2021 Fall AMC 12B Problem 11

The product fails to be divisible by $4$ when it has at most one factor of $2.$ Each die is odd with probability $\tfrac12,$ contributes exactly one factor of $2$ (a $2$ or $6$) with probability $\tfrac13,$ and two factors (a $4$) with probability $\tfrac16.$ All six odd: $\left(\tfrac12\right)^6 = \tfrac{1}{64}.$ Exactly one die a $2$ or $6$ and the rest odd: $6 \cdot \tfrac13 \cdot \left(\tfrac12\right)^5 = \tfrac{1}{16} = \tfrac{4}{64}.$ The complement is $\tfrac{1}{64} + \tfrac{4}{64} = \tfrac{5}{64},$ so the answer is $1 - \tfrac{5}{64} = \tfrac{59}{64}.$ Thus, the correct answer is C.