# Compatible and incompatible events

Let's begin with the following experiment: we throw a dice of six faces and see what the result is. Let's consider the following events $$A = \{ 2, 3 \}$$, $$B = \{ 1 , 2 \}$$, $$C = \{ 5 \}$$.

We observe that if we extract $$2$$, then $$A$$ is satisfied as well as $$B$$. We say that the events are compatible, this means that they can happen simultaneously. On the contrary, events $$B$$ and $$C$$ are incompatible, since the two of them cannot happen simultaneously.

To see when two events are compatible or not, we can observe that $$A$$ and $$B$$ have a common element: $$2$$, therefore they will be compatible. On the contrary, $$A$$ and $$C$$ do not have any common element, and therefore they are incompatible.

We express this by saying that two events $$A$$ and $$B$$ are compatible if:

$$A \cap B = \emptyset$$$and on the contrary, they are incompatible if: $$A \cap B \neq \emptyset$$$

If we have three or more events, we say that they are incompatible two by two if any two events are incompatible (similarly, they are compatible two by two if any two events are compatible). In our case, $$A, B$$ and $$C$$ are not incompatible two by two, since, although $$A$$ and $$C$$, as well as $$B$$ and $$C$$ are incompatible, $$A$$ and $$B$$ are compatible.

## How is this related to complementary events?

In our experiment of throwing a dice, we have our event $$A = \{ 2, 3 \}$$, so let's analyze what happens with its complementary event.

In this case $$\overline{A}=\{1,4,5,6\}$$, since they are all the elementary events that do not satisfy $$A$$.

It turns out that $$A$$ and $$\overline{A}$$ are incompatible, since they cannot happen simultaneously. For any event $$A$$ we calculate its complementary doing $$\overline{A}=\Omega - A$$, then $$A \cap \overline{A}=\emptyset$$, that is to say, two complementary events will always be incompatible.

Let's suppose that $$D=$$"to extract an even number"$$=\{ 2, 4, 6 \}$$. Its complementary event is $$\overline{D}=$$"to extract an odd number"$$=\{ 1, 3, 5 \}$$. Then, $$D\cup \overline{D} =$$"to extract an even or odd number"$$= \{ 1, 2, 3, 4, 5, 6 \} = \Omega$$, that is to say, it is a sure event.

By the definition of a complementary event, this will always happen, since one of the two is always satisifed, and as they are incompatible, either one or the other is satisfied.