
#1
Oct408, 04:24 PM

P: 3

Do you know how to proof
P(A U B U C) = P(A) + P(B) + P(C)  P(A^B)  P(B^C)  P(C^A) + P(A^B^C) ^ is intersection. Do you know how to find P(A U B U C U D) Thank you very much. 



#2
Oct408, 05:11 PM

HW Helper
P: 1,344

Let
[tex] D = B \cup C [/tex] and note that [tex] A \cup B \cup C = A \cup D [/tex] then [tex] \begin{align*} \Pr(A \cup B \cup C) & = \Pr(A \cup D)\\ & = \Pr(A) + \Pr(D)  \Pr(A \cap D) \\ & = \Pr(A) + \Pr(B \cup C)  \Pr(A \cap D)\\ & = \Pr(A) + \Pr(B) + \Pr(C)  \Pr(B \cap C)  \Pr(A \cap D) \end{align*} [/tex] The rest of the proof comes from realizing that [tex] \Pr(A \cap D) = \Pr(A \cap \left(B \cup C\right)) = \Pr((A \cap B) \cup (A \cap C)), [/tex] using the Addition Rule for probability to expand the final term, and being very careful with positive and negative signs. 



#3
Oct408, 07:00 PM

P: 3

Thank you so much Statdad. I would like to ask another question.
How to proof P(A U B) = P(A) + P(B)  P(A ^ B) ? Thank you again. 



#4
Oct408, 07:02 PM

HW Helper
P: 1,344

How to proof P(A U B U C) without using Venn Diagram
This proof isn't needed for the problem you posted above  is there a reason you need it here?




#5
Oct408, 07:06 PM

P: 3

Sorry. I'm just curious. :)




#6
Oct408, 07:14 PM

HW Helper
P: 1,344

No  I was interrupted by someone at the door.
Here is one method  there are others. First, note that [tex] A \cup B = (AB) \cup (A \cap B) \cup (B  A) [/tex] and the three sets on the right are pairwise disjoint. Now [tex] \begin{align*} \Pr(A \cup B) & = \Pr(AB) + \Pr(A \cap B) + \Pr(B  A)\\ & = \left(\Pr(AB) + \Pr(A \cap B) \right) + \left(\Pr(BA) + \Pr(A \cap B)\right)  \Pr(A \cap B) \\ & = \Pr(A) + \Pr(B)  \Pr(A \cap B) \end{align*} [/tex] Again, sorry for the abrupt end to my previous post  I'm getting really tired of our election season. 


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