Kayla rolls four fair $6$-sided dice. What is the probability that at least one of the numbers Kayla rolls is greater than $4$ and at least two of the numbers she rolls are greater than $2$?
$\textbf{(A)}\ \frac{2}{3} \qquad  \textbf{(B)}\ \frac{19}{27} \qquad  \textbf{(C)}\ \frac{59}{81} \qquad  \textbf{(D)}\ \frac{61}{81} \qquad  \textbf{(E)}\ \frac{7}{9}$
We will subtract from one the probability that the first condition is violated and the probability that only the second condition is violated, being careful not to double-count the probability that both conditions are violated.
For the first condition to be violated, all four dice must read $4$ or less, which happens with probability $\left( \frac23 \right)^4 = \frac{16}{81}$.
For the first condition to be met but the second condition to be violated, at least one of the dice must read greater than $4$, but less than two of the dice can read greater than $2$. Therefore, one of the four die must read $5$ or $6$, while the remaining three dice must read $2$ or less, which happens with probability ${4 \choose 1} \left(\frac13\right) \left(\frac13\right)^3 = 4 \cdot \frac13 \cdot \frac{1}{27} = \frac{4}{81}$.
Therefore, the overall probability of meeting both conditions is $1 - \frac{16}{81} - \frac{4}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}$.