I see two ways to think about it...



Draw out a table of all possible sums with each possible die value across the columns and each down the rows.



     1   2   3   4   5   6



1    2   3   4   5   6   7

2    3   4   5   6   7   8

3    4   5   6   7   8   9

4    5   6   7   8   9   10

5    6   7   8   9   10  11

6    7   8   9   10  11  12







Now the first way to solve is to notice there are five(5) sums of 6.

Of these five sums, two(2) have one die that is 2: 

 Pr(one die is 2| sum is 6) = 2/5.



Another way is to apply Bayes' Theorem to the problem:



P(A|B) = P(B|A)*P(A) / P(B)



where

P(A) = P(one die is 2) = 11/36

P(B) = P(sum of dice is 6) = 5/36

P(B|A) = P(sum of dice is 6 given one die is a 2) = 2/11



P(A|B) = P(one die is 2 given sum is 6) = (2/11)(11/36) / (5/36) = 2/5



 

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