conditional probability of dependent random variables

As @BobHanlon points out that the probability is zero for a specific value. But probabilities are not (necessarily) 0 in intervals. So we can get a probability statement for an interval and then take the limit as the size of that interval goes to zero.

p = Probability[X == 1 \[Conditioned] Abs[X + Z - y] <= δ, 
  {X \[Distributed] BernoulliDistribution[1/2], Z \[Distributed] NormalDistribution[]},
  Assumptions -> δ > 0 && y ∈ Reals]

Limit[p, δ -> 0]

So we end up with what you did with pencil and paper.

WARNING: The answer I provided earlier is wrong because, perhaps among other reasons, it equates a PDF with a probability. I provide a better answer immediately below that slightly extends @JimB's correct answer.

  prob = Limit[
  FullSimplify[
   Probability[
    X == 1 \[Conditioned] 
     Abs[X + Z - y] < \[Delta], {X \[Distributed] 
      BernoulliDistribution[f], 
     Z \[Distributed] NormalDistribution[m, s]}, 
    Assumptions -> {y \[Element] Reals, \[Delta] > 0}]], \[Delta] -> 
   0]

The output is:

 (1 - (E^((1 + 2*m - 2*y)/(2*s^2))*(-1 + f))/f)^(-1)

If we now do the following substitution, we get the special case considered in the original post.

prob /. {f -> 1/2, m -> 0, s -> 1}

We get the same answer as the OP did using pencil and paper and @JimB did using his code. It is in a slightly different form, but if you test for equality, you get True.

Also, if anyone wants a numeric answer, I get the following:

prob /. {f -> 1/2, m -> 0, s -> 1, y -> 1} // N

0.622459

---

The following earlier answer is not correct. Sorry for sowing confusion.

Let's try something harder and then do your problem as a special case:

    ydist[f_, m_, s_] := ProbabilityDistribution[{"CDF", 
    FullSimplify[
    Probability[
     X == 1 \[Conditioned] 
      X + Z >= y, {X \[Distributed] BernoulliDistribution[f], 
      Z \[Distributed] 
       NormalDistribution[m, s]}]]}, {y, -\[Infinity], \[Infinity]}]

You can then write

 ydist[f,m,s]

and get an elaborate formula. Or you can plug in the values from your example:

 ydist[1/2,0,1]

From there you can compute a PDF and bunch of other stuff. I would note that the answer produced by this method is more complicated than your paper and pencil version. So, either you have erred, I have erred, or the Wolfram engine has erred. Hmm.