Equivalently, we are told that "this process produces 25% defectives" so for any one item the probability it is defective (D) is 0.25= 1/4, the probability it is non-defective (N) is 075= 3/4.
The probability of NNN, all three are non-defective, since we are also told these are "independent", is (3/4)(3/4)(3/4)= 27/64. With X the number of defective items, P(X= 0)= 27/64.
The probability the first chosen is defective, the other two are non-defective, NND, in that order, is (3/4)(3/4)(1/4)= 9/64, But it is easy to see that the probability of "non-defective, defective, non-defective", NSN, in [b]that[/b ] order is (3/4)(1/4)(0.75)= 9/64 also and similarly "non-defective, non-defective, defective", NND, is also 9/64. P(X= 1)= 9/64+ 9/64+ 9/64= 3(9/64)= 27/64.
The probability the both the first two chosen are defective, the third non-defective, DDN, in that order, is (1/4)(1/4)(3/4)= 3/64. Again, it is easy to see that the probabililties of "defective, non-defective, defective", DND, and "non-defective, defective, defective", NDD are (1/4)(3/4)(1/4)= 3/64, and of (3/4)(1/4)(1/4)= 3/64 also. P(X= 2)= 3/64+ 3/64+ 3/64= 3(3/64)= 9/64.
Finally, the probability of "DDD", all three defective, is (1/4)(1/4)(1/4)= 1/64. P(X= 3)= 1/64.