Negative binomial distribution vs geometric distributiionHow can this question be geometric and also negative binomial. In order to join a group Peter needs 9 invitations.The probability that he receives an invitation on any day is 0.8, independent of other days.He joins as soon as he receives his 9 th invitation. Given that he joined on the 14 th day, find the probability that he receives his first invite on the first day.

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martinmommy26nv8 · Accepted Answer

Step 1How can this question be geometric and also negative binomial.A negative binomial random variable,   X  ∼      N    B    (  r  ,  q  ), is a count of successes occurring (at rate q) before r failures.A (zero-based) geometric random variable,   Y  ∼            G      e      o        0    (  p  ), is a count of failures before the first success, with success rate p.Step 2Hence:   Y  ∼      N    B    (  1  ,  1  −  p  ). That is: A geomemtric distribution with parameter p is a negative binomial distribution with parameters   1  ,  1  −  p

Aliyah Thompson · Accepted Answer

Step 1The geometric distribution is a special case of the binomial negative distribution with parameter   r  =  1.To see why this is the reason, see how they come to be. The geometric distribution represents the number of tries needed to get a positive result in a binomial setting, while the negative binomial represents the number of tries needed to get an arbitrary number r of successes.From here we can deduce what I claimed, and also that the negative binomial of parameter r corresponds to the sum of r independent identically distributed random variables with geometric distribution.Step 2Let&#039;s try to solve your problem! Let       X    i   denote the event &#039;Peter receives an invitation the ith day&#039;. Then each       X    i    ∼  B  e  r  n  (  p  =  0.8  ), and       ∑          i      =      1              13            X    i    ∼  B  i  n  (  p  =  0.8  ,  r  =  13  ) (note that       X    1    4 is fixed to be true since Peter is guaranteed to have received an invitation the last day).Let&#039;s use Bayes:  P  (      X    1    =  1      |        ∑          i      =      1              13            X    i    =  8  )  =  P  (      X    i    =  1  )  ⋅            P      (              ∑                  i          =          1                          13                            X        i            =      8              |                    X        1            =      1      )                      ∑                  i          =          1                          13                            X        i            =      8      Note that   [      ∑          i      =      1              13            X    i    =  8      |        X    1    =  1  ]  ≡  [      ∑          i      =      2              13            X    i    =  7  ]  ∼  B  e  r  n  (  p  =  0.8  ,  r  =  12  )  =  7Thus we know the distribution of each term in the Bayes equation, so we can directly substitute:  P  (      X    1    =  1      |        ∑          i      =      1              13            X    i    =  8  )  =  0.8  ⋅                                                        (                                            12            7                                              )                                          (      0.8              )                  7                    (      1      −      0.8              )                  12          −          7                                                                        (                                            13            8                                              )                                          (      0.8              )                  8                    (      1      −      0.8              )                  13          −          8                      =      8    13  Note this result can be obtained in a more direct way using the following reasoning. The probability of receiving an invitation each day is equal, and we know that from days 1 to 13 he has received 8 invitations. Then the probability of getting an invitation the first day is equal to the number of combinations of 8 days between those 13 including the first between the total number of combinations of 8 days over 13:  P  (      X    1    =  1      |        ∑          i      =      1              13            X    i    =  8  )  =                              (                            12        7                              )                                              (                            13        8                              )                      =      8    13  .So yeah, using the negative binomial or geometric here is not needed, and more complicated than necessary.

How can this question be geometric and also negative binomial. In order to join a group Peter needs 9 invitations.The probability that he receives an invitation on any day is 0.8, independent of other days.

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