2005 AMC 10B Problem 12

1. What does a prime product look like? A prime can only be split as $1 \times$ itself. So if twelve numbers multiply to a prime, eleven of them must be $1$ and the last one must be the prime. For example, $1 \cdot 1 \cdots 1 \cdot 5 = 5$ works, but $2 \cdot 3 = 6$ does not.
2. Primes on a die. The faces are $1$ to $6$; the primes are $2, 3, 5$. (Remember, $1$ is not prime.)
3. Count good outcomes. Choose which die shows the prime: $12$ ways. Choose which prime it shows: $3$ ways. The other eleven dice must all be $1$: only $1$ way. Total: $12 \cdot 3 = 36$ good outcomes.
4. Count all outcomes. Each die has $6$ faces, so there are $6^{12}$ outcomes, all equally likely.
5. Probability. Notice $36 = 6^2$, so
$$\frac{36}{6^{12}} = \frac{6^2}{6^{12}} = \frac{1}{6^{10}} = \left(\frac{1}{6}\right)^{10}.$$

The answer is $\boxed{\textbf{(E)}\ \left(\frac{1}{6}\right)^{10}}$.

"The product is prime" is a very strict condition: it fixes every die but one. Once you describe the good outcomes in words ("eleven $1$'s and one prime"), counting them is two small choices multiplied together. Writing $36$ as $6^2$ lets it cancel neatly with $6^{12}$.

A product of twelve positive integers is prime only if every factor except one equals $1$, and the remaining factor is itself prime. On a die the primes are $2, 3, 5$.

So a favorable roll consists of eleven $1$'s and one prime. Choose which of the $12$ dice shows the prime ($12$ ways) and which prime it shows ($3$ ways): $12 \cdot 3 = 36$ favorable outcomes.

All $6^{12}$ ordered outcomes are equally likely, so
$$P = \frac{36}{6^{12}} = \frac{6^2}{6^{12}} = \frac{1}{6^{10}}.$$

The answer is $\boxed{\textbf{(E)}\ \left(\frac{1}{6}\right)^{10}}$.