What is the probability of having exactly k successes from n coins if all n coins are flipped and there are x successes, then n - x coins are re-flipped to give the additional k − x successes?Example: n = 3 , k = 3 , p = 1 / 2, and let: f ( k ) = ( n k ) p k ( 1 − p ) n − k I tried adding the success probabilities for the following scenarios:All 3 success on the first attempt f ( 3 ) = 0.1252 on the first attempt, then 1 on the second f ( 2 ) ⋅ f ( 1 ) = 0.1406251 on the first attempt, then 2 on the second f ( 1 ) ⋅ f ( 2 ) = 0.1406250 on the first attempt, then 3 on the second f ( 0 ) ⋅ f ( 3 ) = 0.015625I get P ( 3 ) = f ( 3 ) + f ( 2 ) ⋅ f ( 1 ) + f ( 1 ) ⋅ f ( 2 ) + f ( 0 ) ⋅ f ( 3 ) = 0.421875.However, doing the same for k = 2 and k = 1, I get: P ( 2 ) = f ( 2 ) + f ( 1 ) ⋅ f ( 1 ) + f ( 0 ) ⋅ f ( 2 ) = 0.5625 P ( 1 ) = f ( 1 ) + f ( 0 ) ⋅ f ( 1 ) = 0.421875But I know this is wrong because P ( 3 ) + P ( 2 ) + P ( 1 ) &gt; 1

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What is the probability of having exactly k successes from n coins if all n coins are flipped and there are x successes, then n - x coins are re-flipped to give the additional   k  −  x successes?Example:   n  =  3  ,  k  =  3  ,  p  =  1      /    2, and let:  f  (  k  )  =                    (                    n      k                      )                  p    k    (  1  −  p      )          n      −      k      I tried adding the success probabilities for the following scenarios:All 3 success on the first attempt   f  (  3  )  =  0.1252 on the first attempt, then 1 on the second   f  (  2  )  ⋅  f  (  1  )  =  0.1406251 on the first attempt, then 2 on the second   f  (  1  )  ⋅  f  (  2  )  =  0.1406250 on the first attempt, then 3 on the second   f  (  0  )  ⋅  f  (  3  )  =  0.015625I get   P  (  3  )  =  f  (  3  )  +  f  (  2  )  ⋅  f  (  1  )  +  f  (  1  )  ⋅  f  (  2  )  +  f  (  0  )  ⋅  f  (  3  )  =  0.421875.However, doing the same for   k  =  2 and   k  =  1, I get:  P  (  2  )  =  f  (  2  )  +  f  (  1  )  ⋅  f  (  1  )  +  f  (  0  )  ⋅  f  (  2  )  =  0.5625    P  (  1  )  =  f  (  1  )  +  f  (  0  )  ⋅  f  (  1  )  =  0.421875But I know this is wrong because   P  (  3  )  +  P  (  2  )  +  P  (  1  )  &amp;gt;  1

Cameron Benitez · Accepted Answer

Step 1Let   S  =      S          1        +      S          2       where       S          i       denotes the number of successes at the i-th flip.Step 2Then:  P      (    S    =    k    )    =      ∑          x      =      0              k        P      (          S              2              =    k    −    x    ∣          S              1              =    x    )    P      (          S              1              =    x    )    =      ∑          x      =      0              k                          (                            n        −        x                    k        −        x                            )                  p          k      −      x                  (      1      −      p      )              n      −      k                          (                    n      x                      )                  p          x                  (      1      −      p      )              n      −      x        =      ∑          x      =      0              k                          (                            n        −        x                    k        −        x                            )                                (                    n      x                      )                  p          k                  (      1      −      p      )              2      n      −      k      −      x        =                    (                    n      k                      )                  p          k                  (      1      −      p      )              2              (        n        −        k        )                  ∑          x      =      0              k                          (                    k      x                      )                        (      1      −      p      )              k      −      x        =                    (                    n      k                      )                  p          k                  (      1      −      p      )              2              (        n        −        k        )                        (      2      −      p      )              k

equipokypip1 · Accepted Answer

Step 1A coin is a success if and only if it is a success at least once out of two flips; the probability for that is   1  −  (  1  −  p  )  (  1  −  p  )  =  2  p  −      p    2  .Step 2Thus the probability you want is                    (                    n      k                      )              (  2  p  −      p    2        )    k    (  (  1  −  p      )    2        )          n      −      k        =                    (                    n      k                      )                  p    k    (  2  −  p      )    k    (  1  −  p      )          2      (      n      −      k      )          .

Binomial probability with selective reflipping. What is the probability of having exactly k successes from n coins if all n coins are flipped and there are x successes, then n - x coins are re-flipped to give the additional k - x successes?

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