When solving this question I only consider the cases for P(AnB) not P(A or B) as Greg applied in the quiz solution video. I am not sure why he used P(A or B) since the question is asking use to find the range of values when the events A and B both occur. It seems strange to consider a case where with events occur.

Following my logic, q = 0 and p = 0.5 which implies that p + q = 0.5.

I’d appreciate your input on this problem. Thanks!!!

Could you show your work? As a start, try to establish a range for the minimum and maximum values for the intersection.

cylverixx, my logic was as follows:

Based on my understanding that the question requests us to consider a scenario where events A and B both occur, then using P(AnB) I consider three cases.

Case 1: No overlap

P(AnB) = 0

Case 2: Complete Overlap

P(AnB) = 0.5

Case 3: Independent

P(AnB) = P(A) X P(B) = 0.7 X 0.5 = 0.35

This will imply that p = 0 and q = 0.5. Thus, p+q = 0.5

Case 1 can’t exist because no overlap would suggest:

\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B) = 1.2 , but this is invalid because you generally have \mathbb{P}(...) \leq 1.

To minimize the intersection, you want to maximize the union \mathbb{P}(A \cup B). When does this happen and how can you use inclusion-exclusion to then solve for the intersection? This is what you should have for case 1.

Your case 2 is good because you do obtain the maximal value of the intersection.

As for case 3, it’s pointless to consider a different case for when the events are independent because it’s not the limiting case. In general, when you’re looking for the minimum/maximum values for the intersection, you’d only consider case 1 and case 2.

cylverixx, thank for sharing your work. Unfortunately, I don’t agree with your logic. I don’t understand why you’d choose to use P(AuB) for a question that is asking for intersection (P(AnB)). Unless I missing something.

How do you plan on finding the minimum value of \mathbb{P} (A \cap B)? It cannot be 0 for the reason I mentioned above.

If you have this (inclusion-exclusion) to work with:

\mathbb{P} (A \cup B) = \mathbb{P} (A) + \mathbb{P} (B) - \mathbb{P} (A \cap B)

and we slightly rewrite this as:

\mathbb{P} (A \cap B) = \mathbb{P} (A) + \mathbb{P} (B) - \mathbb{P} (A \cup B)

It should now be evident that when we maximize \mathbb{P} (A \cup B), we are in turn minimizing \mathbb{P} (A \cap B) `(which is what we want)`

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