Question 1031741
Choose any two out of the 12 questions, which can be done in C(12,2) = 66 ways.  (The only choice for each one of these is the correct answer.)  After doing this, for each of the remaining 10 incorrectly-answered questions, there are two choices, either the wrong answer or a no answer at all, which can be done in {{{2^10 = 1024}}} ways.

Therefore the probability is {{{(C(12,2)*2^10)/3^12 = (66*1024)/531441 =0.127}}}, to three significant figures.