imc

In order for the product of the numbers to be prime, $11$ of the dice have to be a $1$, and the other die has to be a prime number. There are $3$ prime numbers ($2$, $3$, and $5$), and there is only one $1$, and there are $\left(\genfrac{}{}{0ex}{}{12}{1}\right)$ ways to choose which die will have the prime number, so the probability is $\frac{3}{6}\times {\left(\frac{1}{6}\right)}^{11}\times \left(\genfrac{}{}{0ex}{}{12}{1}\right)=\frac{1}{2}\times {\left(\frac{1}{6}\right)}^{11}\times 12={\left(\frac{1}{6}\right)}^{11}\times 6=\boxed{(\frac{1}{6}{)}^{10}}$.