document.write( "Question 234955: When you toss a pair of dice, each containing the digits 1 through 6, what is the probability that the sum of the values face-up on the cubes will not be 3?
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document.write( "a) 17/18 b) 1/6 c) 5/6 d) 18/36 e) 3/36. \n" );
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Algebra.Com's Answer #173168 by Theo(13342)![]() ![]() You can put this solution on YOUR website! It would be 1 minus the probability that it would be 3.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "You have 2 dice.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Total possible number of combinations would be 6*6 = 36\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Possible ways you could get a 3 would be:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "1 + 2 \n" ); document.write( "2 + 1\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "That's only 2 out of 36.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Possible ways you could not get a 3 would be 34 out of 36.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "34 out of 36 is the same as 17 out of 18.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Your answer would be A).\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |