Jaylin has a wooden cube which is painted blue on the outside. She cuts the cube into $1000$ identical cubes, some of which have some sides

Wait... I made an error... 

The probability is \((5^{394} \times 2^{96}) \times 2^{-392} \times 3^{-470} = 5^{384} \times 2^{-296} \times 3^{-470} = \color{brown}\boxed{-382}\)

Each side of the cube is \(\sqrt[3]{1000} = 10\).

There are 512 cubes inside that have no faces painted, so the probability is \(1^{512} = 1 \)

There are 8 corners, each with a probability of \({3 \over 6} = {1 \over 2} = 2^{-1}\), so the probability is \({1 \over 2}^8 = {1 \over 2^8}\)

There are \(8 \times 12 = 96\) edge pieces (there are 12 edges in a cube, each edge has 8), and each cube has a probability of \({4 \over 6} = {2 \over 3}\), so the probability is \({2 \over 3}^{96} = {2^{96} \over 3^{96}}\)

There are \(8 \times 8 \times 6 = 384\) center cubes, each with a probability of \({5 \over 6} \), so the probability is \(\large{{5 \over 6}^{384} = {5^{384} \over 6^{384}} = {{5^{384}} \over 2^{384} \times 3^{384}}}\). 

So, the probability is \(\large{{1 \over 2^8} \times {2^{96} \over 3^{96}} \times {5^{384} \over 2^{384} \times 3^{384}} = {5^{384} \times 2^{96} \over 2^{392} \times 3^{470}} = {(5^{384} \times 2^{96})} \times 2^{-392} \times 3^{470} = 5^{384} \times 2^{-296} \times 3^{470}} \)

Thus, \(a + b + c = \color{brown}\boxed{558}\)

Since probabilities must be less than 1, I've interpreted the question as follows: