My Math Forum If A, B independent events. Then A and B^C, as well as, A^C and B are independent

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 June 9th, 2015, 09:10 AM #1 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 If A, B independent events. Then A and B^C, as well as, A^C and B are independent Hi, I want to prove the following, but I encounter some problems. Can you give me any suggestions? Let A, B be independent events. Show that $\displaystyle A$ and $\displaystyle B^C$ are independent. I have done some things but without success: So, we have to show that $\displaystyle P(A \cap B^C) = P(A) \cdot P(B^C)$. Let: $\displaystyle P(A) = x \Rightarrow P(A^C) = 1 - P(A) = 1- x, \\ P(B) = y \Rightarrow P(B^C) = 1 - P(B) = 1 - y$ Since A, B are independent events we have: $\displaystyle P(A \cap B) = P(A) \cdot P(B) = x \cdot y$ So here we want to know: "In what way do we can rewrite $\displaystyle P(A \cap B^C)$, from the math data available?" I have tried this, using DeMorgan, but without obtain what I needed: $\displaystyle P[(A^C \cup B)^C] = P[(A^C)^C \cap B^C] = P(A \cap B^C) = 1 - P(A^C \cup B) =$ here I'm not sure if the addition rule can be applied, however: $\displaystyle = 1 - [P(A^C) + P(B) - P(A^C \cap B)]$ but here I don't have any $\displaystyle P(A^C \cap B)$ term to substitute using x,y. So I don't know how to proceed! Can you give any suggestions? Many thanks!
 June 9th, 2015, 01:33 PM #2 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 for clarity I post here another proof, because it is from the following one that I have to prove what I have written in the first post: Suppose $\displaystyle A$ and $\displaystyle B$ are independent events. Then $\displaystyle A^C$ and $\displaystyle B^C$ are independent events. We have to show that $\displaystyle P(A^C \cap B^C) = P(A^C) \cdot P(B^C)$. Let: $\displaystyle P(A) = x \Rightarrow P(A^C) = 1 - P(A) = 1- x, \\ P(B) = y \Rightarrow P(B^C) = 1 - P(B) = 1 - y$ Since A, B are independent events we have: $\displaystyle P(A \cap B) = P(A) \cdot P(B) = x \cdot y$ By the addition rule: $\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) = x + y - xy$ Using DeMorgan law: $\displaystyle P(A^C \cap B^C) = P[(A \cup B)^C] = 1 - P(A \cup B) = 1 - x - y + xy.$ And also: $\displaystyle P(A^C) \cdot P(B^C) = (1-x)(1-y) = 1-x-y+xy$ END. Similarly, I have tried to show in the first post that A and B^C, as well as A^C and B are independent, but without success. Can you give me any helping hand? many thanks! Last edited by beesee; June 9th, 2015 at 01:37 PM.
 June 9th, 2015, 01:52 PM #3 Global Moderator   Joined: May 2007 Posts: 6,586 Thanks: 612 P(A)=P(A∩B)+P(A∩B')=P(A)P(B)+P(A∩B') (B and B' are disjoint and add to entire space) P(A∩B')=P(A)(1-P(B))=P(A)P(B') Thanks from beesee

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### A~B then P(A)~P(B)

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