<PRE><B>Question 1113047</B>

Since the number of outcomes is small, we can enumerate the possible outcomes as a table:

&lt;pre&gt;
            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  
&lt;/pre&gt;

You can see there are 36 outcomes and everything on the diagonal consisting of 7&#39;s and down to the right is greater than 6.  That&#39;s 6+5+4+3+2+1=21, out of 36 outcomes, so Pr(dice total &gt; 6) = 21/36 = 7/12.

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Another approach:

P(first die is a 6 and 2nd die is ANY value) = 1/6 = 6/36
+ P(first die is a 5 and 2nd is 2 or more) = (1/6)(5/6) = 5/36  
+ P(first die is a 4 and 2nd is 3 or more) = (1/6)(4/6) = 4/36
+ P(first die is a 3 and 2nd is 4 or more) = (1/6)(3/6) = 3/36
+ P(first die is a 2 and 2nd is 5 or more) = (1/6)(2/6) = 2/36
+ P(first die is a 1 and 2nd is 6) = (1/6)(1/6) = 1/36

Adding these up gives (6+5+4+3+2+1)/36 = 21/36 = 7/12
</PRE>