Compatible and incompatible events

Problems

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.

See development and solution

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.

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