2022 AMC 12B Problem 12

Sort each die into low $\{1,2\},$ mid $\{3,4\},$ or high $\{5,6\};$ each has probability $\tfrac13,$ so the $3^4 = 81$ category patterns are equally likely. We need at least one high die (a number greater than $4$) and at least two dice that are greater than $2$ (mid or high). The two bad events are having no high die and having at most one non-low die. There are $2^4 = 16$ patterns of the first kind, $9$ of the second kind, and $5$ in their intersection. By inclusion-exclusion the count of good patterns is $81 - 16 - 9 + 5 = 61.$ The probability is $\dfrac{61}{81}.$ Thus, the correct answer is D.