Question 935080
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Given: P(A&#8898;B) = P(A)P(B)
To prove: P(A'&#8898;B') = P(A')P(B')

     First we use DeMorgan's theorem: A'&#8898;B' = (A&#8899;B)'

P(A'&#8898;B') = P[(A&#8899;B)'] = 1-P(A&#8899;B) = 1-[P(A)+P(B)-P(A&#8898;B)] = 

= 1-P(A)-P(B)+P(A&#8898;B) = 1-P(A)-P(B)+P(A)P(B) = 

     Group the first two terms and factor -P(B) out 
     of the last two terms

= [1-P(A)] - P(B)[1-P(A)] =

     [1-P(A)] is a common factor.  Factor it out:

= [1-P(A)][1-P(B)] 

= P(A')P(B') 

Therefore:  P(A'&#8898;B')=P(A')P(B') 

Edwin</pre>