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172                    Fundamentals of Probability and Statistics for Engineers

             Example 6.9. The negative binomial distribution is widely used in waiting-
           time problems. Consider, for example, a car waiting on a ramp to merge into
           freeway traffic. Suppose that it is 5th in line to merge and that the gaps between
           cars on the freeway are such that there is a probability of 0.4 that they are
           large enough for merging. Then, if X  is the waiting time before merging for
           this particular vehicle measured in terms of number of freeway gaps, it has
           a negative binomial distribution with r ˆ  5 and p ˆ  0.4. The mean waiting time
           is, as seen from Equation (6.27),

                                          5
                                 EfXgˆ      ˆ 12:5 gaps:
                                         0:4


           6.2  MULTINOMIAL DISTRIBUTION

           Bernoulli trials can be generalized in several directions. A useful generalization
           is to relax the requirement that there be only two possible outcomes for each
           trial. Let there be r possible outcomes for each trial, denoted by E 1 , E 2 ,..., E r ,
           and let P(E i ) ˆ p i , i ˆ 1, ... , r,  and p 1 ‡ p 2 ‡     ‡ p r ˆ 1.  A typical outcome of
           n trials is a succession of symbols such as:

                                    E 2 E 1 E 3 E 3 E 6 E 2 ... :

             If we let random variable X i , i ˆ 1, 2, ... , r,  represent the number of E i  in a
           sequence of n trials, the joint probability mass function (jpmf) of X 1 , X 2 ,.. . , X r ,
           is given by


                                                 n!
                       p      …k 1 ; k 2 ; ... ; k r †ˆ  p p .. . p ;   …6:30†
                                                        k 1 k 2
                                                                k r
                        X 1 X 2 ...X r                  1  2    r
                                            k 1 !k 2 !... k r !
           where k j ˆ  0, 1, 2, . . . , j ˆ  1, 2, . . . , r, and k 1 ‡ k 2 ‡     ‡ k r ˆ n.
             Proof for Equation 6.30: we want to show that the coefficient in Equation
           (6.30) is equal to the number of ways of placing k 1  letters E 1 , k 2  letters E 2 ,.. .,
           and k r  letters E r  in n boxes. This can be easily verified by writing
                      n!             n   k 1       n   k 1   k 2         k r 1
                                n
                             ˆ                                        :
                  k 1 !k 2 ! ... k r !  k 1  k 2         k r

             The first binomial coefficient is the number of ways of placing k 1  letters E 1
           in  n boxes; the second  is the number of ways of placing k 2  letters E 2  in  the
                          unoccupied boxes; and so on.
           remaining n   k 1







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