Are "most" continuous functions also differentiable?

Let $A$ be a nonempty open subset of $\mathbb{R}$.

Consider a function $f : A \rightarrow \mathbb{R}$. Given that $f$ is continuous, what is the probability that it is differentiable? I suspect it is $0$.

There is a very precise sense in which the answer to your question is "$0$".

Let us denote by $ND$ the set of all continuous nowhere differentiable functions on, say, the interval $[0,1]$, and by $\mathcal C([0,1]$ the Banach space of all continuous functions on $[0,1]$. Then, it can be shown that $SD:=\mathcal C([0,1])\setminus ND$ (the set of all functions which are somewhere differentiable) is a "Haar-null" set, which means that there exists a Borel probability measure $\mu$ on $\mathcal C([0,1])$ such that $$\forall f\in\mathcal C([0,1])\;:\; \mu(f+SD)=0\, .$$

Here, "$\mu(E)=0$" for a possibly non-Borel set $E$ means that $E$ is contained in a Borel set $\widetilde E$ such that $\mu(\widetilde E)=0$. (This remark is needed because $SD$ is not Borel in $\mathcal C([0,1])$).

Thus, in this sense, a continuous function is nowhere differentiable "with probability $1$".

For a proof of this result, this the original paper by Hunt: http://www.ams.org/journals/proc/1994-122-03/S0002-9939-1994-1260170-X/S0002-9939-1994-1260170-X.pdf