# How to proof P(A U B U C) without using Venn Diagram

by ooooo
Tags: diagram, proof, venn
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 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.
 HW Helper P: 1,361 Let $$D = B \cup C$$ and note that $$A \cup B \cup C = A \cup D$$ then \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*} The rest of the proof comes from realizing that $$\Pr(A \cap D) = \Pr(A \cap \left(B \cup C\right)) = \Pr((A \cap B) \cup (A \cap C)),$$ using the Addition Rule for probability to expand the final term, and being very careful with positive and negative signs.
 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.
 HW Helper P: 1,361 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?
 P: 3 Sorry. I'm just curious. :)
 HW Helper P: 1,361 No - I was interrupted by someone at the door. Here is one method - there are others. First, note that $$A \cup B = (A-B) \cup (A \cap B) \cup (B - A)$$ and the three sets on the right are pair-wise disjoint. Now \begin{align*} \Pr(A \cup B) & = \Pr(A-B) + \Pr(A \cap B) + \Pr(B - A)\\ & = \left(\Pr(A-B) + \Pr(A \cap B) \right) + \left(\Pr(B-A) + \Pr(A \cap B)\right) - \Pr(A \cap B) \\ & = \Pr(A) + \Pr(B) - \Pr(A \cap B) \end{align*} Again, sorry for the abrupt end to my previous post - I'm getting really tired of our election season.

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