document.write( "Question 1084318: 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( "2
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document.write( ".
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document.write( "Write your answers as exact fractions. \n" );
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Algebra.Com's Answer #698416 by rothauserc(4718)![]() ![]() You can put this solution on YOUR website! There are 6 * 6 = 36 possible events \n" ); document.write( ": \n" ); document.write( "(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) \n" ); document.write( "(2,1), (2,2), (2,3), (2,4), (2,5), (2,6) \n" ); document.write( "(3,1), (3,2), (3,3), (3,4), (3,5), (3,6) \n" ); document.write( "(4,1), (4,2), (4,3), (4,4), (4,5), (4,6) \n" ); document.write( "(5,1), (5,2), (5,3), (5,4), (5,5), (5,6) \n" ); document.write( "(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \n" ); document.write( ": \n" ); document.write( "A. Sum is > 9 \n" ); document.write( "(4,6), (5,5), (5,6), (6,4), (6,5), (6,6) \n" ); document.write( "There are 6 events whose sum is > 9 \n" ); document.write( "Probability(P) sum is > 9 = 6/36 = 1/6 \n" ); document.write( ": \n" ); document.write( "B. sum is divisible by 2 \n" ); document.write( "There are 36/2 = 18 events whose sum is divisible by 2 \n" ); document.write( "P sum is divisible by 2 is 18/36 = 1/2 \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( " |