Proving two events are independent given that one has a probability of 1

AI Thread Summary
An event A is independent of every event B if P(A) equals 0 or 1. When P(A) equals 1, the relationship simplifies to P(A n B) = P(B), indicating that P(B/A) must equal P(B) for independence. The discussion highlights the challenge of proving P(B/A) = P(B) without assuming independence. The user successfully resolved the problem by employing the complement approach. This demonstrates the principle of independence in probability theory effectively.
LCBlazer07
Messages
3
Reaction score
0
Show that an event A is independent of every event B if P(A)=0 or P(A)=1.


**I was able to prove the first part of this problem that is that the events are independent when P(A)=0. However I am stuck on the part where P(A)=1.


I have this so far:

P(A n B) = P(A)P(B/A)
= (1)P(B/A)
= P(B/A)

If the events are independent then P(A n B) = P(A)P(B) = (PB)

So basically i have to show P(B/A) = P(B)...but I do cannot find a way to do this.
 
Physics news on Phys.org
Try it the other way:
P(A n B) = P(B)P(A/B)
 
Ok, so i started using that suggestion and got the following...

P( A n B) = P(B)P(A/B)
then I was going to say that P(A n B) = P(B), but I how can I prove that P(A/B)=1 without assuming that A and B are independent.
 
thanks for the suggestion, but I was able to solve this problem by using the complement.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top