To solve this problem, we need to calculate the probability that the fourth ball is the first one placed in an already occupied box, given that balls are placed independently and randomly into one of the five boxes.

Let's break down the problem:

1. Initially, all five boxes are empty. When placing the first ball, it will definitely occupy an empty box because all are empty.
2. Similarly, when placing the second ball, there are 4 empty boxes left, so it still has a chance of landing in an empty box. The probability that it lands in an empty box is \(\frac{4}{5}\).
3. For the third ball, assuming the second ball also went into an empty box, there are 3 empty boxes remaining. Hence, the probability that the third ball also finds an empty box is \(\frac{3}{5}\).
4. Now, the fourth ball will be placed. To find the probability that it is the first to land in an occupied box, it should not land in an empty box available. Hence:
   • The probability that the fourth ball goes into an occupied box is simply the complement of it going into an empty box.
   • So, the probability it will not land in an empty box is \(\frac{2}{5}\), because two boxes are already occupied by the first three balls.

So, for the fourth ball to be the first to land in an occupied box, it must follow a sequence where the first three balls find empty boxes and then it lands in an occupied one, which is given by:

\(\left(\frac{4}{5}\right) \times \left(\frac{3}{5}\right) \times \left(\frac{2}{5}\right)\) for the first three balls finding empty boxes, and finally, the probability of landing in an occupied is \(\frac{2}{5}\).

This gives a total probability of:

\(\left(\frac{4}{5}\right) \times \left(\frac{3}{5}\right)^2\)

Thus, the correct answer is \(\frac{4}{5} \left(\frac{3}{5}\right)^2\), which is the correct option.