Problem Analysis #5

[THOUGHT] The problem asks us to find the probability that all edges of a cube are colored with distinct colors when each edge can be colored in 12 different ways (since there are 12 colors).

Firstly, let's calculate the total number of possible colorings for the cube. Each edge has 12 choices, and there are 12 edges, so the initial count would be $12^{12}$. However, this counts configurations where two edges sharing a vertex have the same color, which is not allowed.

To correct for overcounting, we need to divide by the number of ways the colors can be permuted at each vertex such that no two adjacent edges receive the same color. For any given vertex, there are 11 choices (since one color has been used on an edge connected to this vertex). The cube has 8 vertices, so we have $(11!)^8$ configurations where all edges are colored differently.

So the total number of valid colorings is $\frac{12^{12}}{(11!)^8}$.

Now let's calculate the probability:

Probability = (Number of successful outcomes) / (Total possible outcomes) = $\frac{(11!)^8}{12^{12}}$

We can simplify this expression by dividing both numerator and denominator by $(11!)^8$:

Simplified Probability = $\frac{1}{(12/11)^8}$ = $\frac{1}{(1+\frac{1}{11})^8}$ ≈ 0.347

Here, we used the binomial expansion to calculate $(1+\frac{1}{11})^8$.

Since the problem asks for the probability as a reduced fraction $\frac{q}{p}$ and $p+q$, let's express this decimal as a fraction:

Probability ≈ 0.347 = 347/1000

So, p=1000, q=347, and p+q = 1347.

\boxed{1347}