We have an urn with fourteen balls, seven red balls, numbered from $$1$$ to $$7$$, three yellow, numbered from $$8$$ to $$10$$, and four purple, numbered from $$11$$ to $$14$$. Our experiment consists of drawing a ball and seeing its number and color.

- Determine the sample space.
- Consider the events A = "get a number bigger or equal to 9", B = "get an even number." Define which elementary events form $$A$$ and $$B$$.
- Define which elementary events form the events $$C =$$"get a yellow or purple ball", $$D =$$"drawing a purple ball or multiple of three."
- Define which elementary events form the events $$E =$$"draw a red ball with a number less than $$4$$", $$F =$$"draw a yellow ball with a number bigger than $$11$$".

### Development:

1) The sample space is the set of all the possible results. In our case, we have

$$$\Omega=\lbrace R1,R2,R3,R4,R5,R6,R7,Y8,Y9,Y10,V11,V12,V13,V14 \rbrace$$$
where $$R$$ indicates that the ball is red, $$Y$$ that it is yellow, and $$V$$ that is violet.

2) The events that we are interested in are all the elementary events that have a number bigger than or equal to $$9$$. Therefore, $$$A=\lbrace Y9,Y10,V11,V12,V13,V14 \rbrace$$$ In the second case, we are only interested in those that have even numbers. That is, $$$B=\lbrace R2,R4,R6,Y8,Y10,V12,V14 \rbrace$$$

3) Our event is formed by all the elementary events that have a yellow or a violet ball. Those that have yellow ball are $$\lbrace Y8,Y9,Y10 \rbrace$$, and those that have violet ball are $$\lbrace V11, V12, V13, V14 \rbrace$$.

Therefore, our event is $$C=\lbrace Y8,Y9,Y10,V11,V12,V13,V14 \rbrace$$.

Now let's consider event $$D$$. Those that have a violet ball are $$\lbrace V11,V12,V13,V14 \rbrace$$, and multiples of $$3$$, between the numbers $$1$$ and $$14$$, are: $$3,6,9,12$$. Therefore, the balls that we are interested in are $$\lbrace R3,R6,Y9,V12 \rbrace$$. And so, we have $$D=\lbrace R3,R6,Y9,V11,V12,V13,V14 \rbrace$$.

Note that the ball $$V12$$ in fact satisfies both conditions: it is violet and also its number is a multiple of $$3$$. That is not a problem, and we must write in $$D$$ the results that satisfy the conditions, and $$V12$$ satisfies them since it satisfies at least one of two conditions.

4) We have $$8$$ red balls, that are $$\lbrace R1,R2,R3,R4,R5,R6,R7,R8 \rbrace$$. Between them, the balls with the number less than four are $$E=\lbrace R1,R2,R3 \rbrace$$. Also, we might have done it another way: first we choose the balls that have a number less than four (in our case, $$\lbrace R1,R2,R3 \rbrace$$), and as they all are red, our event is again $$E = \lbrace R1, R2, R3 \rbrace$$.

We can start by considering the yellow balls: $$\lbrace Y8, Y9, Y10 \rbrace$$. Between them, there is no a number bigger than $$11$$. Therefore, there is no elementary event that satisfies the condition, that is, $$F=\emptyset$$. In other words, $$F$$ is an impossible event, it can be never satisfied.

### Solution:

- $$\Omega=\lbrace R1,R2,R3,R4,R5,R6,R7,Y8,Y9,Y10,V11,V12,V13,V14 \rbrace$$.
- $$A=\lbrace Y9,Y10,V11,V12,V13,V14 \rbrace$$, $$B=\lbrace R2,R4,R6,Y8,Y10,V12,V14 \rbrace$$.
- $$C=\lbrace Y8,Y9,Y10,V11,V12,V13,V14 \rbrace$$, $$D=\lbrace R3,R6,Y9,V11,V12,V13,V14 \rbrace$$.
- $$E=\lbrace R1,R2,R3 \rbrace$$, $$F=\emptyset$$.