document.write( "Question 626440: A single die with 6 faces numbered 1 through 6 is thrown thrice. What is the probability that the sum of the number appearing after each throw is more than 15?
\n" ); document.write( "A)5/216
\n" ); document.write( "B)5/108
\n" ); document.write( "C)5/36
\n" ); document.write( "D)1/5
\n" ); document.write( "E)1/6 \n" ); document.write( "

Algebra.Com's Answer #394165 by Edwin McCravy(20065) $\"\"$ $\"About$

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The only combinations of rolls with sum exceeding 15 are\r\n" );
document.write( "these 4: \r\n" );
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document.write( "4+6+6 = 16\r\n" );
document.write( "5+5+6 = 16\r\n" );
document.write( "5+6+6 = 17 \r\n" );
document.write( "6+6+6 = 18\r\n" );
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document.write( "However, each of the first 3 rolls have 3!/2! = 3 distinguishable permutations.\r\n" );
document.write( "That's 3×3 or 9 ways. The fourth roll 6+6+6 can only be had in one way. So\r\n" );
document.write( "that's a total of 10 ways the sum can exceed 15.\r\n" );
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document.write( "The number of possible rolls is 6·6·6 = 216\r\n" );
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document.write( "So the desired probability if 10 ways out of 216, or ,\r\n" );
document.write( "which reduces to , choice B).\r\n" );
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document.write( "Edwin

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