Proving Independence: Probability Theory of A and B in a Simple Proof"

Click For Summary

Homework Help Overview

The discussion revolves around proving the independence of the complements of two events, A and B, in the context of probability theory. The original poster seeks assistance with this proof, indicating a deadline for the assignment.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to prove the independence of the complements, having already established some relationships but struggling with the final part. Participants question the definitions of independence being used and clarify the mathematical expression for independence.

Discussion Status

The discussion includes attempts to clarify definitions and approaches to the proof. While the original poster expresses difficulty, there is no explicit consensus on a method, and the conversation remains open-ended with a mention of a solution being posted by the professor.

Contextual Notes

The original poster mentions a deadline for the assignment, indicating a time constraint that may affect the discussion. There is also a reference to using De Morgan's law and various identities, suggesting a range of mathematical tools being considered.

FrogPad
Messages
801
Reaction score
0
If A and B are independent, prove that [itex]\bar A[/itex], [itex]\bar B[/itex] are independent.

Could someone help me start this. It's due tomorrow. I managed to prove that [itex]\bar A [/tex] is independent with [itex]B[/itex] and that [itex]\bar B[/itex] is independent with [itex]A[/itex], but I can't get the last one (the question I put above). Just a little nudge would be good. I've been going in circles trying stuff, from De Morgans law, to every identity I can think of.<br /> <br /> Thanks[/itex]
 
Physics news on Phys.org
What are you using as the definition that A and B are independent?
 
jing said:
What are you using as the definition that A and B are independent?
For A and B being independent,
P[AB]=P[A]P
 
The professor posted the solution. Later :)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K