document.write( "Question 1166529: a pair of dice is tossed. find the probability that one of the dice is 2 if the sum is 6! \n" ); document.write( "

Algebra.Com's Answer #791042 by math_helper(2461) $\"\"$ $\"About$

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document.write( "I see two ways to think about it...\r
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document.write( "Draw out a table of all possible sums with each possible die value across the columns and each down the rows.\r
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document.write( "     1   2   3   4   5   6\r
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document.write( "1    2   3   4   5   6   7
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document.write( "2    3   4   5   6   7   8
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document.write( "3    4   5   6   7   8   9
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document.write( "4    5   6   7   8   9   10
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document.write( "5    6   7   8   9   10  11
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document.write( "6    7   8   9   10  11  12\r
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document.write( "Now the first way to solve is to notice there are five(5) sums of 6.
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document.write( "Of these five sums, two(2) have one die that is 2: 
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document.write( " Pr(one die is 2| sum is 6) = 2/5.\r
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document.write( "Another way is to apply Bayes' Theorem to the problem:\r
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document.write( "P(A|B) = P(B|A)*P(A) / P(B)\r
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document.write( "where
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document.write( "P(A) = P(one die is 2) = 11/36
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document.write( "P(B) = P(sum of dice is 6) = 5/36
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document.write( "P(B|A) = P(sum of dice is 6 given one die is a 2) = 2/11\r
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document.write( "P(A|B) = P(one die is 2 given sum is 6) = (2/11)(11/36) / (5/36) = 2/5\r
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