Problems from Operations among events: union, intersection, difference and complement

In an urn we have white $(W)$ , red $(R)$ , green $(G)$ and black $(B)$ balls. We draw a ball from the urn, and look at what color it is.

Consider the following events:

$A_{1} =$ "draw a white or a red ball".

$A_{2} =$ "draw a ball that is not green".

$A_{3} =$ "draw a black ball".

Describe the results that form each event.

Consider now the following events: $A_{1} \cup A_{3}$ , $A_{2} - A_{1}$ , $\overset{―}{A_{1}} \cap A_{3}$ .

Describe the results that form each event.

See development and solution

Development:

First of all, we have to determine what the sample space is. We already know that the possible results are to extract a white ball $(W)$ , to extract a red ball $(R)$ , to extract a green ball $(G)$ and to extract a black ball $(B)$ . So, $Ω = {W, R, G, B}$ .

The event $A_{1} =$ "extract a white or red ball" is formed by $A_{1} = {W, R}$ . We can see it , considering that $A_{1}$ is the union of "extract a white ball", ${W}$ , and "extract a red ball", ${R}$ .

The event $A_{2} =$ "extract a ball that is not green" is the opposite of the event "extract a green ball" $= {G}$ . So, $A_{2} = \overset{―}{G}$ . And so, we know that we can find $A_{2}$ doing $A_{2} = Ω - {G} = {W, R, G, B} - {G} = {W, R, G}$ .

The event $A_{3} =$ "extract a black ball" $= {B}$ .

Let's consider now the operations between events that arise next:

$A_{1} \cup A_{3} =$ "extract a white or red ball, or to extract a black ball" $= {W, R} \cup {B} = {W, R, B}$ .

$A_{2} - A_{1}$ is the difference between $A_{2}$ and $A_{1}$ . It is formed by all the events that are in $A_{2}$ , but not in $A_{1}$ . And so, $A_{2} - A_{1} = {W, R, G} - {W, R} = {G}$ .

To calculate $\overset{―}{A_{1}} \cap A_{3}$ , first we have to calculate what $\overset{―}{A_{1}}$ is. We have seen that the complementary $A_{1}$ can be found as follows $\overset{―}{A_{1}} = Ω - A_{1} = {W, R, G, B} - {W, R} = {G, B} .$

Now we can calculate the event $\overset{―}{A_{1}} \cap A_{3}$ , formed by all the events that satisfy $\overset{―}{A_{1}}$ and $A_{3}$ . We find it by doing $\overset{―}{A_{1}} \cap A_{3} = {G, B} \cap {B} = {B}$ .

Solution:

$A_{1} = {W, R}$ , $A_{2} = {W, R, G}$ , $A_{3} = {B}$ .

$A_{1} \cup A_{3} = {W, R, B}$ , $A_{2} - A_{1} = {G}$ , $\overset{―}{A_{1}} \cap A_{3} = {B}$ .

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