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           Lesson 4.1 Discrete Probability Distributions: An Introduction
            Random Variable
            A random variable assigns values to the outcomes of a statistical experiment. Upper case letters denote random variables.  
            Example: 
            
              - Let X = the number of cars in a household.
 
              - Let Y = the amount of time (in minutes) a college students waits in line at the cafeteria.
 
             
                        
            Discrete Random Variable
            A discrete random variable is a random variable that assigns countable values to the outcomes of a statistical experiment. Often the phrase "the number of" is used in the description of a discrete random variable. 
            Example:  
            
              - Let X = the number of patients cured from a particular disease.
 
              - Let Y = the number of questions that a statistics student answers correctly on the first exam.
 
                         
            
            Discrete Probability Distribution Function (Discrete PDF)
             
  A discrete probability distribution function (discrete PDF) consists of 
            
              - countable numerical values.
 
              - probabilities for those numerical values. The sum of the probabilites is 1.
 
             
            Example: A controversial drug is given to TWO patients to cure a particular disease.  
            P(a cure for each patient) = 5/6.  
            P(no cure for each patient) = 1/6. 
            
                
                
            
            Let X = the number of patients (out of two) cured. Then, because one patient being cured is independent of the other patient being cured, we have: 
            P(0 patients out of two cured)  
            = P(1st person not cured)*P(2nd person not cured)  
            = (1/6)(1/6) = 1/36 
            P(1 out of two cured)  
            = P(1st person cured)*P(2nd person not cured) +  
            P(2nd person cured)*P(1st person not cured) 
            = (5/6)(1/6) + (1/6)(5/6) = 10/36 
            P(both patients cured)  
            = P(1st person cured)*P(2nd person cured)  
            = (5/6)(5/6) = 25/36 
            We can represent the discrete PDF in a table: 
            
              
                
                  X  | 
                  
                      P(X)
                   | 
                 
                
                  | 
                      0
                   | 
                  
                      P(X=0) = 1/36
                   | 
                 
                
                  1  | 
                  P(X=1) = 10/36  | 
                 
                
                  | 
                      2
                   | 
                  
                      P(X=2) = 25/36
                   | 
                 
               
            
            Notice that (1/36) + (10/36) + (25/36) = 1. 
            Please continue to the next section of this lesson.  
              
            
              
            Up » 4.1 Discrete Probability » 4.2 Expected Value » 4.3 Binomial Probability » 4.4 Poisson Probability 
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