What is the probability that given a bin (out of $n$ bins) that it will receive no balls I'm working on a problem where there are $n$ bins. I want to know the probability that given a bin (Specific bin) that it will receive zero balls. There are $m$ balls thrown at random into the bins.
This is similar to my homework problem, but I'm looking for the formula. I'll input my specific variables after I understand it. thanks
 A: The answer depends on whether the balls are distinguishable.


*

*Suppose that the balls are indistinguishable. Using stars and bars, the number of ways we can put $m$ balls into $n$ bins is equal to
$$
{n+m-1\choose n-1}.
$$ 
The number of ways to put $m$ balls into $n-1$ bins (we select a specific bin and do not put any balls into it) is equal to
$$
{n+m-2\choose n-2}.
$$
The probability that the selected bin will contain $0$ balls is then given by
$$
\frac{n+m-2\choose n-2}{n+m-1\choose n-1}=\frac{n-1}{n+m-1}.
$$

*Suppose that the balls are distinguishable. Then the number of ways we can put $m$ balls into $n$ bins is equal to
$$
n^m.
$$
The number of ways to put $m$ balls into $n-1$ bins is equal to
$$
(n-1)^m.
$$
The probability that the selected bin will contain $0$ balls is then given by
$$
\biggl(\frac{n-1}n\biggr)^m.
$$
A: Hint: What is the probability of the first ball not landing in the bin? What about the second ball? Are the events "the first ball does not go into the selcted bin" and "the second ball does not go into the selected bin" dependent or independent?
A: To further Andre's response, the probability that no balls will be thrown in a certain bin is $\frac{n-1}{n}$. Every time you throw the ball in a bin, you run the risk of getting a ball in, so the probability of having a bin receive zero balls is $(\frac{n-1}{n})^m$.
