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

by ooooo
Tags: diagram, proof, venn
 HW Helper P: 1,372 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.
 HW Helper P: 1,372 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.