document.write( "Question 1084176: An ordinary (fair) die is a cube with the numbers
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document.write( "1
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document.write( " through
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document.write( "6
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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.
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document.write( "Compute the probability of each of the following events:
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document.write( "Event
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document.write( "A
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document.write( ": The sum is greater than
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document.write( "9
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document.write( ".
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document.write( "Event
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document.write( "B
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document.write( ": The sum is divisible by
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document.write( "4
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document.write( ". \n" );
document.write( "
Algebra.Com's Answer #698297 by Boreal(15235) You can put this solution on YOUR website! The probabilities \n" ); document.write( "sum is 2: 1/36 \n" ); document.write( "sum is 3: 2/36 \n" ); document.write( "(others omitting the denominator) \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( "Therefore, greater than 9 is 6/36=1/6 \n" ); document.write( "Divisible (evenly, I assume) by 4, is 3,6,9,12, or 12/36=1/3 \n" ); document.write( " |