One coin chosen between a biased coin and a fair coin, and is tossed n times. Find probability of having gotten the biased coin.

$ \begingroup $ In a different question, I had asked for clearing on the surveil trouble where I wanted to just understand the trouble. now, I have attempted it and wish to know if my solution is right .
Problem statement:
A drawer contains two coins. One is an indifferent coin, which when tossed, is equally probable to turn up heads or tails. The other is a bias coin, which will turn up heads with probability $ phosphorus $ and tails with probability $ 1 − phosphorus $. One coin is selected ( uniformly ) at random from the draftsman. Two experiments are performed :

a ) The selected mint is convulse $ nitrogen $ times. Given that the coin turns up heads $ k $ times and tails $ n − k $ times, what is the probability that the coin is biased ?
b ) The selected coin is tossed repeatedly until it turns up heads $ k $ times. Given that the mint is flip $ n $ times in total, what is the probability that the mint is biased ?
My attempt:
a ) Let $ F $ be the jell of outcomes where I have chosen the fair coin, and $ B $ be the set of outcomes where I have chosen the bias mint. Let $ A_k $ be the determine of outcomes where I tossed $ nitrogen $ times and got $ k $ heads. I need to find $ P ( B|A_k ) $ .
$ $ P ( A_k \cap F ) = \frac { 1 } { 2 } { { north } \choose { potassium } } \frac { 1 } { 2^n } $ $ $ $ P ( A_k \cap B ) = \frac { 1 } { 2 } { { north } \choose { potassium } } p^k ( 1-p ) ^ { n-k } $ $ Since $ F $ and $ B $ partition the sample space, we have
$ $ P ( A_k ) = \frac { 1 } { 2 } { { north } \choose { potassium } } \left \ { p^k ( 1-p ) ^ { n-k } +\frac { 1 } { 2^n } \right \ } $ $
We from Bayes ‘ theorem know that

$ $ P ( B|A_k ) =\frac { P ( A_k|B ) } { P ( A_k ) } $ $
therefore we get
$ $ P ( B|A_k ) =\frac { p^k ( 1-p ) ^ { n-k } } { p^k ( 1-p ) ^ { n-k } +\frac { 1 } { 2^n } } $ $
barn ) In this case, we keep tossing till we get $ k $ heads. now this means that the last flip is a headway. nowadays we know that we had to toss $ normality $ times to get $ k $ heads .
As in ( a ) above, let $ F $ be the bent of outcomes where I have chosen the fairly coin, and $ B $ be the set of outcomes where I have chosen the bias mint. Let $ C_n $ be the event that I had to toss $ north $ times to get $ k $ heads. I need $ P ( B|C_n ) $ .
$ $ P ( C_n \cap F ) = \frac { 1 } { 2 } \times \frac { 1 } { 2 } \times { { n-1 } \choose { k-1 } } \frac { 1 } { 2^ { n-1 } } $ $ $ $ P ( C_n \cap B ) = \frac { 1 } { 2 } \times p \times { { n-1 } \choose { k-1 } } p^ { k-1 } ( 1-p ) ^ { n-k } $ $
again using Bayes ‘ theorem along the lines of what was done in ( a ), we get

$ $ P ( B|C_n ) =\frac { p^k ( 1-p ) ^ { n-k } } { p^k ( 1-p ) ^ { n-k } +\frac { 1 } { 2^n } } $ $
I am a moment skeptic about my answer as the answers to ( a ) and ( bacillus ) are turning out to be the like. I can not find an intuitive explanation as to why that is .
Please provide feedback and let me know if I have solved this interview correctly. In lawsuit there is a error, please point me to it .

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