<PRE><B>Question 1043409</B>
&lt;pre&gt;
Here are all 36 possible rolls with the 7&#39;s and 11&#39;s 
in red:  



(1,1) (1,2) (1,3) (1,4) (1,5) &lt;font color = &quot;red&quot;&gt;&lt;b&gt;(1,6)&lt;/b&gt;&lt;/font&gt;

(2,1) (2,2) (2,3) (2,4) &lt;font color = &quot;red&quot;&gt;&lt;b&gt;(2,5)&lt;/b&gt;&lt;/font&gt; (2,6) 

(3,1) (3,2) (3,3) &lt;font color = &quot;red&quot;&gt;&lt;b&gt;(3,4)&lt;/b&gt;&lt;/font&gt; (3,5) (3,6) 

(4,1) (4,2) &lt;font color = &quot;red&quot;&gt;&lt;b&gt;(4,3)&lt;/b&gt;&lt;/font&gt; (4,4) (4,5) (4,6) 

(5,1) &lt;font color = &quot;red&quot;&gt;&lt;b&gt;(5,2)&lt;/b&gt;&lt;/font&gt; (5,3) (5,4) (5,5) &lt;font color = &quot;red&quot;&gt;&lt;b&gt;(5,6)&lt;/b&gt;&lt;/font&gt; 

&lt;font color = &quot;red&quot;&gt;&lt;b&gt;(6,1)&lt;/b&gt;&lt;/font&gt; (6,2) (6,3) (6,4) &lt;font color = &quot;red&quot;&gt;&lt;b&gt;(6,5)&lt;/b&gt;&lt;/font&gt; (6,6) 

Count &#39;em!  There are 8 red ones, so that&#39;s 8 out of 36,
or a probability of {{{8/36}}} which reduces to {{{2/9}}}.

Edwin&lt;/pre&gt;</PRE>