Definition of probability, sample space and sure and impossible event
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 $11$".
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.
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, $$B=\lbrace R2,R4,R6,Y8,Y10,V12,V14 \rbrace$$
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.
- 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.
- $\Omega=\lbrace R1,R2,R3,R4,R5,R6,R7,Y8,Y9,Y10,V11,V12,V13,V14 \rbrace$.
- $A=\lbrace Y9,Y10,V11,V12,V13,V14 \rbrace$, $B=\lbrace R2,R4,R6,Y8,Y10,V12,V14 \rbrace$.
- $C=\lbrace Y8,Y9,Y10,V11,V12,V13,V14 \rbrace$, $D=\lbrace R3,R6,Y9,V11,V12,V13,V14 \rbrace$.
- $E=\lbrace R1,R2,R3 \rbrace$, $F=\emptyset$.