document.write( "Question 279073: The cable TV industry reports that 59% of homes in the US to cable TV. If you pick four US households at random, what is the probability that at least one of the households is not a cable TV subscriber? \n" ); document.write( "
Algebra.Com's Answer #203044 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The cable TV industry reports that 59% of homes in the US to cable TV. If you pick four US households at random, what is the probability that at least one of the households is not a cable TV subscriber? \n" ); document.write( "------------------------ \n" ); document.write( "Binomial Problem: \n" ); document.write( "n = 4 ; p = 0.59 ; q = 0.41 ; 1<= x <=4 \n" ); document.write( "--------------------------------------------- \n" ); document.write( "P(at least one) = 1 - P(none) \n" ); document.write( "--- \n" ); document.write( "P(at least DISABLED_event_one= 1 - 4C0(0.59)^0*(0.41)^4 = 1-0.41^4 = 0.9717... \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |