document.write( "Question 537280: I need help with this please:\r
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document.write( "Given that a two-digit number is randomly selected, what is the probability that the sum of the digits exceeds five? Express your answer as a common fraction. \n" );
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Algebra.Com's Answer #352786 by josmiceli(19441) You can put this solution on YOUR website! Let P stand for probability \n" ); document.write( "( P that sum of digits exceeds 5 ) + ( P that sum is 5 or less ) = 1 \n" ); document.write( "The probabilities of everything that can possibly happen \n" ); document.write( "must add up to one, therefore: \n" ); document.write( "( P that sum of digits exceeds 5 ) = 1 - ( P that sum is 5 or less ) \n" ); document.write( "The 2 digit numbers that are 5 or less are \n" ); document.write( "10,11,12,13,14 \n" ); document.write( "20,21,22,23 \n" ); document.write( "30,31,32 \n" ); document.write( "40,41 \n" ); document.write( "50 \n" ); document.write( "There are 15 of these numbers \n" ); document.write( "The sum of digits of all the other 2 digit numbers exceed 5 \n" ); document.write( "The tens digit can be 1 - 9, or 9 possible \n" ); document.write( "The units digit can be 0 - 9 or 10 possible \n" ); document.write( "Total possible numbers = \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "P that sum of digits exceeds 5 = 5/6 \n" ); document.write( " \n" ); document.write( " |