Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Post History

#1: Initial revision by user avatar Tim Pederick‭ · 2022-02-27T05:05:31Z (over 2 years ago)
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:

1. Adding probabilities is something special that happens only with mutually exclusive events. Multiplying probabilities is something special that only happens with independent events.
2. 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<sup>†</sup> 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.