No, not quite so. You want the distribution of the count for iid Bernoulli _trials_ until the $k$ th errorfree packet. This count follows a _negative binomial distribution_.
Give an iid errorfree rate of $p$, the probability for an arrangement of exactly $k-1$ errorfree packets among $n-1$ packets and _then_ the $k^{\rm th}$ error free placket exactly on the $n^{\rm th}$ arrival is: ___?
> $$\mathsf P(N=n)~=~\binom{n-1}{k-1}p^k(1-p)^{n-k}$$