Analysis of brute force decoder in a $q$-ary erasure channel
To upper bound the first probability, we can assume without loss of generality that the sent codeword is the zero codeword. After deleting $|J|$ rows, we are left with a random $(n-|J|)\times k$ matrix $A$, and we want to bound the probability that its (right) kernel is non-trivial, that is, that there is a non-zero vector $x$ such that $Ax=0$. For any non-zero vector $x$, $Ax$ is a uniformly random vector (why?), and so the probability that $Ax=0$ is $q^{-(n-|J|)}$. Since there are $q^k-1$ many non-zero vectors $x$, the union bound shows that the probability that one of them satisfies $Ax=0$ is at most $q^k \cdot q^{-(n-|J|)}$. We can slightly improve on this by considering vectors up to scalar products, which reduces the probability by a factor of $q-1$.

The second half involves applying Chernoff's bound (also known as Hoeffding's bound) to show that it is unlikely that there are many erasures, and then applying the bound in the first half. Choose the threshold so that $q^{k-n+|J|} \leq q^{-\epsilon n}$.