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?
Best Answer
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 $4/5*3/5*3/5=36/125$.