How Independence and Mutually Exclusive connected?

What can we infer from knowing that two (or more) events are independent or mutually exclusive?

Independence of events means that $$P(A \cap B) = P(A)P(B)$$

and mutually disjoint events are such that $A \cap B =\emptyset$. Therefore $$P(A \cap B)= 0$$

EDIT: Note that if $P(A)>0$ and $P(B)>0$ then they can't be independent and mutually exclusive

• That's not true in the continuos case. – Michael Hoppe Jul 9 '15 at 13:48

[assuming I understand what you mean]: in the first case, $P(A \cap B) = P(A)P(B)$. In the second, $P(A \cap B) = 0$

• So mutually exclusive $\rightarrow$ dependence? – gbox Jul 9 '15 at 13:41
• If the events are independent, they are not mutually exclusive – Alex Jul 9 '15 at 13:43
• You mean $A\cap B=\emptyset$, not $P(A\cap B)=\emptyset$. – user940 Jul 9 '15 at 13:45
• yes, apologies. Intersect of sets is an empty set; probability is 0. – Alex Jul 9 '15 at 13:50

If two events $A$ and $B$ are mutually exclusive then $$P(A \cap B)= 0$$ and so $$P(A\cup B)=P(A)+P(B)$$

If two events $A$ and $B$ are independent then $$P(A \cap B)= P(A) \times P(B)$$

You might want to supplement these correct answers with some intuition. When events are independent, knowledge of one tells you nothing about the other. When they are mutually exclusive, knowing that one of them happened means the other did not. So when events are mutually exclusive, knowing about one of them tells you a lot about the other - they are not independent.

If two events are independent then in presence of one of them does not guarenteed any information about the other one, meanwhile if events are mutually exclusive, then you might conclude something in presence of one of them.

Actually, two events $A$ and $B$ are independent if $${\rm Pr}(A|B) = {\rm Pr}(A) \quad\text{ and }\quad {\rm Pr}(B|A) = {\rm Pr}(B),$$ which equivalent to $${\rm Pr}(A\cap B) = {\rm Pr}(A)\cdot{\rm Pr}(B).$$

(In words) it means that it doesn't metter do we know any information about $B$, the probability of $A$ is the same in both cases. However, if the events are mutually disjoint, situation is different. For example, if $A$ and $B$ are mutually disjoint, and let's assume that the complement of one is the other: $$A\cap B =\emptyset\quad\text{ and }\quad A\cup B=\Omega,$$ then $${\rm Pr}(A) + {\rm Pr}(B) = {\rm Pr}(\Omega) = 1,$$ and as one can notice that, from knowing the ${\rm Pr}(A)$ we will find out ${\rm Pr}(B)$ and conversely. So, in this case events $A$ and $B$ are depending each other.