Given are vectors $x$ and $y$ with $d (x, y) = k ≤ n$, compute $P \left \{Y = y | X = x \right \}$. Does this probability depend on the concrete choice of the vectors x and y?
I'm little bit confused what should I do with d or Hamming distance. Also what formula should I use? Usually I would use probability mass function but what would n and k be?
If the Hamming distance between $x$ and $y$ is $k$ then there are $k$ positions at which $x$ and $y$ are different.
If $x$ is given and we would like to see $y$ at the output then we would like to see $k$ changes and $n-k$ non-changes.
Assuming that the channel works independently of the input and independently of its own history then the probability we seek is
$$(1-p)^{n-k}p^k$$
because we want $k$ changes and ...
This probability does not depend on $x$, it depends only on its Hamming distance from $y$.