(Axiomatic) Definition of probability and its properties

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)

We will use the rule of Laplace in the first cases, and then we will calculate using the properties that we know.

$P(A)=\dfrac{4}{7}$, since there are four favorable cases out of the seven, and they all are equiprobables.

$P(C)=\dfrac{2}{7}$, as before, applying the rule of Laplace.

$P(D)=0$, since it is the impossible event.

To calculate $P(\overline{C})$, as we already have $P(C)$, we do it accordingly to $P(\overline{C})=1-P(C)=1-\dfrac{2}{7}=\dfrac{5}{7}$.

With the same formula, $P(\overline{D})= 1 - P(D) = 1 - 0 = 1$. Also we might calculate it by reasoning that the opposite of the impossible event is the sure event, which has probability 1 due to axiom 2.

To calculate $P(A\cup\overline{C})$, we must calculate the event $A\cup\overline{C}=\{4,5,6,7\}\cup\{1,2,4,5,7\}=\{1,2,4,5,6,7\}$. For the rule of Laplace $P(A\cup\overline{C})=\dfrac{6}{7}$, since there are $6$ favorable ones out of the $7$ elementary events.

Finally, we can calculate $P(A\cap\overline{C})$ using the formula $P(A\cup\overline{C})=P(A)+P(\overline{C})-P(A\cap\overline{C})$.

By substituting for the values that we know $\dfrac{6}{7}=\dfrac{4}{7}+\dfrac{5}{7}-P(A\cap\overline{C})$. Therefore $$P(A\cap\overline{C})=\dfrac{4}{7}+\dfrac{5}{7}-\dfrac{6}{7}=\dfrac{3}{7}$$