10 2discranvars

# Discrete Random Variables

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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.

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Discrete Random Variables have separate and countable outcomes (usually things we count)
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Continuous Random Variables deal with all Real values within a domain. (usually things we measure).

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## Discrete Probability Distributions

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A Discrete Probability Distribution is the set of possible outcomes and the associated probability for that outcome.
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A Discrete Probability Distribution is often shown as

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

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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.

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### 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.
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Solution

a) .. Display the Probability Distribution for X as a table:
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… … The sample space for the number of tails achieved from three coins

… … … … is $X = \big\{0,\; 1,\; 2,\; 3\big\}$
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… … 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 $Pr(T') = \dfrac{1}{2}$
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… … 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.
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… … 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

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

… … … … $Pr(X \leqslant 2)$

… … … … … $= 1 - Pr(X = 3)$

… … … … … $= 1 - \dfrac{1}{8}$

… … … … … $= \dfrac{7}{8}$
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c) .. Display the Probability Distribution for X as a column graph.

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### 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

… … … … $p(0) = \dfrac{3}{42}$

… … … … $p(1) = \dfrac{8}{42}$

… … … … $p(2)=\dfrac{13}{42}$

… … … … $p(3) = \dfrac{18}{42}$
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… … p(x) meets criteria 1 since all p(x) are in the domain $\big[0,\; 1\big]$
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… … … … $\sum \, p(x) = \dfrac{3}{42} + \dfrac{8}{42} + \dfrac{13}{42} + \dfrac{18}{42}$
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… … … … $\sum \; p(x) = \dfrac{42}{42} = 1$
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… … p(x) meets criteria 2 since the sum of all probabilities is equal to 1.

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… … Hence p(x) is a probability distribution function.

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### 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.
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… … … … $5k + 3k + 2k = 1$
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… … … … $10k = 1$
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… … … … $k = 0.1$

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… … Hence the distribution is:

… … Notice that:

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

… … … … and $\sum p(x) = 1$

… … so this is a probability distribution.

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