Balls-into-bins problem: Lower bound for the bins which exactly have one ball

Question

Here is my problem:

If $n$ balls are thrown into $m$ bins (uniformly at random and independently) and $n < m$, can we prove:

With high probability (w.h.p) there are $\Theta(n)$ bins which exactly have one ball in them.

If we can, how to prove it?

I tried to calculate the expectation of one-ball-bins and then use Chernoff bound to solve it. But it was wrong because the random variables are not independent.

Sorry for my bad expression because I'm not very good at English. :-)

Answer 1

Suppose $m$ and $n$ are large, and let $\lambda = n/m$, where $n$ is the number of balls and $m$ is the number of bins. Then according to Example III.10, page 177, in Analytic Combinatorics by Flajolet and Sedgewick (available online: Analytic Combinatorics pdf), in the limit as $n \to \infty$ with $n/m$ fixed, the average proportion of bins containing $s$ balls is $$e^{-\lambda} \frac{\lambda^s}{s!}$$ so the occupancy of a random bin in a random allocation satisfies a Poisson law in the limit. In the case $s=1$, the average proportion containing one ball reduces to $e^{-\lambda} \lambda$.