Problems from (Axiomatic) Definition of probability and its properties

Development:

a)

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

$D=\mathrm{\varnothing }$, 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.

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

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

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

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

With the same formula, $P(\bar{D})=1-P(D)=1-0=1$. 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 $P(A\cup \bar{C})$, we must calculate the event $A\cup \bar{C}=\{4,5,6,7\}\cup \{1,2,4,5,7\}=\{1,2,4,5,6,7\}$. For the rule of Laplace $P(A\cup \bar{C})={\displaystyle \frac{6}{7}}$, since there are $6$ favorable ones out of the $7$ elementary events.

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

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

Solution:

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

b) $P(A)={\displaystyle \frac{4}{7}}$, $P(C)={\displaystyle \frac{2}{7}}$, $P(D)=0$, $P(\bar{C})={\displaystyle \frac{5}{7}}$, $P(\bar{D})=1$, $P(A\cup \bar{C})={\displaystyle \frac{6}{7}}$, $P(A\cap \bar{C})={\displaystyle \frac{3}{7}}$.

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