To determine the probability that $E$ occurs before $F$, we can ignore
all the (independent) trials on which neither $E$ nor $F$ occurred,
that is, $(E\cup F)^c$ occurred, since we are going to repeat the
experiment until one of $E$ and $F$ does occur. So, look at the
trial of the experiment on which one of $E$ and $F$ has occurred
for the very first time.
We are given that on this trial, the event $E \cup F$ has occurred.
But, we don't yet know which of the two has occurred. So, given the
knowledge that $E \cup F$ has occurred, what is the conditional
probability that it was $E$ that occurred (and so $E$ occurred before $F$
since this is the first time we have seen either $E$ or $F$)?
$$P(E \mid (E \cup F)) = \frac{P(E(E \cup F))}{P(E \cup F)}
= \frac{P(E \cup EF)}{P(E) + P(F) - P(EF)}
= \frac{P(E)}{P(E)+P(F)}$$
since $P(EF) = P(\emptyset) = 0$.
Alternatively, let $G = (E\cup F)^c = E^c \cap F^c$ be the event that neither
$E$ nor $F$ occurs on a trial of the experiment. Note that
$P(G) = 1 - P(E) - P(F)$. Then, the event $E$ occurs
before $F$ if and only if one of the following compound events occurs:
$$
E, (G, E), (G, G, E), \ldots, (\underbrace{G, G, \ldots, G,}_{n-1} E), \ldots
$$
where $(\underbrace{G, G, \ldots, G,}_{n-1} E)$ means $n-1$ trials on which $G$
occurred and then $E$ occurred on the $n$-th trial. The desired probability
is thus
$$P(E ~\text{before}~ F) = P(E) + P(G)P(E) + [P(G)]^2P(E) + \cdots
= \frac{P(E)}{1 - P(G)} = \frac{P(E)}{P(E)+P(F)}.$$
"Mutually exclusive" and "independent" do not mean the same thing: they are different.
"Mutually exclusive events are those that are not dependent upon one another, correct?"
NO:
Two events are mutually exclusive if they cannot both occur. If we flip a coin, we get either a head, or a tail. We cannot get both. That is, the events are mutually exclusive.
Independent events are events where knowing the outcome of one doesn't change the probability of the other. Knowing that it's a sunny day doesn't tell me anything about the outcome of rolling a die. Those "events" are independent of one another.
When events are mutually exclusive, their probabilities add up to the probability that one event (or the other) occurs. In this case, if the $A$ and $B$ were mutually exclusive events, then you are correct, we would need for $P(A) + P(B) = 80$. But what we have, as you point out, is that $\,P(A) + P(B) = 70\neq 80.\;$ So you're right that $A$ and $B$ are not mutually exclusive, and for the right reason - because $P(A) + P(B) \neq P(A\cup B)$ - though you want to be clear about the terminology you use.
Best Answer
What about vicious cycles? For example, suppose you have three binary random variables $A$, $B$, and $C$ that each take values $0$ or $1$. In this example, $A$ and $B$ are mutually exclusive if whenever $A = 0$, $B = 1$, and similarly for $C$.
$A$, $B$, and $C$ cannot all be pairwise mutually exclusive.
However, they can be "globally" mutually exclusive in the sense that, for example, if $A = 0$, then $B = 1$ and $C = 1$, and so on for the other variables.
I think that this counterexample proves that pairwise mutual exclusivity is not equivalent to mutual exclusivity.