Probability of selecting a non-measurable set If you randomly select a subset of $[0,1]$, what is the probability that it will be measurable?
Edit: This question may be unanswerable as asked.  If additional assumptions could be made to make it answerable, I would be interested in learning about them.
 A: Almost the same Question was asked on mathoverflow: https://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable
The conclusion was that the set of all measurable sets was not measurable; Therefore, no Probability can be provided.
A: I am assuming that we have fixed some $\sigma$-field of subsets of the interval and "measureable" subset means a subset from that $\sigma$-field.
Your question is meaningless until we specify what does "randomly select a subset" mean. And to do that we have to choose a $\sigma$-field of subsets of the set $2^{[0,1]}$, say $\mathcal{A}$, and a probability measure $\mathbb{P}:\mathcal{A}\rightarrow [0,1]$. It is easy to see that now when we know what does "randomly select a subset" mean we can answer your question. And what is the answer? Depending on the choice of $\mathcal{A}$ and $\mathbb{P}$ it can be any number between $0$ and $1$, and none of those answers is better than the other. (From the point of view of probability theory)
A: The following topological smallness (rather than measure smallness) result may be of interest.
Let ${\mathbb P}[0,1]$ be the collection of all subsets of $[0,1]$ modulo the equivalence relation $\sim$ defined by $E \sim F \Leftrightarrow {\lambda^{*}(E \Delta F)} = 0,$ where $\lambda^{*}$ is Lebesgue outer measure and $\Delta$ is the symmetric difference operation on sets. The set ${\mathbb P}[0,1]$ can be made into a complete metric space by defining the distance function $d,$ where $d(E,F) = {\lambda^{*} (E \Delta F)}.$ In the paper cited below it is proved that the collection of measurable subsets of $[0,1]$ is a perfect nowhere dense set in ${\mathbb P}[0,1].$
Thus, in this setting, the collection of measurable subsets of $[0,1]$ makes up a very tiny part of the collection of all the subsets of $[0,1].$ Note that in the space ${\mathbb P}[0,1]$ the collection of measurable subsets is not just a first category subset of ${\mathbb P}[0,1]$ (this alone would make the collection a tiny subset of ${\mathbb P}[0,1]$), but in fact the collection of measurable subsets is actually a nowhere dense subset of ${\mathbb P}[0,1]$ (hence my saying the collection is a very tiny subset of ${\mathbb P}[0,1]$).
Nobuyuki Kato, Tadashi Kanzo, and Oharu Shinnosuke, A note on the measure problem, International Journal of Mathematical Education in Science and Technology 19 (1988), 315-318.
