If we distribute $k$ identical balls to $n$ distinct bins, what is the probability a randomly selected bin will contain at least 2 balls, assuming all balls have a uniform probability of being placed in any bin and $k < n$.
Since it's probability, not "number of ways to arrange," and all distributions have equal probability, first I found the total number of distributions by assuming $k$ distinct balls, $n^k$, which means there are $n^{k+1}$ bins across all distributions.
I can figure out that if there are $k < n$ balls, the number of distributions where every bin has at most $1$ ball is $k \choose n$ (we have $k$ balls and each chooses 1 distinct bin), so guaranteed ${k \choose n} n$ bins across all distributions have fewer than $2$ balls. But I'm not sure how to figure out how many of the $n$ bins contain fewer than $2$ balls in the cases where some bins have $2$ or more balls.
I could easily be going about this from the wrong angle completely though.
Best Answer
The probability no balls are in a random bin is $P_0=(\frac{n-1}{n})^k$ and the probability that exactly one ball is in the bin is $P_1=\frac{k}{n}(\frac{n-1}{n})^{k-1}$ What you want is $P_{2+}=1-P_0-P_1$.