with a probability of \frac{1}{5}. A defective item is designated as a success.
- Mean: E(X) = \frac{1}{5}
- Variance: Var(X) = \frac{4}{25}
Opening an email to check for a malicious attachment, with a probability of 0.02 of being malicious. A malicious attachment is designated as a success.
- Mean: E(X) = 0.02
- Variance: Var(X) = 0.0196
Checking if a webpage contains a specific keyword, with a probability of 0.1. The presence of the keyword is designated as a success.
- Mean: E(X) = 0.1
- Variance: Var(X) = 0.09
Detecting a fraudulent transaction among online purchases, with a fraud probability of 0.001. A fraudulent transaction is designated as a success.
- Mean: E(X) = 0.001
- Variance: Var(X) = 0.000999
Observing whether a website visitor makes a purchase, with a conversion probability of 0.07. A purchase is designated as a success.
- Mean: E(X) = 0.07
- Variance: Var(X) = 0.0651

Models the x number of successes in n Bernoulli trials.
Characterizes the sum of multiple independent Bernoulli trials, specifically n such trials.
When a Bernoulli trial is repeated n times, each with an identical probability of success p, the total number of successes, denoted by X, follows a binomial distribution:
X = X1 + X2 + \cdots + Xn \sim \text{Binomial}(n, p) where each Xi \sim \text{Bernoulli}(p).
The binomial distribution emerges as the distribution of the sum of these independent Bernoulli random variables.

Let X be a random variable representing the number of defective items among three items randomly selected from a manufacturing process. Each item is inspected and classified as either defective (designated as a success) or non-defective (a failure).
The random variable X can take integer values from 0 to 3, corresponding to the number of defective items observed.
The eight possible outcomes and their associated values of X are:
- Sample Space: NNN, NND, NDN, DNN, NDD, DND, DDN, DDD
- x: 0, 1, 1, 1, 2, 2, 2, 3

A variable described as the number of successes in a sequence of independent Bernoulli trials follows a binomial distribution.
The probability mass function of the binomial distribution is given by:
P(X) = f(x) = b(x; n, p) = {n \choose x} p^x (1 - p)^{n-x}, \quad x = 0, 1, 2,…, n
The binomial distribution derives its name from the fact that the n + 1 terms in the binomial expansion of ((1 - p) + p)^n correspond to the values of the binomial probability mass function b(x; n, p) for x = 0, 1, 2,…, n.
((1-p)+p)^n = {n \choose 0}(1-p)^n+{n \choose 1}p(1-p)^{n-1}+{n \choose 2}p^2(1-p)^{n-2}+\cdots+{n \choose n}p^n = b(0; n, p)+b(1; n, p)+\cdots+b(n; n, p)
Since p + (1 - p) = 1, it follows that \sum_{x=0}^{n} b(x; n, p) = 1, which confirms a fundamental property of any probability distribution: the total probability must sum to 1.

As part of a business strategy, 20% of new internet service subscribers are randomly selected to receive a special promotion from the provider. A group of 10 neighbors signs up for the service.
- What is the probability that exactly 5 of them receive the special promotion?
- What is the probability that at least 2 of them receive the special promotion?
A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%.
- The inspector randomly picks 20 items from a shipment. What is the probability that there will be at least one defective item among these 20?
- Suppose that the retailer receives 10 shipments in a month and the inspector randomly tests 20 devices per shipment. What is the probability that there will be exactly 3 shipments each containing at least one defective device among the 20 that are selected and tested from the shipment?

Using Binomial Distribution table we can compute for P(X < a) or P(a \le X \le b).

As part of a business strategy, 20% of new internet service subscribers are randomly selected to receive a special promotion from the provider. A group of 10 neighbors signs up for the service.
- What is the probability that at least 4 of them receive the special promotion?
- What is the probability that between 3 and 8 of them (inclusive) receive the special promotion?

The expected value and variance of the binomial distribution are:
E(X) = np
Var(X) = np(1 - p)
If X = X1 + X2 + \cdots + X_n then,

E(X) = E(X1 + X2 + … + Xn) = E(X1) + E(X2) + … + E(Xn) = p + p + … + p = np

Var(X) = Var(X1 + X2 + … + Xn) = Var(X1) + Var(X2) + … + Var(Xn) = p(1 - p) + p(1 - p) + … + p(1 - p) = np(1 - p)

As part of a business strategy, 20% of new internet service subscribers are randomly selected to receive a special promotion from the provider. A group of 10 neighbors signs up for the service.
- Compute for the mean and the variance.
- Use Chebyshev’s theorem to interpret the interval \mu \pm 2$$

An exciting computer game is released. Sixty percent of players complete all the levels. Thirty percent of them will then buy an advanced version of the game. Among 15 users,
- what is the expected number of people who will buy the advanced version?
- what is the probability that at least two people will buy it?

Probability and Statistics for Engineers and Scientists by Walpole, 9th Edition
Probability and Statistics for Computer Scientists by Baron, Michael, 2nd Edition