For question $1$, you can draw a Venn diagram for independent events, however you will not be able to tell if the events are independent by looking at the Venn diagram, as it will just look like a standard Venn diagram. (standard meaning how a Venn diagram would look for $2$ events $A,B$, that are not mutually exclusive)
For question $2$, if $A$, $B$ are $2$ independent events then $P(A\cup B)=P(A)+P(B)-P(A)P(B)$
For question $3$, a mutually exclusive events are necessarily dependent events (assuming the probability of both events is greater than $0$).
Proof:
Recall the following:
Let two events, $X$ and $Y$ be independent. Then it follows that $P(X \cap Y)=P(X)P(Y).$ These events are mutually exclusive if $P(X \cap Y)=0.$ Lastly remember that $P(X)>0$ and that also $P(Y)>0$, as we are discussing probability and it ranges from $0-1$.
So, since we know that $P(X)>0,P(Y)>0$, then it follows that $P(X)P(Y)>0.$ If these events were independent then $P(X)P(Y)=P(X\cap Y)>0$, but this would mean that they aren't mutually exclusive.
Therefore, the events can not be independent and mutually exclusive simultaneously if both their probabilities are more than $0$.
$Q.E.D.$
No, actually if two events are mutually exclusive it is only in a special case that they can be independent. The definitions state that two events $A,B$ are
- mutually exclusive iff: $P(A\cap B)=0$, i.e. $A\cap B =\emptyset$,
- independent iff: $P(A\cap B)=P(A)P(B)$.
Assume know that $A,B$ are both mutually exclusive and independent. Then combining the above yields $$P(A)P(B)=0$$ which implies that at least one of the $P(A)$ and $P(B)$ has to be zero.
Informal examples: The events $A=$ I will roll a $6$ and $B=$ I will roll a $2$ are mutually exclusive. They cannot occur simultaneously. Similarly the events "rain" and "no rain" etc.
On the other hand the events "rain" and "I will roll a $6$" can occur simultaneously (i.e. the occurrence of the one does not exclude the occurrence of the other) but they are obviously independent, i.e. if you are given the information that it is "raining" or not you cannot in anyway infer whether I will roll a $6$ or not and vice versa.
Best Answer
Two mutually exclusive events are neither necessarily independent nor dependent. For example, the events that a coin will come up head or that it will come up tail are exclusive, but not independent, because $P(H \text{ and } T) = 0$, whereas $P(H)P(T) = \frac{1}{4}$. On the other hand, any event $A$ is independent from the empty event $\emptyset$, because $P(\emptyset) = 0$, so $P(A \cap \emptyset) = P(\emptyset) = 0 = P(A)P(\emptyset)$, and $A$ is of course mutually exclusive from the empty event. For an example of two independent events that are not mutually exklusive, suppose you throw a coin two times. The event that the first coin comes up head is independent from the event that the second comes up head, because $$P(X_1 = H \text{ and } X_2 = H) = P(X_1 = H)P(X_2 = H),$$ but the events are not mutually exclusive, because of course $(X_1 = H \text{ and } X_2 = H)$ lies in both events.