For this, i tried solving it by drawing two venn diagrams that are overlapped. And in the middle (the both), would be 0.2 and since A is 0.8, the remainder of A would be 0.6. And since independent events sum to 1, i got b as 0.4.(0.2 of just B and 0.2 from the both a and B).
Can someone tell me why this doesn’t work?
Don’t know where you got this from, but like:
\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)
and this is not related to independent events alone.
So, now you have two unknowns: \mathbb{P}(A \cup B) and \mathbb{P}(B), so with your logic the question is not uniquely solvable.
Basically, independent events would imply (in your case):
\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)
I’m very confused. Do independent events not always add to 1 especially if they have a both? I remember solving questions where I knew the both and P(A) and I would draw venn diagrams to find what P(B) based on 1 - both - P(A). Can you guys help clarify this for me?
Don’t ping people unnecessarily.
What do you mean?
Sorry about that.
I have a couple questions:
In this case, the overlap isn’t equal to P(A) \times P(B). You can also have a 0.5, 0.2 and 0.3 case for P(A), P(A and B) and P(B) respectively in that case.
In the question, it says that A and B are independent and the probability of A and B BOTH occurring is 0.2 so since both is calculated as P(A) x P(B) for independent, shouldn’t the overlap be P(A) * P(B) = 0.2?
What is P(A) and P(B)?
It doesn’t give you the individual values for each in this question but it says BOTH P(A) and P(B) is 0.2 which is why Greg solved it using the P(A) x P(B)= 0.2 which I understand.
What I don’t understand is why the BOTH part is not the overlap in the venn diagram I drew?
Hey, @hiiie28, I think I understand why are you getting confused. This happened to me too. Basically-
The probability of P(AUB) not equaling 1 for independent events occurs when the union of the events A and B does not cover the entire sample space. This happens in cases where the events A and B are not complementary (they do not account for all possible outcomes together).
Example:
The probability P(A∪B) in case of independent will not equal 1 when:
In your solution, it could also be that there is a probability of neither but you just assumed that neither doesn’t exist so the right approach would be to use the formula P(A).P(B) to find respective probabilities in case of independent events.
I think it would be helpful if leaderboard could confirm if my interpretation is correct or not. Thank you!
The key idea is “do not cover all outcomes” indeed.
Thank you so much for your detailed response! I think I am starting to understand now 
I’m curious though: would there be a situation where the sum of probabilities is 2? because one event could have a sum up to 1.
The sum of probabilities P(A)+P(B) can exceed 1 because it only represents the combined likelihood of each individual event occurring, without considering any possible overlap or mutual exclusivity between the events but the P(AUB) can never exceed 1 because it subtracts the overlap so if question says that sum of P(A)+P(B) is 1 there is a high possibility that they are independent events with overlap because in that situation P(AUB) would never exceed 1.
Why should this relate to independent events?
Also it’s just 0 \leq \mathbb{P}(...) \leq 1
What I am trying to say for this specific case is that is P(A) + P(B) exceed 1 (should have mentioned this instead of P(A)+P(B) is 1) it is likely that they could be independent events because there exists overlap between them and after subtracting that either could equal to 1 or less than 1 but never greater than 1. So, I agree with you that it is always a range from 0-1 to account for other possibilities like mutually exclusive or complete overall but in some questions they do say that P(A) + P(B) is greater than 1 which implies that there exists some overlap so that is what I was referring to.
This is how I have interpreted why it is not necessary for P(AUB) to equal one for independent case. Would be happy to know if there is a different caveat to this as I am still in a learning stage but I am no way implying that this solely relates to independent events.
Why are we even talking about independent events?
\mathbb{P} (A \cap B) = \mathbb{P}(A) + \mathbb{P} (B) - \mathbb{P} (A \cap B) and since we have that 0 \leq \mathbb{P}(A \cup B) \leq 1 then nothing is stopping \mathbb{P}(A) + \mathbb{P} (B) \geq 1.
Because the person who asked the question, asked about the situation where the sum of probabilities is 2 in other words greater than 1 and this can only happen when:
Sure, nothing is stopping P(A) + P(B) to exceed 1 but is is always helpful to explain with specific examples that is why I explained it like that. It is not very intuitive for a lot of people like me so I prefer talking about specific examples
No, my only qualm is that bringing up independent events when talking about \mathbb{P}(A) + \mathbb{P}(B) \geq 1 is just irrelevant because they aren’t linked at all.
In particular,
This part is not relevant and just causes confusion for OP for no reason. I mean they’re already confused about this so this is just doing more harm than good.
Here’s a pictorial argument on why the “likelihood” of events being independent doesn’t hinge on \mathbb{P}(A) + \mathbb{P}(B) \geq 1.
To start off, all possible probabilities lie in the purple figure:
This is the relevant 3D-figure for \mathbb{P}(A) + \mathbb{P}(B) > 1:
Finally, this is the “chance” for two events being independent:
As you might be able to intuit, there’s very little chance for a randomly selected 3d point to lie inside the thin red slice.
In fact, what’s important here is that the thin “red thin slice” is just as rarely pickable whether you’re in the purple chunk or within the blue chunk.
Essentially, there is no reason to discuss independent events when considering \mathbb{P}(A) + \mathbb{P}(B) \geq 1. Consequently, I don’t think the examples are particularly meaningful in this context either.
This explanation is not very intuitive for me. I am not the best at math so I understand better with specific examples. I was simply trying to say that if there are cases where P(A) + P(B) ≥1. These events can be independent or overlapping events because there exists a common intersection which is not subtracted from the same. The probability of independent events can also be less than 1. They don’t always have to be greater than 1 but if the sum of two probabilities is greater than 1 then they could be independent or overlapping. This is the most intuitive explanation for me but if the other person is able to understand your explanation better that is great. What I still don’t understand from your explanation is that why can’t one of the possibilities be of independent events if its greater than 1. If you have time and can explain with specific examples that would be great but I understand if you don’t want to.
