Example 1 (National College Entrance Examination for Science, 2004) It is known that among 8 teams, 3 are weak teams. The 8 teams are divided into groups $A$ and $B$ by drawing lots, with 4 teams in each group.
(1) The probability that one of the groups $A$ or $B$ has exactly two weak teams;
(2) The probability that group $A$ has at least two weak teams.
Solution (1) The total number of ways to divide is $\frac{C_{8}^{4} C_{4}^{4}}{2!}$, among which there are $C_{5}^{1}$ ways for all weak teams to be in the same group; therefore, the probability that all three weak teams are in the same group is: $P_{0}=\frac{C_{5}^{1}}{\left(C_{8}^{4} \times C_{4}^{4}\right) / 2!}=\frac{1}{7}$, so the probability that one group has exactly two weak teams is: $P=1-P_{0}=\frac{6}{7}$.
(2) The probability that at least two weak teams are in one of the groups A or B is 1. Since the probability of having two weak teams in group A and group B is the same, so $P=\frac{1}{2}$.