When two events $A$ and $B$ have no result in common I and my friends just had a little discussion whether the events are independent or dependent if they have no outcome in common. I thought that they have to be independent. When two events are independent, then $P(A)=P(A\mid B)$. Is the information given in the question enough to establish this?
If you look at it as a Venn diagram, then if there is no overlap between A and B, then they are independent. But my friend objected and said that this depends on the sample space. 
So let's say we have two events: $P(A)=6/12$, $P(A\mid B)=2/4$, and $P(B)=4/12$, then obviously $P(A\mid B)$ equals $P(A)$. But for me, something smells fishy here. 
I know this is more a stochastic question that a statistic question, but maybe someone can help.
 A: It is true that if $P(A)=P(A\mid B)$ then the two events are intependent.
Now we know that: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Note the intersection in the numerator. If the two events didn't have "anything in common" then we would have $P(A\cap B)=0$ since the set $A\cap B$ would be empty. However independence works in the space of probabilities. Two events are called independent if $P(A\cap B)=P(A)P(B)$, ie the probability of having both events occuring is equal to the probability of the first one occuring times the probability of the second one. The fact that event $A$ occured doesn't tell us anything about event $B$.
If on the other hand $P(A\cap B)=0$ this simply says that the two events cannot happen at the same time: they are disjoint.
Example 1.
What is the probability of throwing a die and getting both 1 and 3 at the same time? Since the two events are disjoint, we have that this probability is zero.
Example 2.
What is the probability of throwing a die two times and getting 1 in the first time and 2 in the second? The two events are not disjoint; the fact that one happened doesn't exclude the other. However they are independent: the fact that one happened tells us nothing about the other. The probability of getting 1 in the first throw is 1/6, the probability of getting 2 in the second throw is 1/6. Then (abusing notation a bit): $P(2\mid 1)= \frac{P(1 \cap 2)}{P(1)} = \frac{ \frac{1}{36}}{\frac{1}{6}}=\frac{1}{6}=P(2)$.
A: You are confusing dependence with mutual-exclusivity.  Two events $A$ and $B$ are independent if and only if $P(A \cap B) = P(A)P(B)$.  They are disjoint if and only if $A \cap B = \emptyset$.
Consider a normally-distributed random variable $X$ with mean 0 and variance 1.
Here is a pair of events for each of the four categories:


*

*the events $0 < X < 1$ and $1 < X < 2$ are dependent and disjoint

*the events $X = 0$ and $1 < X < 2$ are independent and disjoint

*the events $0 < X < 1$ and $0 < X < 2$ are dependent and not disjoint, and

*the events $X < 0$ and $\lvert X \rvert < 3$ are independent and not disjoint.



If events $A$ and $B$ are independent and disjoint then
\begin{align}
0 &= P(A \cap B) \\\\
&= P(A)P(B) \\\\
&\implies P(A) = 0 \vee P(B) = 0.
\end{align}
A: Let $A$ and $B$ denote two events defined on a sample space $\Omega$.
The formal definition of independent events is as follows.
Definition: $A$ and $B$ are said to be (stochastically)
mutually independent events if
$$P(A\cap B) = P(A)P(B).$$
It is easily shown that any one of the four relations 
shown below implies the other three:
$$\begin{align*}
P(A\cap B) &= P(A)P(B)\\
P(A^c\cap B) &= P(A^c)P(B)\\
P(A\cap B^c) &= P(A)P(B^c)\\
P(A^c\cap B^c) &= P(A^c)P(B^c)
\end{align*}$$
and so if $A$ and $B$ are mutually independent events, then so
are $A^c$ and $B$ mutually independent events, as are
$A$ and $B^c$, and $A^c$ and $B^c$.
Now, if $P(B) > 0$ so that we can write $P(A \mid B)$ as $P(A\cap B)/P(B)$,
then $P(A\mid B)$ equals $P(A)$, and this is often taken as the 
colloquial meaning (or definition) of independence. $A$ and $B$ are
independent events if knowing that $B$ has occurred does not
change our estimate of 
the probability of $A$.  Put another way, the posterior probability
$P(A\mid B)$ is the same as the prior probability $P(A)$.
The asymmetry in the colloquial definition even leads
people to say $A$ is independent of $B$ (which
can make beginners wonder whether $B$ is independent of $A$
or not), but the
formal definition makes it clear that independence is
a mutual property: one cannot have $A$ independent of $B$
but $B$ dependent on $A$.

Turning to the OP's question, if $0 < P(A), P(B) < 1$,
then mutual independence and mutual exclusion are mutually
exclusive properties. If one property holds, the other cannot.
Of course, the most commonly encountered case is
that neither property holds. Said out loud and clear


*

*If $A$ and $B$ are mutually independent
events, then they cannot be mutually exclusive events. 

*If $A$ and $B$ are mutually exclusive
events, then they cannot be mutually independent events. 
In the first case, note that mutual independence 
implies that $P(A\cap B) = P(A)P(B) > 0$ and so the intersection
of $A$ and $B$ has positive probability.  In the second case,
$P(A\cap B) = 0$ cannot equal $P(A)P(B)$ since neither
$P(A)$ nor $P(B)$ is $0$ by assumption and so their product
is a positive number.
As a corollary, note that $A$ and $A$ cannot be a pair of
mutually independent events. and nor can $A$ and $A^c$
be mutually independent events.

Much of the discussion in the comments has centered on the
rare cases when $P(A)$ or $P(B)$ happen to equal $0$ or $1$.
First note that since
$$P(A \cap \Omega) = P(A) = P(A)P(\Omega)$$
and so $A$ and the certain event $\Omega$ are independent 
events for all choices of $A$.  Similarly, since
$$P(A \cap \emptyset) = P(\emptyset) = 0 = P(A)P(\emptyset),$$
$A$ and the impossible event $\emptyset$ are independent 
events for all choices of $A$.  More generally, if $B$
is an event of probability $0$ (not necessarily the impossible
event), then since $A\cap B$ is a subset of $B$ and hence also
has probability $0$, we can generalize to 
$P(A \cap B) = 0 = P(A)P(B)$ and so


*

*Any event of probability $0$ is independent of all events
(including itself and its complement). If $B$ is an event
of probability $0$, then $B$ and $B^c$ are independent
events that are mutually exclusive. 


If $B$ is an event of probability $0$, then $B^c$ is
an event of probability $1$. Since $B$ and $A$ are
independent events for all choices of $A$, so also
are  $B^c$ and $A$ independent events for all choices of $A$. 
Thus, we have


*

*Any event of probability $1$ is independent of all events
(including itself and its complement). If $A$ is an event
of probability $1$, then $A$ and $A^c$ are independent
events that are mutually exclusive. 


Note that, as @NeilG has pointed out in his answer,
if $A$ and $B$ are independent events that are mutually
exclusive, then at least one of $A$ and $B$ must be
an event of probability $0$.
We also have an anticorollary:  $A$ and $A$ are mutually independent
events if and only if $P(A)$ equals either $0$ or $1$.
$A$ and $A^c$ are mutually independent
events if and only if one of $P(A)$ and $P(A^c)$ equals $0$ 
(and the other equals $1$.)
A: No, events with no result in common are not independent if the events come from the same sample space. 
An example: Throw a single fair die. Let event A be 'throw is a 1', and event B be 'throw is a 2'. Then $P(A) = P(B) = 1/6$, but $P(A|B) = P(B|A) = 0$, as the die throw can't be both 1 and 2.
A: Independence can be thought of as "If event A occurs it tells you nothing about the probability of event B occurring".
However, if two events share nothing in common then clearly if I know that one event occurred then I know that the other event COULDN'T have occurred.  So I am getting information about the other event so there the two events can't be independent.
Note that this was an intuitive explanation and it's possible for B to be empty in which case A and B can be mutually exclusive and still be independent as explained in some of the other answers.
A: mutually exclusive events..  natural blonde hair, and black skin.    If I know someone has black skin, I know they will not have naturally blonde hair.  Therefore, the characteristic of hair color and skin color are dependent.. Knowing that someone has black skin tells me information about what their hair will be like.. Their hair will NOT be blonde.. These things are dependent.  Mutually exclusive implies dependence!
independent events... IQ and shoe size.. If you thought someone was a size 8 shoe, would it change your opinion about their IQ if I told you their shoe size was actually a 10?  No.. Shoe size and IQ (likely) have no relationship.  (Assume it's true.)  Shoe size and IQ are independent of one another.  
I hope that helps.
