SOLUTION: If two fair dice are rolled, find the probability that the sum of the dice is 7, given that the sum is greater than 3.

We start with the sample space of all 36 
possible rolls:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

The next step is to reduce the sample space by looking 
at what is given.  These words,starting with the word
"given":

tell us to eliminate what is NOT given from the sample 
space, which is this reduced sample space which contains
only those rolls whose sums are greater than 3:

            (1,3) (1,4) (1,5) (1,6)

      (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we have a reduced sample space which only
has (count them!) 33 outcomes.

Now using that reduced sample space, we turn to
the first part:

I will color those rolls with sum 7 red:

            (1,3) (1,4) (1,5) (1,6)

      (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Count the red ones.  There are 6.  That's 6 out of 33, or
a probability of 6/33 which reduces to 2/11.

Edwin