Problems from 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 $11$ ".

See development and solution

Development:

1) The sample space is the set of all the possible results. In our case, we have
$Ω = {R 1, R 2, R 3, R 4, R 5, R 6, R 7, Y 8, Y 9, Y 10, V 11, V 12, V 13, V 14}$ 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, $A = {Y 9, Y 10, V 11, V 12, V 13, V 14}$ In the second case, we are only interested in those that have even numbers. That is, $B = {R 2, R 4, R 6, Y 8, Y 10, V 12, V 14}$

3) Our event is formed by all the elementary events that have a yellow or a violet ball. Those that have yellow ball are ${Y 8, Y 9, Y 10}$ , and those that have violet ball are ${V 11, V 12, V 13, V 14}$ .

Therefore, our event is $C = {Y 8, Y 9, Y 10, V 11, V 12, V 13, V 14}$ .

Now let's consider event $D$ . Those that have a violet ball are ${V 11, V 12, V 13, V 14}$ , and multiples of $3$ , between the numbers $1$ and $14$ , are: $3, 6, 9, 12$ . Therefore, the balls that we are interested in are ${R 3, R 6, Y 9, V 12}$ . And so, we have $D = {R 3, R 6, Y 9, V 11, V 12, V 13, V 14}$ .

Note that the ball $V 12$ 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 $V 12$ satisfies them since it satisfies at least one of two conditions.

4) We have $8$ red balls, that are ${R 1, R 2, R 3, R 4, R 5, R 6, R 7, R 8}$ . Between them, the balls with the number less than four are $E = {R 1, R 2, R 3}$ . Also, we might have done it another way: first we choose the balls that have a number less than four (in our case, ${R 1, R 2, R 3}$ ), and as they all are red, our event is again $E = {R 1, R 2, R 3}$ .

We can start by considering the yellow balls: ${Y 8, Y 9, Y 10}$ . 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:

$Ω = {R 1, R 2, R 3, R 4, R 5, R 6, R 7, Y 8, Y 9, Y 10, V 11, V 12, V 13, V 14}$ .
$A = {Y 9, Y 10, V 11, V 12, V 13, V 14}$ , $B = {R 2, R 4, R 6, Y 8, Y 10, V 12, V 14}$ .
$C = {Y 8, Y 9, Y 10, V 11, V 12, V 13, V 14}$ , $D = {R 3, R 6, Y 9, V 11, V 12, V 13, V 14}$ .
$E = {R 1, R 2, R 3}$ , $F = \emptyset$ .

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