As for your original problem, I think it can be solved as follows. We use the

principle of inclusion-exclusion. The probability of getting A1,A2,B1,B2 black regardless of all others is $\displaystyle (0.05)^4$. There are 49 of these 2x2 squares, so take $\displaystyle 49\cdot(0.05)^4$. But we're not done yet because we counted duplicates. Now the probability of getting A1,A2,A3,B1,B2,B3 is $\displaystyle (0.05)^6$. Considering just the rectangles with these dimensions and this orientation, we have $\displaystyle 6\cdot7$ of them, giving $\displaystyle 42\cdot(0.05)^6$. Next, consider that the probability of getting A1,A2,B1,B2,C1,C2 is the same; combining all this, we arrive at the intermediate result

$\displaystyle 49\cdot(0.05)^4-84 \cdot(0.05)^6$

Then we would proceed to calculate the probability of getting arrangements leading to three 2x2 squares, using a "+" instead of a "-" for the next term, etc., all the way to 49.

It's possible there is some mistake in my reasoning, although at the moment everything seems sound. To counteract the possibility of a mistake, I would write a Monte Carlo and see how close the results were. If they were quite close, then it would increase my confidence that I didn't make a mistake. Another option is to try out the idea on smaller boards (e.g., 4x4) and see if the results make sense.

I'm assuming you want an exact answer. Because if you're just after an approximation, then you can use Monte Carlo, which will give you more precision the longer you run it, but if you need a lot of precision then this could take a prohibitively long time.