Advanced independence

You may recall that certain rules of probability only hold for independent events, and others only hold for dependent events. You may also recall that I said: if the situation doesn’t let you use a rule, you should change the way you are looking at the situation so that the rules apply again.

That all works… but there is also another option.

Here’s the deal: breaking the rules is bad because when you break the rules, you wind up overcounting or undercounting. But in some cases, the easiest thing to do is break the rules, and just keep track of the overcounting or undercounting, then correct for it at the end.

For example, let’s say that there’s a 1/3 chance or rain on Tuesday and a 1/3 chance or rain on Wednesday, and you want to know the chance that it will rain on either or both days.

We’ve already solved this problem by carefully reframing it in terms of independent and dependent events, of course. But our new way of solving it works well too. It goes like this:

There is a 1/3 chance of rain on Tuesday, and a 1/3 chance of rain on Wednesday. If we want the total chance that either or both happens, we can start by adding 1/3 + 1/3. That’s because we want the chance of rain on Tuesday or Wednesday, and or means add.

Now, of course, we are breaking a rule here… but here’s how we deal with it. We notice that the problem happens when it rains on both days. The issue is that we counted rain on Tuesday as part of the first 1/3, and we counted rain on Wednesday as part of the second 1/3 chance. If it rains on both days, then we’ve double-counted.

If that seems confusing, try thinking of it this way. We want it to rain on either day or both. When it rained on Tuesday, we got what we wanted. When it rained on Wednesday, we got what we wanted again. We got what we wanted twice. But we were only trying to count one win, not two.

So to get the right answer, we start with the 2/3 we calculated earlier, then we notice that the chance of rain on both days (1/9) got counted twice. So we have to subtract one of those 1/9th’s from our total. We get 2/3 - 1/9 = 6/9 - 1/9 = 5/9, which is the same answer we got earlier by being much more careful with the rules.

(You might also find it helpful to look at the first few paragraphs of the Wikipedia entry on the Inclusion-Exclusion Principle.)

See if you can apply the same idea:

How many whole numbers from $1$ through $46$ are divisible by either $3$ or $5$ or both?

A. 18
B. 21
C. 24
D. 25
E. 27