Balanced allocation-Hash table- overflow probability

My question is related to this:
Hash-Table in Practice

In [1] page 7, it is said that if we throw $n$ balls into $k$ bins, then each bin contains at most $\frac{n}{k}+O(\sqrt[2]{(\frac{n}{k})\log k}+\log k)$ elements with a high probability.

Question 1: Why is $O()$ used in the above estimation?

Question 2: Does it mean the probability that a bin contains more than the above value is negligible?

[1]. http://www.pinkas.net/PAPERS/FNP04.pdf

A more formal statement of the claim is as follows. There is a constant $C > 0$ and for each $k$, a function $\epsilon(n)$ satisfying $\lim_{n\to\infty} \epsilon(n) = 0$, such that if you throw $n$ balls into $k$ bins, then with probability at least $1-\epsilon(n)$ the contents of each bin is at most $ \frac{n}{k} + C(\sqrt{(\frac{n}{k})\log k}+\log k)$.

StackExchange Computer Science Q#49027, answer score: 3