Let E be the event of an even number of successes. u n : Probability of E occurring at the nth trial not necessarily for the first time f n : Probability of E occurring at the nth trial for the first timeLet U(x) and F(x) be the corresponding probability generating functions and from that we have the equation U ( x ) = 1 / ( 1 − F ( x ) )

Question

Let E be the event of an even number of successes.      u    n  : Probability of E occurring at the nth trial not necessarily for the first time      f    n  : Probability of E occurring at the nth trial for the first timeLet U(x) and F(x) be the corresponding probability generating functions and from that we have the equation  U  (  x  )  =  1      /    (  1  −  F  (  x  )  )

apopihvj · Accepted Answer

Step 1Here&#039;s a way to do by using a Markov chain:                     A                            =                  (                                                    q                                            p                                            0                                                                    0                                            q                                            p                                                                    q                                            p                                            0                                              )                    Answer will be the third column of the first row in An, and the generating function for all entries can be obtained directly from the matrix by computing                                           (            I            −            x                        A            )                                −            1                              Step 2Thus, the gf for the required entry is:                    G        (        x        )                            =        −                                            p                              2                                                    x                              2                                                                                        (                                  p                                      2                                                  −                                  q                                      2                                                  )                                                    x                              2                                      +            2                        q            x            −            1                              and by partial fractions,                     G        (        x        )                            =                              p            2                                q            −            p                          +                  p          2                          (                      1                          1              −              x                                −                      1                                          (                q                −                p                )                                            (                1                −                (                q                −                p                )                                x                )                                              )                    Hence, the solution for       u    n   is given by   [      x    n    ], which is                              u          n                                    =                  p          2                          (          1          −          (          q          −          p                      )                          n              −              1                                )

ajakanvao · Accepted Answer

Step 1Let       u    n   be the probability that n Bernoulli trials result in an even number of successes. This occurs if an initial failure is followed by an even number of successes, or an initial success is followed by an odd number of successes. Therefore       u    0    =  1 and for   n  ≥  1      u    n    =  q      u          n      −      1        +  p  (  1  −      u          n      −      1        )  .Multiplying by       s    n   and adding over n we see that the generating function satisfies  U  (  s  )  =  1  +  q  s  U  (  s  )  +  p  s  (  1  −  s      )          −      1        −  p  s  U  (  s  )or  2  U  (  s  )  =  [  1  −  s      ]          −      1        +  [  1  −  (  q  −  p  )  s      ]          −      1        .Expanding the right hand side using geometric series we find that the coefficients satisfy       u    n    =            1      2        +                    (        q        −        p                  )          n                    2        .

Answered question

Answer & Explanation

New Questions in High school geometry