1) This is just a simple combination problem in disguise, because any combination of three cards for Jon automatically leaves Ryan with the other nine. So it's just:
nCr = n! / ( r! (n - r)! )
where n = 12 and r = 3. So...
= 12! / ( 3! (12 - 3)! ) = 479,001,600 / ( 6 * 362880 )
= 479,001,600 / 2,177,280 = 220
If it's really worded exactly as you stated, then you have to multiply by two, for an answer of 440, because for each of the 220 combinations, it could be Jon with three cards and Ryan with nine, or Ryan with three cards and Jon with nine, and those are different ways of dividing the cards. But I would consider that in the realm of "trick questions" of the kind that rarely appear on CLEP, DANTES or ECE exams.
2) There are 36 possible combinations of two throws of a die, of which 11 have a 1.
1-1
1-2
1-3
1-4
1-5
1-6
2-1
3-1
4-1
5-1
6-1
So the answer is 11/36. Technically the "right" way to solve it is to figure the probability of it
not happening, and subtract that result from one. So you multiply the probabilities of two "I didn't roll a one" throws (5/6 * 5/6) and subtract the result from one. (1 - 5/6*5/6 = 1 - 25/36 = 11/36)
The reason you have to do it that way, is that multiplying two probabilities gives the combined probability of both events happening -- an AND condition. But that doesn't work for an OR condition, so you have to convert that into (NOT A) AND (NOT B), and then subtract the result from 1 (absolute certainty). I hope that makes some sense.
-Gary-