document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #568601 by Edwin McCravy(20060)\"\" \"About 
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Given:  P(A∩B) = P(A)P(B)\r\n" );
document.write( "Prove:  P(A∩B') = P(A)P(B')\r\n" );
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document.write( "P(A∩B) = P(A)P(B)      given\r\n" );
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document.write( "We will use the fact that P(X) = P(X∩Y)+P(X∩Y')\r\n" );
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document.write( "P(A∩B) = [P(A∩B')+P(A∩B)][P(A∩B)+P(A'∩B)]\r\n" );
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document.write( "and that P(A∩B)+P(A'∩B)+P(A∩B')+P(A'∩B') = 1, since one of those\r\n" );
document.write( "must occur, we can multiply it on the left side without changing\r\n" );
document.write( "the value:\r\n" );
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document.write( "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)]\r\n" );
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document.write( "Multiplying all that out:\r\n" );
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document.write( "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)  \r\n" );
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document.write( "Simplifying:\r\n" );
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document.write( "P(A∩B)P(A'∩B') = P(A∩B')P(A'∩B)\r\n" );
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document.write( "We want to prove:\r\n" );
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document.write( "P(A∩B') ≟ P(A)P(B')\r\n" );
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document.write( "P(A∩B') ≟ [P(A∩B')+P(A∩B)][P(A∩B')+P(A'∩B')]\r\n" );
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document.write( "and since P(A∩B)+P(A'∩B)+P(A∩B')+P(A'∩B') = 1, since one of those\r\n" );
document.write( "must occur, we can multiply it on the left side:\r\n" );
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document.write( "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')]    \r\n" );
document.write( "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')\r\n" );
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document.write( "So to prove that, we take\r\n" );
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document.write( "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), \r\n" );
document.write( "proved above\r\n" );
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document.write( "and add P(A∩B')²+P(A∩B')P(A∩B)+P(A∩B')P(A'∩B') to both sides:\r\n" );
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document.write( "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')\r\n" );
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document.write( "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')  \r\n" );
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document.write( "rearrange to reverse the above steps\r\n" );
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document.write( "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')]\r\n" );
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document.write( " since P(A∩B')+P(A∩B)+P(A'∩B)+P(A'∩B') = 1\r\n" );
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document.write( "P(A∩B')*1 = [P(A∩B')+P(A∩B)][P(A∩B')+P(A'∩B')]   \r\n" );
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document.write( "P(A∩B') = P(A)P(B')     \r\n" );
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document.write( "which is what we had to prove.\r\n" );
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document.write( "Edwin
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