document.write( "Question 1084318: An ordinary (fair) die is a cube with the numbers
\n" ); document.write( "1
\n" ); document.write( " through
\n" ); document.write( "6
\n" ); document.write( " on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.
\n" ); document.write( "Compute the probability of each of the following events:
\n" ); document.write( "Event
\n" ); document.write( "A
\n" ); document.write( ": The sum is greater than
\n" ); document.write( "9
\n" ); document.write( ".
\n" ); document.write( "Event
\n" ); document.write( "B
\n" ); document.write( ": The sum is divisible by
\n" ); document.write( "2
\n" ); document.write( ".
\n" ); document.write( "Write your answers as exact fractions. \n" ); document.write( "

Algebra.Com's Answer #698422 by Boreal(15235) $\"\"$ $\"About$

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It can be shown that the sum is greater than 9 (10, 11, 12) is 6/36 or 1/6. Those 6 possibilities are 4-6,5-5,6-4 and 5-6, 6-5, and 6-6
\n" ); document.write( "The sum is divisible by 2:
\n" ); document.write( "2: 1-1, or 1 way
\n" ); document.write( "4: 1-3, 2-2-, 3-1: 3 ways
\n" ); document.write( "6: 1-5, 2-4,3-3,4,2,5-1: 5 ways
\n" ); document.write( "8:also 5 ways 2-6, 6-2,3-5,5-3, 4-4
\n" ); document.write( "10: 3 ways
\n" ); document.write( "12: 1 way
\n" ); document.write( "That is a total of 18 ways out of 36 or 1/2.
\n" ); document.write( " \n" ); document.write( "

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