document.write( "Question 619237: Just over half (0.6) of the homes built in the new upper-bracket snootyville subdivision included a room designed specifically to be a home theater. Four snootyville homes are selected at random.
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document.write( "what is the chance that all four homes will have the home theater room?\r
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document.write( "What is the chance that non of the homes in the sample of four will have the home theater room? \n" );
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Algebra.Com's Answer #389491 by Theo(13342)![]() ![]() You can put this solution on YOUR website! p = .6 \n" ); document.write( "q = .4 \n" ); document.write( "p = probability they have a theater room. \n" ); document.write( "q = probability they don't. \n" ); document.write( "4 selected at random. \n" ); document.write( "probablity all 4 have a theater room = .6^4 = .1296 \n" ); document.write( "all of the probabilities are: \n" ); document.write( "p(0) = .6^0 * .4^4 * 4C0 = .0256 \n" ); document.write( "p(1) = .6^1 * .4^3 * 4C1 = .1536 \n" ); document.write( "p(2) = .6^2 * .4^2 * 4C2 = .3456 \n" ); document.write( "p(3) = .6^3 * .4^1 * 4C3 = .3456 \n" ); document.write( "p(4) = .6^4 * .4^0 * 4C4 = .1296 \n" ); document.write( "sum of all probabilities is equal to 1 \n" ); document.write( "nCx is equal to n! / (x! * (n-x)!) \n" ); document.write( " \n" ); document.write( " |