Question about the definition of independent events
If you have two events A and B that are independent, then it is said that $p(A)p(B)=p(A\cap B)$, and illustrated in a venn diagram as two areas that do not overlap.
The opposite goes for dependent events A and B, where if A is dependent on B, then $p(A|B)p(B)=p(A\cap B)$ and in the venn diagram, the areas representing these events do overlap.
My questions are:
- Are there any instances where event A does overlap B but is not dependent on B?
- Are there instances where A does not overlap with B, but is dependent on B?
My guess here is that there are no such cases, where if one instance is satisfied then it by definition calls for the other. For instance, regarding question 1, if A overlaps B, then a dependence between the two exists. For question 2, if A does not overlap B, then it cannot be dependent on it.
Are these ideas correct, or are there any situations where this might not work?
Solution 1:
I think you are confusing independent events with disjoint events.
Independent events are defined by $P(A\cap B) = P(A)P(B)$. $A$ and $B$ will overlap in the Venn diagram, except in the case of $P(A) = 0$ or $P(B) = 0$.
Disjoint events are given by $P(A\cap B) = 0$, meaning it is impossible for the events to occur together. Here the Venn diagram areas have no overlap.
Solution 2:
"Are there any instances where event $A$ does overlap $B$ but is not dependent on $B$?"
Yes, consider choosing a ball out of a set of two balls, one red and one blue, with replacement, twice. Let the event that the first ball chosen is red be $A$, that the second ball is red be $B$. (We can define the events more rigorously using a probability space, but I think you want the intuition).
Then $A$ is independent of $B$ since $P(A) = \frac{1}{2} = P(A|B)$. But the events do "overlap", i.e. the intersection is non-empty ($A\cap B \neq \emptyset$). Also, $P(A \cap B) = \frac{1}{4} > 0$.
Any two events that are mutually exclusive and which both have non-zero probability of occuring are dependent on each other e.g. choosing a blue red ball (let this event be $C$) and a red ball on the first choice. $P(A|C) = \frac{P(A\cap C)}{P(C)} = 0 \neq \frac{1}{2} = P(A)$