Probability of two successes, possible geometric distribution problemLet Y be the number of trials required to observe r successes. List the outcome space for the variable Y and show that the probabilities for Y are giver by the formula P ( Y = k ) = ( k − 1 ) ( p 2 ) ( 1 − p ) k − 2 .Okay I know the outcome space is Y := 2 , 3 , 4 , . . . , the first two trials could be successes or you could never get two successes. The problem looks geometric but I dont undertand the extra ( k − 1 ) ⋅ p in front.

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Probability of two successes, possible geometric distribution problemLet Y be the number of trials required to observe r successes. List the outcome space for the variable Y and show that the probabilities for Y are giver by the formula   P  (  Y  =  k  )  =  (  k  −  1  )  (      p    2    )  (  1  −  p      )          k      −      2      .Okay I know the outcome space is   Y  :=      2    ,    3    ,    4    ,    .    .    .  , the first two trials could be successes or you could never get two successes. The problem looks geometric but I dont undertand the extra   (  k  −  1  )  ⋅  p in front.

Steinherrjm · Accepted Answer

Step 1Assuming independence of trials and that each trial has probability p of success and also that   r  =  2, as in the title:  P  [  Y  =  k  ] is the probability that the second success occurred on the k&#039;th trial. If the second success occurred on the k&#039;th trial, then there was exactly one success in the first   k  −  1 trials and the k&#039;th trial was a success.Now, there are   k  −  1 ways to have exactly one success in the first   k  −  1 trials (for example, one way is that the second trial was a success and the other   k  −  2 trials were failures), each of which having probability   (  1  −  p      )          k      −      2        p. So the probability that exactly one success occurred in the first   k  −  1 trials isP(exactly one success in the first                     (1)                    P        (                                            exactly one success                               in the  first                 k                −                1                 trials                                              )          =          (          k          −          1          )          (          1          −          p                      )                          k              −              2                                p          .                    Step 2The probability of a success on the k&#039;th trial is                     (2)                    P        (                                             success on the                              k                &#039;th trial                                              )          =          p                    The probability that the second success occurred on the k&#039;th trial is the product of (1) and (2):  P  [  Y  =  k  ]  =            (          (  k  −  1  )  (  1  −  p      )          k      −      2        p              )        ⋅  p  =  (  k  −  1  )      p    2    (  1  −  p      )          k      −      2        .

Let Y be the number of trials required to observe r successes. List the outcome space for the variable Y and show that the probabilities for Y are giver by the formula P(Y=k)=(k-1)(p^2)(1-p)^{k-2}.

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