We have an urn with seven balls numbered from 1 to 7. Our experiment consists in extracting a ball and observing what number it has.

a) Determine the sample space, and the events $$A =$$ "to extract a number equal to or greater than $$4$$", $$B =$$ "to extract an even number", $$C =$$ "to extract a multiple number of $$3$$", $$D =$$ "to extract a number greater than $$8$$" that is to say , $$A, B, C$$ and $$D$$ are expressed as sets of possible results.

b) Are the sets $$A$$, $$B$$, and $$\overline{C}$$ compatible two by two? Explain why.

### Development:

a)

The sample space is the set of all possible results. In our case, we have seven numbered balls, thus $$\Omega=\{1,2,3,4,5,6,7\}$$, that is to say, to extract ball $$1$$, to extract ball $$2$$, etc.

We can only extract balls between $$1$$ and $$7$$. Therefore, $$A= \{4, 5, 6, 7\}$$, which are the balls equal to or greater than $$4$$.

$$B= \{2, 4, 6\}$$, since it corresponds to the even numbers that exist between $$1$$ and $$7$$.

$$C= \{3, 6\}$$, the multiples of $$3$$ between $$1$$ and $$7$$.

$$D=\emptyset$$, that is to say, $$D$$ is an impossible event, since we only have numbers from $$1$$ to $$7$$, and therefore, we can never extract a ball with a number greater than $$8$$.

b)

First, we must calculate $$\overline{C}$$. The opposite set to $$C$$ is the set of all possible results that are not in $$C$$. In our case $$\overline{C}=\Omega-C=\{1,2,4,5,7\}$$.

Now, let's consider all the possible couplings between $$A, B$$ and $$\overline{C}$$. There are the following: $$A$$ and $$B$$, $$A$$ and $$\overline{C}$$, $$B$$ and $$\overline{C}$$. We have seen that $$A=\{4, 5, 6, 7\}$$, $$B=\{2, 4, 6\}$$, $$\overline{C}=\Omega-C=\{1,2,4,5,7\}$$.

$$A$$ and $$B$$ are compatible, since they can happen simultaneously. The common elements between $$A$$ and $$B$$ are $$\{ 2 , 4 \}$$, or in other words $$A\cap B=\{2,4\}$$. Namely, if $$2$$ comes up, or also if $$4$$ turns up, $$A$$ is satisfied as well as $$B$$.

$$A$$ and $$\overline{C}$$ are also compatible. If $$4, 5$$, or $$7$$ are turned up, both $$A$$ and $$\overline{C}$$ are satisfied.

$$B$$ and $$\overline{C}$$ are compatible. With $$2$$ or $$4$$, $$B$$ happens as much as $$\overline{C}$$.

Namely $$A, B$$ and $$\overline{C}$$ are two by two compatible. Note that, if we extract ball $$4$$, three events are satisfied simultaneously, or put aother way, three events are compatible: $$A\cap B \cap \overline{C}=\{4\}$$.

And so, we know that the three events will be two by two compatible. Nevertheless, this does not work the other way around: what might occur is that three events were two by two compatible, but we could not find any elementary event that would make the three of them be simultaneously satisfied.

### Solution:

a) $$\Omega=\{1,2,3,4,5,6,7\}$$, $$A= \{4, 5, 6, 7\}$$, $$B= \{2, 4, 6\}$$, $$C= \{3, 6\}$$, $$D=\emptyset$$.

b) They are two by two compatible, given that $$A$$ and $$B$$, $$A$$ and $$\overline{C}$$, $$B$$ and $$\overline{C}$$ are compatible.