Probability of building an Invertible Matrix If we build a 10X10 matrix,randomly filling with 1's and 0's, is it more likely to be invertible or singular?
First of all until we have 10 1's its not going to be invertible.
With 10 1's on the diagonal, we can make many lower and upper triangular matrices, each of which will be invertible.  
But I'm not finding a way to proceed.  
 A: Over $\mathrm{GF}(2)$, the number of invertible $n \times n$ $(0,1)$-matrix is given by $$\prod_{k=1}^n (1-2^{-k}).$$  See this math.SE answer for the argument.  Hence the probability of invertibility is approximately $0.289$.
The answer is much harder over the rationals (or equivalently reals).  There's some asymptotic results discussed on the above-linked page.  It seems the exact number is not known for $10 \times 10$ since it's not listed on the OEIS: http://oeis.org/A046747.
Experimentally, you can be fairly certain that the probability is $>0.5$ (i.e., more likely to be invertible).  The following figure plots an estimate of the probability as the number of samples increases:

A: Rebecca gave a good answer. Let $P_d$ be the probability that your $d\times d\;\;$ $0-1$ matrix is SINGULAR. Then it is known (Kahn,Komlos,Szemeredi) that $P_d$ tends to $0$ when $d$ tends to $\infty$. Tao and Vu give an estimate (2005) $P_d=(3/4+o(1))^d$. Yet, for $d=10$ (a small number), the various known  estimates are absolutely useless. Numerical experiments show that $P_{10}\approx 0.297$, but the exact value seems to be unknown. 
Yet, the possible values of $\det(A)$ when $A$ is a $10\times 10\;\;$ $0-1$ matrix are known (Orrick) ; cf. http://mypage.iu.edu/~worrick/slides/SpectrumANU.pdf
