$c=T/N$ is the (uniform) capacity of the bins; $W=T-B$ is the number of white balls.
It suffices to compute the complementary probability, which is that at least one of the $N$ bins contains
only, which is exactly $c$, white balls. The balls are randomly assigned, so one can assign first the white
balls and fill up the bins with the black balls (all to the same capacity) a posteriori. That means one needs only to consider the ways of partitioning $W$ balls into $N$ bins. I can only think of a recursive approach. After placing $w$ white balls, $0\le w\le W$, there are $f_0$ bins with no white balls, $f_1$ bins with one white ball etc. That state is written down as $0^{f_0}1^{f_1}\cdots$, the notation of weak partitions, where $\sum_{i\ge 0}f_i= N$, $\sum_{i\ge 0} if_i=w$. To each of these states a probability $p$ is assigned, and we are interested in the probability $\sum_{k\ge 1} p(0^. 1^. \cdots c^k)$ that one or more bins contain only white balls, once $w=W$. Initially, before a white ball is placed, $w=0$, the state is $0^N$, which means all bins are empty, and $p(0^N)=1$. Placement of an additional ball disperses a state $0^{f_0}1^{f_1}\cdots i^{f_i}(i+1)^{f_{i+1}}\cdots c^{f_c}$
to the states $0^{f_0}1^{f_1}\cdots i^{f_i-1}(i+1)^{f_{i+1}+1} \cdots c^{f_c}$; for every positive exponent $f_i\ge 1$, $i<c$ in the state with $w$ balls, the state with $w+1$ balls has one of the exponents transferred to the next higher index. One can visualize that with a Hasse diagram of a poset of the states, where the state $0^N$ is at level 0, and the states at level $w+1$ are all those that can be generated this way from the states at level $w$. A state of probablity $p$ contributes $p\times f_i/\sum_{j<c} f_j$ to the state at the next level, because there are $f_i$ bins that could be receiving the next ball out of a total of $\sum_{j<c} f_j$. A recursive program of that kind can also solve What is the probability that throwing $m$ balls at random in $n$ urns at least one urn contains $c$ elements? .
