First, a note: while the question is tagged “probability”, the quoted text talks about the “number of options”, which is equally applicable to counting and combinatorial problems. In what follows, I’ll talk about probability, but this too is equally applicable to counting.
I don’t much like the way this textbook presents this information. Maybe it’s just the result of taking one paragraph in isolation—maybe it’s clearer in context—but it seems to give two false impressions:
- Adding probabilities is something special that happens only with mutually exclusive events. Multiplying probabilities is something special that only happens with independent events.
- These two calculations (adding mutually exclusive probabilities, and multiplying independent probabilities) are used in the same kinds of situations.
To the second point first. Adding probabilities is something we do to find the union, when we have two things and we want the probability of this or that (or both) happening. Multiplying probabilities, in contrast, is what we do to find the intersection, when we want to know the probability of this and that both happening.
Moreover, adding probabilities is (part of) what we do for any union/“or” problem. The complication is that adding up the two probabilities double-counts the chance of them both happening, so we need to subtract that from the sum:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
What’s special about mutually exclusive events is that they can’t both happen. That last term becomes zero. So, our calculation simplifies to just an addition:
$$P(A \cup B) = P(A) + P(B) \iff P(A \cap B) = 0$$
Likewise, multiplying probabilities is something we can do for any intersection/“and” problem. The complication this time is that the second† probability needs to be made conditional on the first:
$$P(A \cap B) = P(A) P(B | A)$$
What’s special about independent events is that their probabilities don’t change if they’re made conditional on each other. So, our calculation can use the regular probability in place of the conditional one (since they’re equal):
$$P(A \cap B) = P(A) P(B) \iff P(B | A) = P(B)$$
† Or we can make the first conditional on the second. Conditional probability (and independence) doesn’t care which order the two events happened in. Or if they were simultaneous, for that matter.