Can Intersecting Events in Probability be Independent?

AI Thread Summary
The discussion centers on the independence of events A and B in probability theory, particularly when considering their intersections and unions. It is established that A and A∪B are independent if P(A∪B) equals 1, which is a special case of independent events. The conversation also highlights that having a non-empty intersection between A and B does not automatically imply dependence; proving independence requires additional numerical probability values. The participants emphasize the need to consider exceptional cases, such as when B is empty or when A and B do not overlap. Overall, the analysis points to the complexity of determining independence in intersecting events.
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##P(A|A∩B) = \frac{P(A∩(A∩B))}{P(A∩B)} = \frac{P(A∩B)}{P(A∩B)} = 1##
So given the the event "A and B" as the sample space, the probability of A occurring is 1.

##P(A|A∪B) = \frac{P(A∩(A∪B))}{P(A∪B)} = \frac{P(A)}{P(A∪B)}##
Those two events are independent if and only if the probability of "A or B" occurring is 1, in which case the conditional probability of A equals the probability of A.

Is my reasoning correct?
 
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you mean that A and A∪B are independent if P(A∪B)=1 ? yes, this is a special case of independent events. There is one other special case which can mean that A and A∪B are independent. You are trying to make the last equation go from
P(A|A∪B) = \frac{P(A)}{P(A∪B)}
to become:
P(A|A∪B) = P(A)
right? and you did this by setting P(A∪B)=1. But there is another way also.
 
MohammedRady97 said:
##P(A|A∩B) = \frac{P(A∩(A∩B))}{P(A∩B)} = \frac{P(A∩B)}{P(A∩B)} = 1##
So given the the event "A and B" as the sample space, the probability of A occurring is 1.
We'd have to consider the case B \cap A = \emptyset

##P(A|A∪B) = \frac{P(A∩(A∪B))}{P(A∪B)} = \frac{P(A)}{P(A∪B)}##
Those two events are independent
Which two events? A and A \cup B ?

if and only if the probability of "A or B" occurring is 1

Suppose B = \emptyset.

I think your results are interesting and worth perfecting by taking care of the exceptional cases.
 
Stephen Tashi said:
We'd have to consider the case B \cap A = \emptysetWhich two events? A and A \cup B ?
Suppose B = \emptyset.

I think your results are interesting and worth perfecting by taking care of the exceptional cases.

What about ##P(A|B)##? My gut tells me that since A and B intersect, A and B must be dependent. This is wrong, of course, but how do I prove the dependence (or independence) of two intersecting events?
 
MohammedRady97 said:
What about ##P(A|B)##?

I don't know what your are asking.

My gut tells me that since A and B intersect, A and B must be dependent. This is wrong, of course, but how do I prove the dependence (or independence) of two intersecting events?

You can't prove the dependence or independence of two events that have a non-empty intersection just from knowing the intersection is non-empty. It's a quantitative question that depends on the numerical values of the probabilities involved. If you want to prove a result, you'll have to add more assumptions - something beyond just knowing that the intersection is non-empty.
 
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