Hi all, how would the solution to this problem change if probability A = 0.6? Thereby, the mutually exclusive case is no longer applicable? Thanks!
Well you can start by noting that the following holds from our axioms:
\mathbb{P}(A \cup B) = \mathbb{P} (A) + \mathbb{P} (B) - \mathbb{P} (A \cap B) \leq 1
This can be equivalently written as follows:
\mathbb{P} (A \cap B) \geq \mathbb{P} (A) + \mathbb{P} (B) -1
You also know that \mathbb{P} (...) \geq 0, and so it’s fair to surmise that the lower bound would be 0 if the individual probabilities add to less than (or equal to) 1. Otherwise, the minimum would be the “amount” by which the sum of the individual probabilities exceeds 1.
Formally put, it looks like this:
\mathbb{P} (A \cap B) \geq \operatorname{max}\{\mathbb{P} (A) + \mathbb{P} (B) - 1, 0\}
For the maximum (upper bound), you may know that \mathbb{P} (A \cap B) \leq \mathbb{P}(A) and \mathbb{P} (A \cap B) \leq \mathbb{P}(B). This is because A \cap B is a region “inside” both A and B simultaneously, as you may be able to deduce from a 2-set Venn diagram. Basic inference from here tells you that if a region is present in both A and B, its maximum size is limited by the smaller of the two sets.
For example, if a sphere is fit inside two cubes, A and B (with possibly different dimensions), it makes sense that the diameter of the largest possible sphere cannot exceed the side length of the smaller cube.
As such, you have:
\mathbb{P} (A \cap B) \leq \operatorname{min} \{\mathbb{P}(A), \mathbb{P}(B)\}
Putting everything together leads you to:
\boxed{\operatorname{max}\{\mathbb{P} (A) + \mathbb{P} (B) - 1, 0\}\leq \mathbb{P} (A \cap B) \leq \operatorname{min} \{\mathbb{P}(A), \mathbb{P}(B)\}}
This is now a result you can use for any questions of this sort. You can also imagine using this logic to bound the union as well.