Variables are defined as characteristics that changes or vary over time and/or for different individuals under consideration. Variables are of different types.

A variable $x$ is a random variable if the value that it assumes, corresponding to the outcome of an experiment, is a chance or random event. Random variables are classified as either discrete or continuous, according to the value that $x$ can assume. In this section we will be discussing more about Discrete Random variables.

Let us suppose we are blindfolded and asked to pick a cube from among many of different colors. What is the probability that it will be purple? This is a problem, which involves probability or chance and statistics or data samples. The color picked at any point is represented by a discrete random variable.
Discrete random variables have a finite number possible values or an infinite number of values that can be counted like 1, 2, 3.
Discrete random variables define and represent numerous statistical problems. These are not merely conceptual entities. It have practical application in the real world.
The probability distribution for a discrete variable is a formula, table, or graph that gives the possible values of x, and the probability p(x) associated with each value of $x$.

The discrete random variables have the countable values for their probability. The variance value for the discrete random variable is calculated with the help of the mean. The random variables are said to discrete random variable when the sum of their probability is one.

The formula used for discrete random variable includes mean, variance and the standard deviation. The formula used are
Mean = $\sum xp(x)$

Variance = $p(x)(x^{2})-(\sum xp(x))^{2}$

Standard deviation = $\sqrt{variance}$

In this above formula $x$ denotes the given set of values and $p(x)$ denotes the required probability function for the given discrete random values in the random variables.

Solved Example

Question: Resolve the mean, variance and standard deviation of a discrete variable for the given probability values.
 x 10  20  30  40  50 
 p(x) 0.24  0.15  0.18  0.09  0.34 


Solution:
 
Mean = $\sum xp(x)$
Mean = $10(0.24)+20(0.15)+30(0.18)+40(0.09)+50(0.34)$
Mean = $31.4$
Variance = $p(x)(x^{2})-(\sum xp(x))^{2}$   
Variance = $((0.24) (10)^{2} + (0.15) (20)^{2} + (0.18) (30)^{2} + (0.09) (40)^{2} + (0.34) (50)^{2}) - (31.4)^{2}$ 

Variance = 254.04

Standard deviation = $\sqrt{variance}$ 

Standard deviation = $\sqrt{254.04}$ 

Standard deviation = 15.9386

The mean value is 31.4, variance is 254.04 and the standard deviation is 15.9386.

 

Several important discrete random variable are derived from the concept of a Bernoulli sequence of trials.
A Bernoulli trial is a random experiment in which there are only two possible outcomes, usually called success or failure, with respective probabilities say $p$ and $q$, where $p+q=1$. We assume $0<p<1$, for otherwise the results are trivial. The sequence of such trials is a Bernoulli sequence if the trials are independent and the probability of success is constant from trial to trial.
Consider a Bernoulli sequence of $n$ trails where the probability of success on each trial is $p$. The random variable $X$, that counts the number of success in the $n$ trial is called binomial random variable with parameters $n$ and $p$.
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Consider the sequence of independent identical trials. Assume that each trial can result in a success or a failure. Each trial has a probability $p$ of success and $q = p-1$ of failure. Let X be the number of trials up to and including the first success. Then X is called a Geometric Random Variable.
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The random variable $n$ is said to have a Poisson distribution with mean $\lambda$ if 
P( X : λ) = $\frac{e^{-\lambda }\lambda ^{X}}{X!}$ for $X$ = 0,1,....
The Poisson distribution appears when we count the number of occurrences of the events that have small probability and are independent.   
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