Math

$P[\text{at least 1 yellow bull in bag 1}]=\\ 1-P[\text{no yellow bulls in bag 1}] = \\ 1 - \dfrac{\sum \limits_{w=3}^{10}\dbinom{10}{w}\dbinom{8}{11-w}}{\dbinom{32}{11}}=\dfrac{206719}{206770}$

I'll post #2 when I figure it out

$P[\text{white bull in group of leftovers}] = \\ 1 - P[\text{all white bulls in bag 1 or bag 2}] = \\ \text{(think of bag 1 and bag 2 as one big bag here)} \\ 1 - \dfrac{\sum \limits_{y=4}^{12}~\dbinom{14}{y}\dbinom{8}{12-y}}{\dbinom{32}{22}}=\dfrac{2456369}{2481240}$

$P[\text{at least 1 black bull in each bag}] = \\ 1 - P[\text{no black bull in bag 1 or no black bull in bag 2 or both}] = \\ \begin{align*} 1 \\ &- P[\text{no black bull in bag 1}]\\ &-P[\text{no black bull in bag 2}]\\ &+P[\text{no black bull in either bag}] \end{align*}$

$\text{As the problem is symmetric w/respect to the bags}\\ P[\text{no black bull in bag 1}] = P[\text{no black bull in bag 2}]\\ P[\text{no black bull in bag 1}] =\\ \dfrac{\sum \limits_{w=0}^{10}~\dbinom{10}{w}\dbinom{14}{11-w}}{\dbinom{32}{11}} =\dfrac{2261}{116870}$

$P[\text{no black bull in either bag}] = \\ \dfrac{\sum \limits_{w=0}^2~\dbinom{10}{w}\dbinom{14}{2-w}}{\dbinom{32}{10}}=\dfrac{1}{233740}$

$P[\text{at least 1 black bull in both bags}] = \\ 1-2\dfrac{2261}{116870}+\dfrac{1}{233740} = \dfrac{224697}{233740}$