document.write( "Question 475441: two digits are randomly selected without repition; compute the probability that their sum is is odd \n" ); document.write( "
Algebra.Com's Answer #326041 by Edwin McCravy(20064)\"\" \"About 
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document.write( "EVEN + EVEN = EVEN\r\n" );
document.write( "EVEN + ODD = ODD\r\n" );
document.write( "ODD + EVEN = ODD\r\n" );
document.write( "ODD + ODD = EVEN\r\n" );
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document.write( "So to have their sum odd, one has to be odd and the other even.\r\n" );
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document.write( "The digits are\r\n" );
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document.write( "0,1,2,3,4,5,6,7,8,9\r\n" );
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document.write( "5 are ODD and 5 are EVEN.\r\n" );
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document.write( "After we have chosen the first digit, there are only 9 digits \r\n" );
document.write( "left to choose from:\r\n" );
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document.write( "P[(odd 1st AND even 2nd) OR (even 1st and odd 2nd)] =\r\n" );
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document.write( "AND indicates multiplication and OR indicates ADDITION:\r\n" );
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document.write( "  P(odd 1st)×P(even 2nd) + P(even 1st)×P(odd 2nd) =\r\n" );
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document.write( "             (5/10)(5/9) + (5/10)(5/9) =\r\n" );
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document.write( "              (1/2)(5/9) + (1/2)(5/9) =\r\n" );
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document.write( "                   5/18  + 5/18 =\r\n" );
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document.write( "                       10/18 =\r\n" );
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document.write( "                        5/9\r\n" );
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document.write( "Edwin
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