N balls and M boxes, probability that there is at least 1 box contains at least 2 balls 
I have $N$ balls and $M$ boxes. Balls are thrown to the boxes at random.
  What is the probability that there is at least 1 box contains at least 2 balls?

Thank you very much
 A: We first make the probability model explicit. Imagine the balls are thrown one at a time, and that a ball is equally likely to land in any of the boxes. So our set of outcomes is the set of all ordered $N$-tuples $(a_1,a_2,\dots, a_N)$, where each $a_i$ is an integer between $1$ and $M$. We are assuming all outcomes are equally likely.  
Note that there are $M^N$ outcomes.
We find the probability $p$ of the complementary event that all boxes have at most one ball. Then our required probability is $1-p$.
So we count the number of outcomes in which the balls land in distinct boxes. Assume that $M\ge N$.  
The first ball has $M$ choices. For every such choice, the second ball has $M-1$ choices. For every choice we make for where the first two balls land, there are $M-2$ choices for the third, and so on. So the number of outcomes in which all the balls land in distinct boxes is $M(M-1)(M-2)\cdots(M-N+1)$. It follows that 
$$p=   \frac{M(M-1)(M-2)\cdots (M-N+1)}{M^N}.$$
 The answer to the posted question is then $1-p$.
Remark: This is a well-known problem, usually called the Birthday Problem.  There is a large literature.  Note that the two other answers currently given are different from the one above. That is because the implicit model they use is that all solutions $(x_1,x_2,\dots,x_M)$ of $x_1+x_2+\cdots+x_M=N$, where $x_i$ is the number of balls in Box $i$, are equally likely. That will not be the case if balls are "thrown" according to the model we described.
A: The complementary event $A$ is that all $M$ boxes contain $0$ or $1$ balls. There are $\binom{M}{N}$ possibilities for this (assuming this to be zero if $N>M$). Overall, we have $R=\binom{N+M-1}{M-1}$ possibilities to distribute $N$ balls on $M$ boxes, so $P(A)=\frac{\binom{M}{N}}{R}$ and thus the probability of your event is $1-P(A)$.
