10 2discranvars

Discrete Random Variables

.

For this topic, we will use:

  • Capital letters (usually X, Y, Z) to represent random variables.
  • 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 the set of possible outcomes and the associated probability for that outcome.
.

A Discrete Probability Distribution is often shown as

  • a table … or
  • a graph … or
  • an equation

.

To be a defined probability distribution, the distribution must meet the following criteria:

1 .. Each probability must be in the domain $\big[0, \; 1\big]$

2 .. The sum of all probabilities must be equal to 1.

.

Example 1

Let X be the number of tails achieved from tossing three coins.

a) .. Display the Probability Distribution for X as a table:

b) .. Hence calculate $Pr(X \leqslant 2)$

c) .. Display the Probability Distribution for X as a column graph.
.

Solution

a) .. Display the Probability Distribution for X as a table:
.

… … The sample space for the number of tails achieved from three coins

… … … … is $X = \big\{0,\; 1,\; 2,\; 3\big\}$
.

… … The probability of getting a tail on one coin is $Pr(T) = \dfrac{1}{2}$

… … The probability of not getting a tail on one coin is $Pr(T') = \dfrac{1}{2}$
.

… … The probability of getting 0 tails on three coins is:

… … … … $Pr(X=0) = Pr(T') \times Pr(T') \times Pr(T')$

… … … … $Pr(X=0) = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}$

… … … … $Pr(X=0) = \dfrac{1}{8}$

… … In a similar way, we can calculate the probabilities of getting 1 or 2 or 3 tails.
.

10.2tbl1a.gif

.

… … Note This table meets the criteria for a probability distribution:

… … … 1 .. All probabilities are in the domain: $\big[0, \; 1\big]$

… … … 2 .. The sum of the probabilities is 1

… … … … … $\dfrac{1}{8} + \dfrac{3}{8} + \dfrac{3}{8} + \dfrac{1}{8} = 1$
.

b) .. Hence calculate $Pr(X \leqslant 2)$

… … … … $Pr(X \leqslant 2)$

… … … … … $= Pr(X = 0) + Pr(X = 1) + Pr(X = 2)$

… … … … … $= \dfrac{1}{8} + \dfrac{3}{8} + \dfrac{3}{8}$

… … … … … $= \dfrac{7}{8}$

… … … OR

… … … … $Pr(X \leqslant 2)$

… … … … … $= 1 - Pr(X = 3)$

… … … … … $= 1 - \dfrac{1}{8}$

… … … … … $= \dfrac{7}{8}$
.

c) .. Display the Probability Distribution for X as a column graph.

10.2tbl1b.gif

.

Example 2

Show that the following function, p(x), is a probability distribution function:

… … $p(x) = \dfrac{1}{42} \Big( 5x+3 \Big) \; \textit{ where } x \in \big\{ 0,\; 1,\; 2,\; 3 \big\}$

Solution

… … … … $p(0) = \dfrac{3}{42}$

… … … … $p(1) = \dfrac{8}{42}$

… … … … $p(2)=\dfrac{13}{42}$

… … … … $p(3) = \dfrac{18}{42}$
.

… … p(x) meets criteria 1 since all p(x) are in the domain $\big[0,\; 1\big]$
.

… … … … $\sum \, p(x) = \dfrac{3}{42} + \dfrac{8}{42} + \dfrac{13}{42} + \dfrac{18}{42}$
.

… … … … $\sum \; p(x) = \dfrac{42}{42} = 1$
.

… … 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:

10.2eg3tbl1.gif

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:

10.2eg3tbl2.gif

… … Notice that:

… … … … all probabilities are in $\big[0, \; 1\big]$

… … … … and $\sum p(x) = 1$

… … so this is a probability distribution.

.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License