document.write( "Question 1097831: An ordinary (fair) die is a cube with the numbers
\n" ); document.write( "1 through 6 on the sides (represented by painted spots). Imagine that
\n" ); document.write( "such a die is rolled twice in succession and that the face values of
\n" ); document.write( "the two rolls are added together. This sum is recorded as the outcome
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\n" ); document.write( "each of the following events:
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Algebra.Com's Answer #712245 by Edwin McCravy(20060)\"\" \"About 
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document.write( "Here are all the possible rolls and their sums.  Count them.  There are 36 of them.\r\n" );
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document.write( "(1,1, 1+1=2) (1,2, 1+2=3) (1,3, 1+3=4) (1,4, 1+4=5)  (1,5, 1+5=6)  (1,6, 1+6=7)\r\n" );
document.write( "(2,1, 2+1=3) (2,2, 2+2=4) (2,3, 2+3=5) (2,4, 2+4=6)  (2,5, 2+5=7)  (2,6, 2+6=8)\r\n" );
document.write( "(3,1, 3+1=4) (3,2, 3+2=5) (3,3, 3+3=6) (3,4, 3+4=7)  (3,5, 3+5=8)  (3,6, 3+6=9)\r\n" );
document.write( "(4,1, 4+1=5) (4,2, 4+2=6) (4,3, 4+3=7) (4,4, 4+4=8)  (4,5, 4+5=9)  (4,6, 4+6=10)\r\n" );
document.write( "(5,1, 5+1=6) (5,2, 5+2=7) (5,3, 5+3=8) (5,4, 5+4=9)  (5,5, 5+5=10) (5,6, 5+6=11)\r\n" );
document.write( "(6,1, 6+1=7) (6,2, 6+2=8) (6,3, 6+3=9) (6,4, 6+4=10) (6,5, 6+5=11) (6,6, 6+6=12)\r\n" );
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document.write( "For the probability that the sum is greater than 9, I will color the\r\n" );
document.write( "ones that have sum greater than 9 red:\r\n" );
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document.write( "(1,1, 1+1=2) (1,2, 1+2=3) (1,3, 1+3=4) (1,4, 1+4=5)  (1,5, 1+5=6)  (1,6, 1+6=7)\r\n" );
document.write( "(2,1, 2+1=3) (2,2, 2+2=4) (2,3, 2+3=5) (2,4, 2+4=6)  (2,5, 2+5=7)  (2,6, 2+6=8)\r\n" );
document.write( "(3,1, 3+1=4) (3,2, 3+2=5) (3,3, 3+3=6) (3,4, 3+4=7)  (3,5, 3+5=8)  (3,6, 3+6=9)\r\n" );
document.write( "(4,1, 4+1=5) (4,2, 4+2=6) (4,3, 4+3=7) (4,4, 4+4=8)  (4,5, 4+5=9)  (4,6, 4+6=10)\r\n" );
document.write( "(5,1, 5+1=6) (5,2, 5+2=7) (5,3, 5+3=8) (5,4, 5+4=9)  (5,5, 5+5=10) (5,6, 5+6=11)\r\n" );
document.write( "(6,1, 6+1=7) (6,2, 6+2=8) (6,3, 6+3=9) (6,4, 6+4=10) (6,5, 6+5=11) (6,6, 6+6=12)\r\n" );
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document.write( "Count the red ones.  There are 6 of them. that's 6 out of 36, so the\r\n" );
document.write( "probability is the fraction with the number that have sum greater than\r\n" );
document.write( "9 as the numerator and the total number of possibilities as the \r\n" );
document.write( "denominator.  That fraction is 6/36 which reduces to 1/6.\r\n" );
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document.write( "For the probability that the sum is not divisible by 6, I will color the\r\n" );
document.write( "ones with sums that are not  divisible by 6 red (only 6 and 12 are divisible\r\n" );
document.write( "by 6, so most of them are colored red):\r\n" );
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document.write( "(1,1, 1+1=2) (1,2, 1+2=3) (1,3, 1+3=4) (1,4, 1+4=5)  (1,5, 1+5=6)  (1,6, 1+6=7)\r\n" );
document.write( "(2,1, 2+1=3) (2,2, 2+2=4) (2,3, 2+3=5) (2,4, 2+4=6)  (2,5, 2+5=7)  (2,6, 2+6=8)\r\n" );
document.write( "(3,1, 3+1=4) (3,2, 3+2=5) (3,3, 3+3=6) (3,4, 3+4=7)  (3,5, 3+5=8)  (3,6, 3+6=9)\r\n" );
document.write( "(4,1, 4+1=5) (4,2, 4+2=6) (4,3, 4+3=7) (4,4, 4+4=8)  (4,5, 4+5=9)  (4,6, 4+6=10)\r\n" );
document.write( "(5,1, 5+1=6) (5,2, 5+2=7) (5,3, 5+3=8) (5,4, 5+4=9)  (5,5, 5+5=10) (5,6, 5+6=11)\r\n" );
document.write( "(6,1, 6+1=7) (6,2, 6+2=8) (6,3, 6+3=9) (6,4, 6+4=10) (6,5, 6+5=11) (1,5, 6+6=12)\r\n" );
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document.write( "Count the red ones.  There are 30 of them. that's 30 out of 36, so the\r\n" );
document.write( "probability is the fraction with the number that have sum not divisible\r\n" );
document.write( "by 6 as the numerator and the total number of possibilities as the \r\n" );
document.write( "denominator.  That fraction is 30/36 which reduces to 5/6.\r\n" );
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document.write( "Edwin
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