If we do not pick the balls at once we need to be a little careful with how we understand the problem. Here is where the terms with replacement and without replacement come into play.

(i) If we pick the 4 balls with replacement, then we put the ball back after each pick (so the same exact ball could be picked more than once). Then each of the picks is independent of the others, and this is discussed more in chapter 9.

(ii) If we pick the 4 balls without replacement, then in fact the probability is the same whether we view the balls all chosen at once or whether we pick the balls one by one (in order). If we pick each ball one by one (in order) then actually both the numerator and denominator are multiplied by $4!$ (since we're doing with order) so the final probability is not affected.