N balls distributed in N boxes This question is related to the following question:
n distinguishable balls into n boxes
Suppose that $n$ balls are distributed in $n$ boxes. I have two items 
a) Given that box 1 is empty what is the probability that only one box is empty.
b) Given that only one box is empty what is the probability that this box is box 1.
If $A$ is the event "box 1 is empty" and  $B=$"only one box is empty" then we are looking for 
$$ 
P(B|A)= \frac{P(A\cap B)}{P(A)}~~~\text{and}~~P(A|B)= \frac{P(A\cap B)}{P(B)}
$$ 
I have no idea how to describe the event $B=$"only one box is empty" in order to calculate the probability. 
Can someone help me with this?
Parcial Solution: For item b) note that  $A\subset B $ so $A\cap B=A$ and it is possible to show that 
$$
\#A= [(n-1)n!]/2n^n~~\text{and that}~~\#B= n(n-1)\binom{n}{2}(n-2)! 
$$
therefore:
$$
P(A|B)=\dfrac{[(n-1)n!]/2n^n}{n(n-1)\binom{n}{2}(n-2)!}
$$
Is this ok!?
 A: For a, you are told that all the balls go into bins $2$ through $n$.  You are asking the chance that all bins $2$ through $n$ have at least one ball.  If we consider the balls labeled, there are $(n-1)^{n}$ ways to distribute the balls.  To have all the bins with at least one ball, there are ${n-1 }$ ways to select the bin with two balls, $n \choose 2$ ways to choose the balls in that bin, and $(n-2)!$ ways to arrange the other balls, so the chance is $$\frac{(n-1){n \choose 2}(n-2)!}{(n-1)^{n}}=\frac {n!(n-1)}{2(n-1)^{n}}=\frac {n!}{2(n-1)^{(n-1)}}$$ 
For b, the bins are equivalent, so it is $\frac 1n$
A: ( PART A )
Each desirable outcome is a surjection from an n set to a n-1 set. The total number of outcomes is the number of functions from an n set to an n-1 set.
There are n-1 surjections ; since we know the remaining n-1 boxes are non-empty go ahead and put a ball in each one. Then we'd only have one ball left and n-1 boxes to choose from for that remaining ball.
(n-1)^n is the total number of functions from an n set to a n-1 set.
Therefore, the desired probability is (n-1)/[(n-1)^n] which equals (n-1)^(1-n).
(PART B)
No matter what box the probability it is the only empty box is the same. These events are obviously mutually exclusive and their probabilities sum to 1. Therefore, the desired probability is 1/n. 
