2021 AMC Fall 10B Problem 14

"Divisible by $4$" happens in many ways, but failing happens in only a few, so count the failures and subtract.

1. Sort each die by how many 2s it gives. A product is divisible by $4$ when it has at least two factors of $2$.
- $1, 3, 5$: no factor of $2$. Chance $\frac36 = \frac12$.
- $2, 6$: exactly one factor of $2$. Chance $\frac26 = \frac13$.
- $4 = 2 \times 2$: two factors of $2$. Chance $\frac16$.
2. When does the product fail? When the six dice together give zero or one factor of $2$.
3. Zero factors of 2. All six dice are odd:
$$\left(\tfrac12\right)^6 = \tfrac{1}{64}.$$
4. Exactly one factor of 2. One die shows $2$ or $6$, and the other five are odd. The special die can be any of the $6$ dice:
$$6 \times \tfrac13 \times \left(\tfrac12\right)^5 = \tfrac{6}{96} = \tfrac{1}{16} = \tfrac{4}{64}.$$
5. Chance of failing: $\frac{1}{64} + \frac{4}{64} = \frac{5}{64}$.
6. Chance of success:
$$1 - \tfrac{5}{64} = \tfrac{59}{64}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{59}{64}}$.

Whether a product is divisible by $4$ depends only on how many factors of $2$ it has in total, so each die only needs to be sorted into "zero, one, or two 2s." When the event you want is big and messy, the opposite event is often small and easy to count. Count it and subtract from $1$.