In statistics and probability theory, a discrete probability distribution is a distribution characterized by a probability mass function. This distribution is commonly used in computer programs which help to make equal probability random selections between a number of choices. The most common applications of discrete probability distribution are binomial distribution, Poisson distribution, geometric distribution and Bernoulli distribution.
Any random variable is called discrete random variable which is the part of discrete distribution. A random variable can take two types of values, either fix numbers that is discrete values or a range that is continuous type of values. In continuous type data, the values can lie anywhere within the specified range. For example: the number of apples in the basket is discrete while the time needed to drive from school to home is continuous.So the probability distribution over a random variable ‘X’ where ‘X’ takes discrete values, is commonly said to be discrete probability distribution.


When we say that the probability distribution of an experiment is discrete then the sum of probabilities of all possible values of the random variable must be equal to 1. That is if X is a discrete random variable, then,

$\sum _e P(X=e)=1$

Here, ‘e’ is the set of all values that the variable X can take.

For Example: consider the event of tossing two coins, $S$ = ${HH, TH, HT, TT}$. Let us consider the event e Y to of occurrence of a tail. Now clearly Y = 0, 1, 2 only, that is discrete values only.

For $Y$ = 0, that is ${HH}$, $P (Y)$ = $\frac{1}{4}$

For $Y$ = 1, that is ${TH, HT}$, $P (Y)$ = $\frac{2}{4}$

For $Y$ = 2, that is ${TT}$, $P (Y)$ = $\frac{1}{4}$

On adding all three we get $\frac{1}{4}$ + $\frac{2}{4}$ + $\frac{1}{4}$ = 1.

Thus we have proved our formula using a very common example.

The discrete probability distribution can always be represented in the form of a table as below:
$Y$            $P (Y)$

0        $\frac{1}{4}$ = 0.25

1        $\frac{2}{4}$ = 0.50

2        $\frac{1}{4}$ = 0.25

For any discrete probability distribution we can always find the mean or the expected value by:

$\sum _e P(X=e)$

In above example, the expected value = 0 + $\frac{2}{4}$ + $\frac{2}{4}$ = 1. But it is not necessary to have expected value equal to 1. It can be


Example 1:  Find the expected value of the following discrete distribution.

$Y$    $P (Y)$
0        0.30
1        0.20
2        0.25
3        0.15
4        0.10


$Y$    $P (Y)$   $Y P (Y)$
0        0.30        0
1        0.20        0.20
2        0.25        0.50
3        0.15        0.45
4        0.10        0.40
So expected value = 0 + 0.20 + 0.50 + 0.45 + 0.40 = 1.55

Example 2: We flip a coin 10 times. Find the probability that 6 heads are obtained.

We solve this using binomial distribution.

A binomial distribution is expressed by B (n, p) where n is the number of trials made, k is the number of successes out of n trials and p is the probability of a success in each trial. So (1 – p) will be the probability of failure in each trial. Then, the binomial distribution is calculated as below.

 $p(X=k) = \binom{n}{k}p^k . (1-p)^{n-k}$
The term $\binom{n}{k}$ is known as the binomial coefficient and is calculated as:

$\frac{n!}{((k!) (n-k)!)}$

Here, n = 10, k = 6, p = $\frac{1}{2}$ = 0.5. So, 1 – p = 0.5

Using this we get, P(X = 6) = 0.2051