Question 1098933
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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":
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given that the sum is greater than 3.
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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:</pre>find the probability that the sum of the dice is 7<pre>
I will color those rolls with sum 7 red:

            (1,3) (1,4) (1,5) <font color="red"><b>(1,6)</b></font>

      (2,2) (2,3) (2,4) <font color="red"><b>(2,5)</b></font> (2,6)

(3,1) (3,2) (3,3) <font color="red"><b>(3,4)</b></font> (3,5) (3,6) 

(4,1) (4,2) <font color="red"><b>(4,3)</b></font> (4,4) (4,5) (4,6) 

(5,1) <font color="red"><b>(5,2)</b></font> (5,3) (5,4) (5,5) (5,6)

<font color="red"><b>(6,1)</b></font> (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</pre>