<PRE><B>Question 1075570</B>
 Roll two dice once. Let A be the event that the sum 
of the two dice is 10... 
a) P(A)=
&lt;pre&gt;
Here are all 36 possible dice rolls with the
ones with sum 10 colored red:

(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) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(4,6)&lt;/b&gt;&lt;/font&gt; 

(5,1) (5,2) (5,3) (5,4) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(5,5)&lt;/b&gt;&lt;/font&gt; (5,6)

(6,1) (6,2) (6,3) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(6,4)&lt;/b&gt;&lt;/font&gt; (6,5) (6,6)

Count the red ones, there are 3.  

So P(A) is 3 out of 36 = 3/36 = 1/12
&lt;/pre&gt;
...and B be the event that at least 
one of the dice is odd.
b) P(B)=  
&lt;pre&gt;
Here are all 36 possible dice rolls with the 
ones that have at least one of the dice odd in red:

&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)&lt;/b&gt;&lt;/font&gt;

&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(2,1)&lt;/b&gt;&lt;/font&gt; (2,2) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(2,3)&lt;/b&gt;&lt;/font&gt; (2,4) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(2,5)&lt;/b&gt;&lt;/font&gt; (2,6)

&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)&lt;/b&gt;&lt;/font&gt; 

&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(4,1)&lt;/b&gt;&lt;/font&gt; (4,2) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(4,3)&lt;/b&gt;&lt;/font&gt; (4,4) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(4,5)&lt;/b&gt;&lt;/font&gt; (4,6) 

&lt;font color=&quot;red&quot;&gt;&lt;b&gt;(5,1) (5,2) (5,3) (5,4) (5,5) (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) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(6,3)&lt;/b&gt;&lt;/font&gt; (6,4) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(6,5)&lt;/b&gt;&lt;/font&gt; (6,6)

Count the red ones, there are 27.  

So P(B) is 27 out of 36 = 27/36 = 3/4
&lt;/pre&gt;
c) Find the probability that the sum of the two dice is at least 9.
&lt;pre&gt;

Here are all 36 possible dice rolls with the
ones with sum of the dice at least 9 colored red:

(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) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(3,6)&lt;/b&gt;&lt;/font&gt; 

(4,1) (4,2) (4,3) (4,4) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(4,5) (4,6)&lt;/b&gt;&lt;/font&gt; 

(5,1) (5,2) (5,3) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(5,4) (5,5) (5,6)&lt;/b&gt;&lt;/font&gt;

(6,1) (6,2) &lt;font color=&quot;red&quot;&gt;&lt;b&gt;(6,3) (6,4) (6,5) (6,6)&lt;/b&gt;&lt;/font&gt;

Count the red ones, there are 10.  

So P(sum&amp;#8805;9) is 10 out of 36 = 10/36 = 5/18.

Edwin&lt;/pre&gt; </PRE>