document.write( "Question 1072890: An ordinary (fair) die is a cube with the numbers 1 through 6 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 A: The sum is greater than 9 .
\n" ); document.write( "Event B: The sum is not divisible by 5.
\n" ); document.write( "Write your answers as exact fractions. \n" ); document.write( "

Algebra.Com's Answer #687820 by Fombitz(32388) $\"\"$ $\"About$

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Look at the possible outcomes, there are 36 of them starting with (1,1) and going to (6,6). The sums range from 2 to 12 with the following distribution,
\n" ); document.write( "2,1
\n" ); document.write( "3,2
\n" ); document.write( "4,3
\n" ); document.write( "5,4
\n" ); document.write( "6,5
\n" ); document.write( "7,6
\n" ); document.write( "8,5
\n" ); document.write( "9,4
\n" ); document.write( "10,3
\n" ); document.write( "11,2
\n" ); document.write( "12,1
\n" ); document.write( "So the number of outcomes greater than 9 is $\"3%2B2%2B1=6\"$
\n" ); document.write( " $\"P%28A%29=6%2F36=1%2F6\"$
\n" ); document.write( "Only the sum of 5 is divisible by 5, it occurs twice. $\"5%2B5=10\"$ so then $\"36-10=26\"$ of the outcomes are not divisible by 5.
\n" ); document.write( " $\"P%28B%29=26%2F36=13%2F18\"$ \n" ); document.write( "

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