What is the probability of getting a sum of 22 or more when four dice are thrown?

Sum is 24: The only possibility is (6,6,6,6) -> 1 way

Sum is 23: The only possibility is (6,6,6,5) -> $$\frac{4!}{3!}$$ = 4 ways

Sum is 22: There are two possibilities, i.e. (6,6,6,4) and (6,6,5,5)

(6,6,6,4) -> $$\frac{4!}{3!}$$ = 4 ways

(6,6,5,5) -> $$\frac{4!}{2!2!}$$ = 6 ways

Total number of ways satisfying the given condition = 1 + 4 + 4 + 6 = 15

Total number of possibilities = (6)(6)(6)(6)

Required probability = $$\frac{15}{1296}$$ = $$\frac{5}{432}$$

The answer is option A.