Here are all 36 ways the two dice can fall

(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)



Now I will list them again with all the rolls 
whose sum is greater than or equal to 10 red
and the other black:

(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)


There a 6 red ones out of the 36. 

The probability of rolling a sum greater than or equal to 10 therefore
is  or .

The probability of rolling a sum of 9 or lower is 


The complement event is to roll a sum of 9 or less twice,

which is P(black roll 1st AND black roll 2nd).  Since AND
means multiply, the complement event is

P(black roll 1st)*P(black roll 2nd) = 

So the desired answer is 1 minus that:

Answer = 

------------------

There is a longer way to do this than to this which does not involve
the complemnent event.  Let's do it that way as a check:


P[(red 1st AND black 2nd) OR (black 1st and red second) OR (red first AND red 2nd)]

Since AND mean "multiply" and OR means "add", the above equals:

P(red 1st)*P(black 2nd) + P(black 1st)*P(red second) + P(red first)*P(red 2nd)

         

                 

                          

Same answer.

Edwin