document.write( "Question 935211: If A and B are independent events, show that A and B' are also independent events, where B' is the complementary event of B. \n" ); document.write( "

Algebra.Com's Answer #568365 by Edwin McCravy(20059) $\"\"$ $\"About$

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document.write( "Given:  P(AᑎB) = P(A)P(B)\r\n" );
document.write( "Prove:  P(AᑎB') ≟ P(A)P(B') \r\n" );
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document.write( "Let P(AᑎB')=w\r\n" );
document.write( "Let P(AᑎB)=x\r\n" );
document.write( "Let P(A'ᑎB)=y\r\n" );
document.write( "Let P(A'ᑎB')=z\r\n" );
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document.write( "Then x+y+w+z = 1\r\n" );
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document.write( "Draw a Venn diagram with the probability of each region written\r\n" );
document.write( "in each region:\r\n" );
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document.write( "     P(AᑎB) = P(A)P(B)      given\r\n" );
document.write( "          x = (w+x)(x+y)    \r\n" );
document.write( " x(w+x+y+z) = (w+x)(x+y)    since w+x+y+z = 1\r\n" );
document.write( "wx+x²+xy+xz = wx+wy+x²+xy   multiplying out\r\n" );
document.write( "         xz = wy\r\n" );
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document.write( "We want to prove:\r\n" );
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document.write( "    P(AᑎB') ≟ P(A)P(B')\r\n" );
document.write( "          w ≟ (w+x)(w+z)\r\n" );
document.write( " w(w+x+y+z) ≟ (w+x)(w+z)    since w+x+y+z = 1  \r\n" );
document.write( "w²+wx+wy+wz ≟ w²+wz+wx+xz\r\n" );
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document.write( "So to prove that, we take\r\n" );
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document.write( "         wy = xz            same as xz = wy , proved above\r\n" );
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document.write( "                            and add w²+wx+wz to both sides:\r\n" );
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document.write( "wy+w²+wx+wz = xz+w²+wx+wz   \r\n" );
document.write( "w²+wx+wy+wz = w²+wz+wx+xz   rearrange to reverse the above steps\r\n" );
document.write( " w(w+x+y+z) = (w+x)(w+z)\r\n" );
document.write( "          w = (w+x)(w+z)    since w+x+y+z = 1\r\n" );
document.write( "    P(AᑎB') = P(A)P(B')     which is what we had to prove.\r\n" );
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document.write( "Edwin

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