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1. Notions of Probability 33

1.7.1 Discrete Distributions
In this subsection, we include some standard discrete distributions. A few of
these appear repeatedly throughout the text.
The Bernoulli Distribution: This is perhaps one of the simplest possible
discrete random variables. We say that a random variable X has the Bernoulli
(p) distribution if and only if its pmf is given by

where 0 < p < 1. Here, p is often referred to as a parameter. In applications, one
may collect dichotomous data, for example simply record whether an item is
defective (x = 0) or non-defective (x = 1), whether an individual is married (x =
0) or unmarried (x = 1), or whether a vaccine works (x = 1) or does not work (x
= 0), and so on. In each situation, p stands for P(X = 1) and 1 – p stands for P(X
= 0).
The Binomial Distribution: We say that a discrete random variable X has
the Binomial(n, p) distribution if and only if its pmf is given by

where 0 < p < 1. Here again p is referred to as a parameter. Observe that the
Bernoulli (p) distribution is same as the Binomial(1, p) distribution.
The Binomial(n, p) distribution arises as follows. Consider repeating the
Bernoulli experiment independently n times where each time one observes the
outcome (0 or 1) where p = P(X = 1) remains the same throughout. Let us
obtain the expression for P(X = x). Consider n distinct positions in a row
where each position will be filled by the number 1 or 0. We want to find the
probability of observing x many 1’s, and hence additionally exactly (n – x)
many 0’s. The probability of any such particular sequence, for example the
first x many 1’s followed by (n – x) many 0’s or the first 0 followed by x
x
n–x
many 1’s and then (n – x – 1) many 0’s would each be p (1 – p) . But, then
there are ways to fill the x positions each with the number 1 and n – x
positions each with the number 0, out of the total n positions. This verifies the
form of the pmf in (1.7.2).
Recall the requirement in the part (ii) in (1.5.3) which demands that all the
probabilities given by (1.7.2) must add up to one. In order to verify this directly,
let us proceed as follows. We simply use the binomial expansion to write