If I pick any of these three: 6,8,10, for my even first number, I can
successfully pick any of these five: 11,12,13,14,15, for my second 'greater than
10' number.
That's (3)(5) = 15 ways to pick successfully.
If I pick 12 or 14 for my even first number, I can successfully pick my second
'greater than 10' number 4 ways. [That is, with 12, I can pick 11,13,14,15, or
with 14, I can pick 11,12,13,15]
That's (2)(4) = 8 more ways to pick successfully.
So there are 15+8=23 ways to pick successfully.
There are (11)(10)=110 ways to pick any first number, then any remaining second
number.
So the desired probability is 23/110 = 0.2090909...
Edwin