• Currently going through a video on Counting Minimum Cuts by Tim Roughgarden.
• $$(A_{i},B_{i}) = \big((A_{1},B_{1}), ..., (A_{t},B_{t})\big) \forall i \in \Bbb{R}$$
• $$P\big((A_{i},B_{i})\big) \geq \frac{1}{\begin{pmatrix} n \\ 2 \end{pmatrix}} = p$$, which I interpret as the lower bound on the probability of having at least one minimal cut.
• In the problem set that follows two answers A and B are highlighted as being correct. I understand why A is correct; but am puzzled why B is also marked as correct.
• A: For every graph $$G$$ with $$n$$ nodes and every min cut $$(A,B)$$ (I am assuming same thing as $$(A_{i},B_{i})$$) $$P\big((A,B)\big) \geq p$$.
• B There exists a graph $$G$$ with $$n$$ nodes and a min cut $$(A,B)$$ (again assuming same thing as $$(A_{i},B_{i})$$) of $$G$$ such that $$P\big((A,B)\big) \leq p$$.

I don't understand what you mean by "$$(A_i,B_i) = ((A_1,B_1),\ldots,(A_t,B_t))$$", an obviously false statement. Perhaps you meant $$(A_i,B_i) \in \{(A_1,B_1),\ldots,(A_t,B_t)\}$$?
I don't quite understand your interpretation of the statement $$P((A_i,B_i)) \geq p := 1/\binom{n}{2}$$. Here is the correct interpretation:
For any minimum cut $$C$$, the probability that Karger's algorithm outputs $$C$$ is at least $$p := 1/\binom{n}{2}$$.
For B, you need to give an example of a graph which satisfies $$P(C) \leq 1/\binom{n}{2}$$ for all cuts $$C$$. One such example is a cycle, an example you were probably shown in class.