I got this wrong thinking that if two events are independent, that they must add to 1.
But just wanted to get some clarification on this: Is it that A + B - both = 1? so A + B including the both can add to > 1?
This is not consequential as in the first claim doesn’t the second
Yeah sure.
\mathbb{P} (A \cup B) = \mathbb{P}(A) + \mathbb{P} (B) - \mathbb{P} (A \cap B)
Thanks. @Leaderboard can you double confirm that this is correct? That my logic is correct?
Including the both? Yes.
So let me get this right.
If they are for sure independent, the both is A x B.
And if two events have abit of overlap, it is not for sure independent? it COULD be independent?
And if mutually exclusive, there is no overlap at all.
Am I correct?
Yes.
This questions confuses me at the conceptual level. If P(A) = 1 and P(B) = 1 wouldn’t this mean that they have complete overlap and therefore not independent. I don’t see how two events that are independent can both have a 100% probability of occurring. Isn’t that just the same event and therefore not independent.
You just have to check independence from the definition, which is:
\mathbb{P} (A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)
If you can show that the product of the probabilities of those 2 events is equal to the intersection then you’re allowed to surmise that said two events are independent.
Not necessarily, cuz here’s an example where they aren’t the same events:
Let event A be “There was a point in my life where i was 2 years old” and event B be “There was a point in my life where I was 5 years old”. I’m hoping you can tie things together from here.
Even if they were the same events, they are completely deterministic and so independent with itself.
You have that P(A \cap A) = P(A)^2 and so P(A \cap A) = P(A) \cdot P(A) holds.