# The Discrete Uniform Distribution

Contents

## Summary

• The values of a discrete random variable are obtained by counting, thus making it known as countable
• Uniform distribution simply means that when all of the random variable occur with equal probability
• A random variable with probability density function is $=\quad P(X\quad =\quad { x }_{ n })\quad =\quad \frac { 1 }{ n\quad +\quad 1 }$
• the expectation and variance of the data we use the following formulas
1. $E(x)\quad =\quad \frac { 1 }{ n\quad +\quad 1 } \quad +\quad \frac { 2 }{ n\quad +\quad 1 } \quad +\quad \frac { 3 }{ n\quad +\quad 1 } \quad .......\frac { n }{ n\quad +\quad 1 }$
2. $Variance\quad =\quad \frac { n(n\quad +\quad 2) }{ 12 }$

You must already be aware of the fact that there are two types of random variables, discrete random variables and continuous random variables. We have already discussed continuous random variable in this chapter we will discuss about the discrete random variable. The values of a discrete random variable are obtained by counting, thus making it known as countable.

You should by now also be aware of these two basic characteristics of a A discrete probability distribution function (PDF).

Each probability is between zero and one, inclusive.

The sum of all the probabilities is always equal to one.

Now that discrete random variable is clear. Lets, discuss uniform distribution.

Uniform distribution simply means that when all of the random variable occur with equal probability. Suppose that the random variable X can assume n different values. Also consider n is constant. Then we can say that: $P(X\quad =\quad { x }_{ n })\quad =\quad \frac { 1 }{ n\quad +\quad 1 }$

For all values of ${ x }_{ n }$ = 0, 1, 2…. N

P(X = x) = 0 for other values of ${ x }_{ n }$

#### Example #1

Q. Suppose a fair die is rolled. What is the probability that the die will land on 5?

Solution:

We are already aware that a die has 6 sides, thus when a die is rolled we already know that the outcome will be one of the following {1, 2, 3, 4, 5, 6}.

We also know that each possible outcome is a random variable (X), and since the die is fair all the outcomes have equal chances to occur ( $\frac { 1 }{ 6 }$).

Thus, we have a uniform distribution.

Therefore we can say: $P(X\quad =\quad 5)\quad =\quad \frac { 1 }{ 6 }$

In order to find the expectation and variance of the data we use the following formulas: $E(X)\quad =\quad 1\quad \times \quad P(X\quad =\quad 1)\quad +\quad 2\quad \times \quad P(X\quad =\quad 2)\quad +\quad ........\quad +\quad n\quad \times \quad P(X\quad =\quad n)$

We can thus write it as $\left[ P(X\quad =\quad { x }_{ n })\quad =\quad \frac { 1 }{ n\quad +\quad 1 } \right]$. $E(x)\quad =\quad \frac { 1 }{ n\quad +\quad 1 } \quad +\quad \frac { 2 }{ n\quad +\quad 1 } \quad +\quad \frac { 3 }{ n\quad +\quad 1 } \quad .......\frac { n }{ n\quad +\quad 1 }$ $Variance\quad =\quad \frac { n(n\quad +\quad 2) }{ 12 }$

#### Example #2

Q. If we repeat the dice rolling experiment from above (Example #1) and we change the question a little bit from what is the probability that it will land on 5 to what is the probability that the die will land on a number that is smaller than 5?

Solution:

When already know that we have a uniform distribution as we discussed before that when a die is rolled, there are 6 possible outcomes that it can land on {1, 2, 3, 4, 5, 6}, with equal chances on landing on either number (if the die is fair of course).

The probability that the die will land on a number smaller than 5 is equal to:

P( X < 5 ) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) $P(X\quad <\quad 5)\quad =\quad \frac { 1 }{ 6 } \quad +\quad \frac { 1 }{ 6 } \quad +\quad \frac { 1 }{ 6 } \quad +\frac { 1 }{ 6 } \quad =\quad \frac { 4 }{ 6 } \quad =\quad \frac { 2 }{ 3 }$

Note: We did not take P(X < 5) because it’s not smaller than equal to 5. If the question said $P(X\quad \le \quad 5)$ only then we would take P(X = 5).