SOLUTION: If A and B are two independent events, then show that A and B' are also independent, where B' is the complementary event of B.

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Question 935079: If A and B are two independent events, then show that A and B' are also independent, where B' is the complementary event of B.
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
Given:  P(A∩B) = P(A)P(B)
Prove:  P(A∩B') = P(A)P(B')

P(A∩B) = P(A)P(B)      given

We will use the fact that P(X) = P(X∩Y)+P(X∩Y')

P(A∩B) = [P(A∩B')+P(A∩B)][P(A∩B)+P(A'∩B)]

and that P(A∩B)+P(A'∩B)+P(A∩B')+P(A'∩B') = 1, since one of those
must occur, we can multiply it on the left side without changing
the value:

P(A∩B)[P(A∩B')+P(A∩B)+P(A'∩B)+P(A'∩B')] = [P(A∩B')+P(A∩B)][P(A∩B)+P(A'∩B)]

Multiplying all that out:

P(A∩B')P(A∩B)+P(A∩B)²+P(A∩B)P(A'∩B)+P(A∩B)P(A'∩B') = P(A∩B')P(A∩B)+P(A∩B')P(A'∩B)+P(A∩B)²+P(A∩B)P(A'∩B)  

Simplifying:
         
P(A∩B)P(A'∩B') = P(A∩B')P(A'∩B)

We want to prove:

P(A∩B') ≟ P(A)P(B')
          
P(A∩B') ≟ [P(A∩B')+P(A∩B)][P(A∩B')+P(A'∩B')]

and since P(A∩B)+P(A'∩B)+P(A∩B')+P(A'∩B') = 1, since one of those
must occur, we can multiply it on the left side:
 
P(A∩B')[P(A∩B')+P(A∩B)+P(A'∩B)+P(A'∩B')] ≟ [P(A∩B')+P(A∩B)][P(A∩B')+P(A'∩B')]    
P(A∩B')²+P(A∩B')P(A∩B)+P(A∩B')P(A'∩B)+P(A∩B')P(A'∩B') ≟ P(A∩B')²+P(A∩B')P(A'∩B')+P(A∩B')P(A∩B)+P(A∩B)P(A'∩B')

So to prove that, we take

P(A∩B')P(A'∩B) = P(A∩B)P(A'∩B') same as P(A∩B)P(A'∩B') = P(A∩B')P(A'∩B), 
proved above

and add P(A∩B')²+P(A∩B')P(A∩B)+P(A∩B')P(A'∩B') to both sides:

P(A∩B')P(A'∩B)+P(A∩B')²+P(A∩B')P(A∩B)+P(A∩B')P(A'∩B') = P(A∩B)P(A'∩B')+P(A∩B')²+P(A∩B')P(A∩B)+P(A∩B')P(A'∩B')

P(A∩B')²+P(A∩B')P(A∩B)+P(A∩B')P(A'∩B)+P(A∩B')P(A'∩B') = P(A∩B')²+P(A∩B')P(A'∩B')+P(A∩B')P(A∩B)+P(A∩B)P(A'∩B')  

rearrange to reverse the above steps
 
P(A∩B')[P(A∩B')+P(A∩B)+P(A'∩B)+P(A'∩B')] = [P(A∩B')+P(A∩B)][P(A∩B')+P(A'∩B')]

 since P(A∩B')+P(A∩B)+P(A'∩B)+P(A'∩B') = 1

P(A∩B')*1 = [P(A∩B')+P(A∩B)][P(A∩B')+P(A'∩B')]   
    
P(A∩B') = P(A)P(B')     

which is what we had to prove.

Edwin
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