# Combining Non-Disjoint Probabilities

In my previous post on probability, I talked about how you need to be careful about covering cases. To understand what I mean by that, it’s good to see some examples.

And we can do that while also introducing an important concept which I haven’t discussed yet. I’ve frequently talked about independence, but equally important is the idea of disjointness.

Two events are independent when they have no ability to influence one another. So two coin flips are independent. Two events are disjoint when they can’t possibly occur together. Flipping a coin, the event “rolled a head” and the event “rolled a tail” are disjoint: if you rolled a head, you can’t roll a tail, and vice versa.

So let’s think about something abstract for a moment. Let’s suppose that we’ve got two events, A and B. We know that the probability of A is 1/3 and the probability of B is also 1/3. What’s the probability of A or B?

Naively, we could say that it’s P(A) + P(B). But that’s not necessarily true. It depends on whether or not the two events are disjoint.

Suppose that it turns out that the probability space we’re working in is rolling a six sided die. There are three basic scenarios that we could have:

1. Scenario 1: A is the event “rolled 1 or 2”, and B is “rolled 3 or 4”. That is, A and B are disjoint.
2. Scenario 2: A is the event “rolled 1 or 2”, and B is “rolled 2 or 3”. A and B are different, but they overlap.
3. Scenario 3: A is the event “rolled 1 or 2”, and B is the event “rolled 1 or 2”. A and B are really just different names for the same event.

In scenario one, we’ve got disjoint events. So P(A or B) is P(A) + P(B). One way of checking that that makes sense is to look at how the probability of events work out. P(A) is 1/3. P(B) is 1/3. The probability of neither A nor B – that is, the probability of rolling either 5 or 6 – is 1/3. The sum is 1, as it should be.

But suppose that we looked at scenario 2. If we made a mistake and added them as if they were disjoint, how would things add up? P(A) is 1/3. P(B) is 1/3. P(neither A nor B) = P(4 or 5 or 6) = 1/2. The total of these three probabilities is 1/3 + 1/3 + 1/2 = 7/6. So just from that addition, we can see that there’s a problem, and we did something wrong.

If we know that A and B overlap, then we need to do something a bit more complicated to combine probabilities. The general equation is: $P(A cup B) = P(A) + P(B) - P(A cap B)$

Using that equation, we’d get the right result. P(A) = 1/3; P(B) =
1/3; P(A and B) = 1/6. So the probability of A or B is 1/3 + 1/3 – 1/6 = 1/2. And P(neither A nor B) = P(4 or 5 or 6) = 1/2. The total is 1, as it should be.

From here, we’ll finally start moving in to some more interesting stuff. Next post, I’ll look at how to use our probability axioms to analyze the probability of winning a game of craps. That will take us through a bunch of applications of the basic rules, as well as an interesting example of working through a limit case.

And then it’s on to combinatorics, which is the main tool that we’ll use for figuring out how many cases there are, and what they are, which as we’ve seen is an essential skill for probability.

## 2 thoughts on “Combining Non-Disjoint Probabilities”

1. Rubén Barroso

It would have helped to explain the last equation with a Venn diagram, to show that we were counting twice the probabilities of the shared events between A and B.

2. Sam

Thank you for your interesting post. It would be interesting to extend your analysis to 3 events, i.e. A, B, and C and further generalize it to n events where n is an arbitrary integer.