What is the probability of the matrix being Singular

To be singular, we need $a=0$ or $d=0$ or $f=0$.

To be non-singular, we need $a=d=f=1$.

Hence, the probaility is $$1-\frac{1}{2^3}=\frac{7}{8}.$$

Note that the values that $b,c,e$ takes are irrelevant. Number of ways to get a singular matrix is $7 \times 8=56$.

It looks basically correct but I'd word it (fully) as follows:

Any given square matrix (over a field) is singular if and only if its determinant is zero. The given matrix is a triangular one and thus its determinant is simply the product of the elements in the main diagonal. Then the matrix is singular iff

$$\;adf=0\iff a=0\,\vee\,d=0\;\vee\, f=0$$

and etc.