Why 1. multiply the number of independent options? 2. add the number of exclusive options?
Ironically, this textbook highlights understanding over memorization, but it doesn't expatiate the two WHY's in the question title!
When faced with a series of independent choices, one after the other, we multiply the number of options at each step. When faced with exclusive options (meaning we can't choose more than one), we add the number of options.
Make sure you see the difference. Don't memorize it — understand it.
David Patrick, BS Math & Computer Science, MS Math (Carnegie Mellon), PhD Math (MIT). Introduction to Counting & Probability (2005), p 29.
This multiplication is commonly pictured as a Tree Diagram, but Tree Diagrams don't expound it.
3 answers
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Look at the sub-picture with the beef burger, and all the nodes which extend from it. There are four options.
Now look at the sub-picture with the chicken burger. Same number of options.
In fact, for each one of those sub-pictures, there are four options, one for each of the different choices of burger. Well that is just exactly what multiplication counts: adding the same quantity a specified number of times. Since each sub-picture is associated with 4 options and there are 3 sub-pictures, you can count the number of options as 3*4.
Now, if you didn't directly look at the four options, you could have counted the intermediate stage where you fries or salad. Then you would have 3 sub-pictures corresponding to choice of burger, and 2options extending from that, a total of 6 options. Then you could continue the logic, seeing that each burger-side choice then is associated with two options for a drink. 6 sub-pictures each extending to 2 options means 6*2 now counts the number of options.
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You might find it more helpful to instead consider only two choices at a time and draw a grid:
| beef | chicken | fish |
------+---------+---------+---------+
| beef | chicken | fish |
fries | + | + | + |
| fries | fries | fries |
------+---------+---------+---------+
| beef | chicken | fish |
salad | + | + | + |
| salad | salad | salad |
------+---------+---------+---------+
But the real takeaway here is that textbooks below post-grad level (and maybe even at post-grad level) are intended for use with a teacher. Online Q&A is not suited for methodically teaching through a textbook. You might do better with something like Khan Academy, Udacity, Udemy, etc.
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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.
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