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 $1$.)
