4 balls are placed successively into 4 cells, all 4^4 arrangements being equally probable. Given that the first 2 balls are in different cells (event B), what is the probability that one cell contains exactly 3 balls (event A)?
P(A|B) = P(A intersection B) / P(B), given P(B) != 0 (Conditional Probability Thereom)
The Attempt at a Solution
My reasoning is: The first ball will be placed for sure in an empty cell, the second ball has a probability of 3/4 to be placed in an empty cell (because there are only 3 cells empty), so P(B) = 3/4. Since we know B has occurred, the 3rd ball has 2/4 probability to be placed in one of the cells having already 1 ball, and the 4th ball has 1/4 probability to hit the cell with 2 balls. So we have P(A|B) = 1/8.
I don't really know if it is correct or not, the problem is that I m supposed to solve it using Conditional Probability. How can I apply it to solve this problem?