Can a probabilistic Turing Machine compute an uncomputable number?
Consider the following reasonable definition for a Turing machine computing an irrational number in $[0,1]$.


  A Turing machine computes an irrational $r \in [0,1]$ if, on input $n$, it outputs the first $n$ digits (after the decimal) of the binary representation of $r$.


One can think of many extensions of this definition for probabilistic Turing machines. Here is a very permissive one.


  A probabilistic Turing machine computes an irrational $r \in [0,1]$ if, on input $n$, (1) it outputs the first $n$ digits of $r$ with probability $p$; (2) it outputs any other string with probability less than $p$; (3) it never halts with probability less than $p$.


Under this definition, it is not immediately clear whether everything that you can compute is indeed computable (under the sense of the first definition).

However, there are some modifications that do allow us to conclude that the resulting number is computable, for example:


We can insist that the machine always halt.
We can insist that $p &gt; 1/2$.


Other modifications are not necessarily enough. For example, does it help if we assume that the non-halting probability tend to $0$ with $n$?

Summarizing, it might depend on the model.