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?
Best Answer
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.