2021 Fall AMC 10B #14

Problem:

Una rolls $6$ standard $6$-sided dice simultaneously and calculates the product of the $6$ numbers obtained. What is the probability that the product is divisible by $4?$

$\mathbf{\text{(A)}}\frac{3}{4}\mathbf{\text{(B)}}\frac{57}{64}\mathbf{\text{(C)}}\frac{59}{64}\mathbf{\text{(D)}}\frac{187}{192}\mathbf{\text{(E)}}\frac{63}{64}$

Solution:

We use complementary probability. Note that the total number of ways is $6^6$. Also, if the product is not divisible by $4$, then a $4$ cannot be rolled in any of the six rolls. Similarly, no two multiples of $2$ can be rolled. This leaves us with two cases:

Case 1: No multiple of 2 is rolled

In this case, the only numbers rolled could be $1,3,5$. The number of ways for this is $3^6$.

Case 2: Exactly one multiple of 2 is rolled

In this case, our choices for the even number to be rolled are $2$ and $6$. There are $2$ ways to pick which one out of these two appears and $6$ ways to pick a spot to roll this number. Then, there are $3^5$ ways for the remaining $5$ spots, because only odd numbers can appear on these rolls. This gives us $2\cdot 6\cdot 3^5$ ways.

So, the probability that the product is not divisible by $4$ is $$\frac{3^6+2\cdot 6\cdot 3^5}{6^6}=\frac{3+2\cdot 6}{2^6\cdot 3}=\frac{15}{192}=\frac{5}{64}$$ Thus the probability that it is divisible by $4$ is $$1-\frac{5}{64}=\frac{59}{64}⟶\boxed{{\displaystyle \mathbf{\text{(C)}}}}$$