document.write( "Question 537280: I need help with this please:\r
\n" ); document.write( "\n" ); 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.
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Algebra.Com's Answer #352786 by josmiceli(19441)\"\" \"About 
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 = \"+9%2A10+=+90+\"
\n" ); document.write( "\"+15%2F90+\" = P that sum is 5 or less
\n" ); document.write( "\"+1+-+15%2F90+=+90%2F90+-+15%2F90+\"
\n" ); document.write( "\"+1+-+15%2F90+=+75%2F90+\"
\n" ); document.write( "\"+75%2F90+=+5%2F6+\"
\n" ); document.write( "P that sum of digits exceeds 5 = 5/6
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