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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