# Problems from (Axiomatic) Definition of probability and its properties

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) Calculate $$P(A), P(C), P(D), P(\overline{C}), P(\overline{D}), P(A\cup\overline{C}), P(A\cap\overline{C})$$

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### 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)

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 $$P(A\cap\overline{C})=\dfrac{4}{7}+\dfrac{5}{7}-\dfrac{6}{7}=\dfrac{3}{7}$$\$

### 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) $$P(A)=\dfrac{4}{7}$$, $$P(C)=\dfrac{2}{7}$$, $$P(D)=0$$, $$P(\overline{C})=\dfrac{5}{7}$$, $$P(\overline{D})= 1$$, $$P(A\cup\overline{C})=\dfrac{6}{7}$$, $$P(A\cap\overline{C})=\dfrac{3}{7}$$.

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