Probability Q: Show A & B Independent

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To show that events A and B are independent given that P(A|B) = P(A|B^c), one must demonstrate that P(A ∩ B) = P(A)P(B). The discussion clarifies that the definition of independence can vary, but the key relationship involves conditional probabilities. The relationship P(A) = P(A|B)P(B) + P(A|B^c)P(B^c) is essential for the proof. Participants emphasize the importance of correctly interpreting the problem statement and applying the definitions of independence accurately. Understanding these concepts is crucial for solving the problem effectively.
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Homework Statement



Suppose
P(A|B) = P(A|B^c)
(P(B) > 0 and P(B^c) > 0 both understood). Show that A and B are independent.

Homework Equations


The Attempt at a Solution



I don't know where to go from here. Thanks for any help

\frac{P(A \bigcap B)}{P(B)} = \frac{P(A \bigcap B^{c})}{P(B^{c})}
 
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Please restate (in fact, go back are reread) the problem. As you have it now it says "Suppose A and B are independent. Show that A and B are independent"!
 
Ok sorry about that I fixed that. Thanks for helping me.
 
Okay, good. Now what is your definition of "independent events"?
 
P(A intersection B) = P(A)P(B)

not sure what to do from here because it's conditional probabilities.
 
Good! So you are given that P(A|B)= P(A|B^c) and want to prove that P(A\cap B)= P(A)P(B).

You should know, then, that P(A)= P(A|B)P(B)+ P(A|B^c)B^c, for any A and B. Further, P(B^c)= 1- P(B).

(The reason I asked about the definition was that some texts use "P(A)= P(A|B)" as the definition of "A and B are independent. Of course it is easy to show that is equivalent to your "P(A\cap <br /> B)= P(A)P(B)".)
 

Thread 'Find the polar form of a complex number'

The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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