Simple probability problem of placing balls in 5 boxes 
There are five empty boxes. Balls are placed independently one after another in randomly selected boxes. Find the probability that the fourth ball is the first to be placed in an occupied box.

We only need to consider the first 5 balls. The total possibilities are $5^5$. I have been trying various cases but cant seem to pinpoint the crux of the strategy any help?
 A: The first ball is always placed in unoccupied box.
The second ball should be placed in one of remaining 4 boxes, and the probability for this is $4/5$.
The third ball should be placed in one of remaining 3 boxes, and the probability for this is $3/5$.
The fourth ball should be placed in one of occupied 3 boxes, and the probability for this is $3/5$.
All above events must occur so the final probability is $4/5*3/5*3/5=36/125$.
A: For the fourth ball to be the first to be placed in an occupied box, there are


*

*$5$ choices for the first ball.

*$4$ choices for the second ball.

*$3$ choices for the third ball.

*$3$ choices for the fourth ball.


Thus, the probability is
$$\frac{(5)(4)(3)(3)}{(5)(5)(5)(5)}$$
A: $$\frac45\times\frac35\times\frac35$$
The first factor denotes the probability that the second ball is placed in a box that is not occupied.
The second factor denotes the probability that the third ball is placed in a box that is not occupied under condition that the second ball is placed in a box not occupied.
The third factor denotes the probability that the fourth ball is placed in a box that is occupied under condition that the second ball and third ball are both placed in a box not occupied.
