Probability of throwing balls into bins. Suppose we have $M$ balls placed randomly into $N$ boxes, wherein each ball has an equal chance of landing in each bin. How would we go about finding the expected number of balls in the first box? I assumed we could use a binomial distribution, wherein we would regard a ball being thrown into the first box as a success, and the probability of success is $1/N$. Hence, from this, the expected value of balls in the first box is $M/N$. Is this a correct approach?

One way to verify the result:

Let $E_N(M)$ be the expected number of balls in the first bin given $N$ bins and $M$ balls, where $E_N(0) = 0$. When allocating a single ball, there is a $\frac{1}{N}$ chance of it going to any particular fixed bin. For a given trial, there are two outcomes: Either a ball goes to the first bin and we add one to the desired total, or it goes somewhere else and we add nothing. Either way, we start over with $M-1$ balls.

$$E_N(M) = \frac{1}{N}(1 + E_N(M-1)) + \frac{N-1}{N}(E_N(M-1)) = \frac{1}{N}+E_N(M-1)$$

In other words:

$$E_N(M) = \sum_{k=1}^{M} \frac{1}{N} = \frac{M}{N}$$