Seven balls are distributed randomly in seven cells. If exactly two cells are empty, show that the (conditional) probability of a triple occupancy of some cells equals $1/4$.
Let $H$ be the event that exactly two cells are empty. Let $A$ be the event that there is a triple occupancy.
I need to compute $P(A|H)$.
2 cells are empty, so a triple occupancy must occur in the remaining 5 cells. Thus, there are five ways to do it. Once the place for a triple occupancy is chosen, the rest 4 cells must contain 1 ball per each.
the number of ways to distribute 7 balls into 5 cells without any cell being empty is ${6 \choose 4}$.
Therefore, my answer is that $P(A|H) = 1/3$, and this is wrong. Can you point out where my logic is wrong?
The problem is that you haven't considered the order of the balls. It's true that there are $5$ patterns (i.e. just counting how many balls are in each cell) with a triple and $10$ patterns without a triple, but each of the patterns with a triple occurs in $\binom 73\times 4!$ ways, since there are this many options for which three balls go in the triple and order the rest. Similarly each pattern without a triple occurs in $\binom 72\times \binom 52\times 3!$ ways.
If we distribute each ball at random, it is the ways with balls labelled which are equally likely, not the patterns.