2005 AMC 10B Problem 12

The product is prime exactly when one die shows a prime ($2,$ $3,$ or $5$) and the other eleven all show $1.$ The probability that any single die is the prime one is $\dfrac36 = \dfrac12,$ and each of the other eleven shows $1$ with probability $\dfrac16.$ Accounting for which of the twelve dice is prime, the probability is $$\begin{aligned} 12 \cdot \dfrac12 \cdot \left(\dfrac16\right)^{11} &= 6 \cdot \left(\dfrac16\right)^{11} \\ &= \left(\dfrac16\right)^{10}. \end{aligned}$$ Thus, E is the correct answer.