1709af17-7e2d-46f9-b19c-23060577e46c9de777f0-602c-4ea9-a300-d43471defb24bf9c8cfb-3123-438a-a941-8945554badac6.3.4In 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: $$A_1\cup A_3$$, $$A_2-A_1$$, $$\overline{A_1}\cap A_3$$. Describe the results that form each event.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, $$\Omega=\lbrace W,R,G,B \rbrace$$. The event $$A_1=$$"extract a white or red ball" is formed by $$A_1= \lbrace W, R \rbrace$$. We can see it , considering that $$A_1$$ is the union of "extract a white ball", $$\lbrace W \rbrace$$, and "extract a red ball", $$\lbrace R \rbrace$$. The event $$A_2 =$$"extract a ball that is not green" is the opposite of the event "extract a green ball"$$= \lbrace G \rbrace$$. So, $$A_2=\overline{G}$$. And so, we know that we can find $$A_2$$ doing $$A_2=\Omega-\lbrace G \rbrace=\lbrace W,R,G,B \rbrace-\lbrace G \rbrace=\lbrace W,R,G \rbrace$$. The event $$A_3=$$"extract a black ball"$$=\lbrace B \rbrace$$. 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"$$= \lbrace W,R \rbrace \cup \lbrace B \rbrace=\lbrace W,R,B \rbrace $$. $$A_2-A_1$$ 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, $$A_2-A_1=\lbrace W,R,G \rbrace-\lbrace W,R \rbrace=\lbrace G \rbrace$$. To calculate $$\overline{A_1}\cap A_3$$, first we have to calculate what $$\overline{A_1}$$ is. We have seen that the complementary $$A_1$$ can be found as follows $$\overline{A_1}=\Omega - A_1 = \lbrace W,R,G,B \rbrace - \lbrace W,R \rbrace = \lbrace G,B \rbrace.$$ Now we can calculate the event $$\overline{A_1} \cap A_3$$, formed by all the events that satisfy $$\overline{A_1}$$ and $$A_3$$. We find it by doing $$\overline{A_1}\cap A_3= \lbrace G,B \rbrace \cap \lbrace B \rbrace = \lbrace B \rbrace$$.$$A_1=\lbrace W,R \rbrace$$, $$A_2=\lbrace W,R,G \rbrace$$, $$A_3=\lbrace B \rbrace$$. $$A_1 \cup A_3 = \lbrace W,R,B \rbrace$$, $$A_2-A_1=\lbrace G \rbrace$$, $$\overline{A_1}\cap A_3=\lbrace B \rbrace$$.7/18/16, 11:44 AM7/18/16, 11:44 AM