Probability of at least $2$ balls in random bin when distributing $K$ identical balls to $N$ distinct bins?

Question

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.

Answer 1

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 $P_{2+}=1-P_0-P_1$.

Answer 2

Total number of ways by which $k$ identical balls can be put into $N$ boxes is ${N-1+k \choose k}$, by partitioning argument.

Number of ways a particular bin will contain 0 balls = $N-2+k \choose k$, by leaving the bin out (event $A_1$)

Number of ways a particular bin will contain exactly 1 ball = $N-2+k-1 \choose k-1$, by keeping a ball into the bin and leaving the bin out (event $A_2$)

By classical definition of probability, with $P(A^c)=1-p(A)$ and the above two events being mutually exclusive, with $P(A)=P(A_1 \bigcup A_2) = P(A_1) + P(A_2)$, the required probability will be = 1 - $\frac{{N-2+k \choose k} + {N-2+k - 1 \choose k-1}}{N-1+k \choose k}$