Bernoulli Trial and Binomial Distribution
Abstract
In this class, we will study the concept of Bernoulli trials and their implications in probability theory. We begin with a detailed definition of Bernoulli trials, then address the concept of independence between events. After clarifying these ideas, we apply the binomial theorem to understand how repeating a Bernoulli trial produces results with a binomial distribution. Finally, practical exercises are proposed to apply and reinforce these concepts.
LEARNING OBJECTIVES:
By the end of this class, students will be able to:
- Identify the main characteristics of Bernoulli trials, including the independence between attempts.
- Correctly apply the notation for binomial events derived from Bernoulli trials.
- Distinguish between different forms of independence (2-independence, 3-independence, n-independence) and understand their relationship and application in Bernoulli trials.
- Understand the relationship between the Bernoulli trial and the binomial theorem, and how this relationship can be used to calculate the probability of a series of successes and failures.
- Apply the binomial (or Bernoulli) distribution to calculate the probability of a certain number of successes in a series of attempts.
TABLE OF CONTENTS:
The Bernoulli Trial
Different Forms of Independence
The Bernoulli Trial and the Binomial Theorem
The Binomial (or Bernoulli) Distribution and Probability Distributions
Exercises:
The Bernoulli Trial
A Bernoulli trial is a random experiment with a certain probability of success p. If a Bernoulli trial is repeated n times identically and independently, we obtain Bernoulli events: A certain number k of successes out of n attempts. These are also known as binomial events and are represented using the notation
\Large \displaystyle Bi(n;k;p)
Another important feature of Bernoulli trials is that all attempts are independent of each other.
EXAMPLE: A 6-sided die is repeatedly rolled. Examples of Bernoulli-type events for this experiment are:
- Getting 3 aces out of 5 attempts: represented by Bi(5;3;1/6)
- Getting 7 even numbers out of 12 attempts: represented by Bi(12;7;1/3)
Different Forms of Independence
The independence between attempts in a Bernoulli trial is not exactly the same independence we have already reviewed; it is a much more restricted version. To explain this difference, let’s examine the types of independence between events
2-independence
The independence we already know is the one between two events. We call it “2-independence”. In these terms, we say that events A and B are 2-independent if
P(A\cap B) = P(A)P(B)
3-independence
Similarly, 3-independence between three events A, B and C is defined through the relationship
P(A\cap B\cap C) = P(A)P(B)P(C)
It is important to highlight that 2-independence between A, B and C does not necessarily imply 3-independence, although the converse is true.
n-independence in Bernoulli Trials
Proceeding similarly to the previous definitions, n-independence between a collection of events A_1, \cdots, A_n is defined through the relationship
\Large \displaystyle P\left(\bigcap_{i=1}^n A_i\right) = \prod_{i=1}^n P(A_i)
And similarly, we have that:
| (n-1)-independence does not necessarily imply n-independence |
| n-independence \Longrightarrow (n-1)-independence |
The n repetitions performed in the Bernoulli trial are n-independent.
The Bernoulli Trial and the Binomial Theorem
Consider a success and failure experiment with a probability of success p; in each attempt, there will consequently be a probability 1-p of failure. It is clear that the probability of either a success or a failure occurring in each attempt is 1; and since all attempts are independent, the probability of either success or failure occurring in the n attempts will be 1^n. From this, we have:
\Large \displaystyle 1 = 1^n = [p + (1-p)]^n = \sum_{k=0}^n {{n}\choose{k}} p^k(1-p)^{n-k}
In the last equality, the Newton’s Binomial Theorem has been applied, and the terms within the summation can be interpreted as follows:
- \displaystyle {{n}\choose{k}}: the number of ways in which k successes can occur out of n attempts
- p^k: The probability of k independent successes occurring
- (1-p)^{n-k}: The probability of n-k independent failures occurring
By combining these elements as they appear in the sum, we obtain: the probability of getting k successes out of n attempts; or equivalently, the probability of getting n-k failures out of n attempts.
If we separate each term of the sum, we have the probabilities of getting:
| \displaystyle {{n}\choose{0}} p^0(1-p)^{n-0} = (1-p)^n | 0 successes out of n attempts |
| \displaystyle {{n}\choose{1}} p^1(1-p)^{n-1} = n p(1-p)^{n-1} | 1 success out of n attempts |
| \displaystyle {{n}\choose{2}} p^2(1-p)^{n-2} | 2 successes out of n attempts |
| \vdots | \vdots |
| \displaystyle {{n}\choose{k}} p^k(1-p)^{n-k} | k successes out of n attempts |
| \vdots | \vdots |
| \displaystyle {{n}\choose{n-1}} p^{n-1}(1-p)^{n-(n-1)} = n p^{n-1}(1-p) | n-1 successes out of n attempts |
| \displaystyle {{n}\choose{n}} p^{n}(1-p)^{0} = p^{n} | n successes out of n attempts |
And the sum of all these, as we have already seen, is “1”. Showing that all possibilities have been covered.
From this, the probability of the Bernoulli event is defined:
\displaystyle\Large \color{blue}{P(Bi(n;k;p)) = {{n}\choose{k}}p^k(1-p)^{n-k}}
Or we also say that the number of successes X has a binomial distribution:
\color{blue}{\Large \displaystyle X\sim Bi(n;p) \longmapsto P(X=x) = {{n}\choose{x}}p^x(1-p)^{n-x}}
The Binomial (or Bernoulli) Distribution and Probability Distributions
Through the binomial distribution, we begin to have the first notions of probability distributions and random variables. In this case, the (discrete) random variable is associated with the number of successes, and its probability distribution is given by the terms of the binomial theorem
{\Large \displaystyle P(X=x) = {{n}\choose{x}}p^x(1-p)^{n-x}}
Exercises:
- A fair 6-sided die is rolled 5 times. Calculate the probability of getting an even number 3 times.
- A coin is tossed 10 times. Calculate the probability of getting 0 to 10 heads and make a graph showing the probability for each outcome. How will the graph look if the number of tosses is increased, and the probability of getting a number of heads from 0 to that number of tosses is examined? An Excel spreadsheet may be useful here.
- A raffle drum has a quantity s of balls, where r are gold and the rest are white. All are mixed and one is drawn at random, winning when a gold one is drawn. If this experiment is repeated identically 20 times, estimate the most probable number of wins for each possible value of 0\leq r\leq s. An Excel spreadsheet may also be useful here.