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.
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