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 $4?$
$\textbf{(A)}\: \frac34\qquad\textbf{(B)} \: \frac{57}{64}\qquad\textbf{(C)} \: \frac{59}{64}\qquad\textbf{(D)} \: \frac{187}{192}\qquad\textbf{(E)} \: \frac{63}{64}$
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 $$1 - \frac{5}{64} = \frac{59}{64} \longrightarrow \boxed{\textbf{(C)}}$$
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