10B-discrete-rvs


 * Discrete Random Variables **

Use capital letters (usually X, Y, Z) for the random variables.

Use lower case letters (x, y, z) for the outcomes associated with that random variable.


 * Discrete Random Variables ** have separate and countable outcomes (usually things we count)


 * Continuous Random Variables ** deal with all Real values within a domain. (usually things we measure).


 * Discrete Probability Distributions **

A ** Discrete Probability Distribution ** is often shown as a table of possible outcomes and the associated probability.

To be a defined probability distribution, the distribution must meet the following criteria:
 * 1) Each probability must be in the domain [0, 1]
 * 2) The sum of all probabilities must be equal to 1.


 * Example 1 **

Let X be the number of tails from tossing three coins. The Probability Distribution for X is:

**Example 2**

Show that the following function, p(x), is a probability distribution function: math . \qquad p(x) = \dfrac{1}{42} \Big( 5x+3 \Big) \; \textit{ where } x \in \{ 0,1,2,3 \} \qquad. math

__**Solution**__

math . \qquad p(0) = \dfrac{3}{42}, \quad p(1) = \dfrac{8}{42}, \quad p(2)=\dfrac{13}{42}, \quad p(3) = \dfrac{18}{42} \qquad. math


 * p(x) meets criteria 1 since all p(x) are in the domain [0, 1] **

math \\ . \qquad \sum \, p(x) = \dfrac{3}{42} + \dfrac{8}{42} + \dfrac{13}{42} + \dfrac{18}{42} \qquad. \\ . \\ . \qquad \sum \; p(x) = \dfrac{42}{42} = 1 math


 * p(x) meets criteria 2 since the sum of all probabilities is equal to 1. **

Hence p(x) is a probability distribution function.


 * Example 3 **

Given the following information, find k so that the table is a probability distribution:


 * Solution:**
 * For this to be a probability distribution function, the sum of the probabilities must be 1. **

... ... 5k + 3k + 2k = 1 ... ... 10k = 1 ... ... k = 0.1

Hence the distribution is:


 * {Notice that all probabilities are in [0, 1] and the sum is 1 so this is a probability distribution} . **

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