document.write( "Question 376288: Cable Television In 2006, 86% of U.S. households had
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document.write( "cable TV. Choose 3 households at random. Find the
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document.write( "probability that
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document.write( "a. None of the 3 households had cable TV
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document.write( "b. All 3 households had cable TV
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document.write( "c. At least 1 of the 3 households had cable TV \n" );
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Algebra.Com's Answer #267725 by edjones(8007)![]() ![]() You can put this solution on YOUR website! Let x=.86, y=.14 \n" ); document.write( "(x+y)^3 Binomial expansion. \n" ); document.write( ". \n" ); document.write( "a) y^3=.0027 \n" ); document.write( ". \n" ); document.write( "b) x^3=.6361 \n" ); document.write( ". \n" ); document.write( "c)1-y^3=1-.0027=.9973 \n" ); document.write( ". \n" ); document.write( "Ed \n" ); document.write( " |