Question 981781
Ask the reverse -- what is the probability of rolling at most one 9 in ten tries: 
probability of zero 9's 
(11/12)^10 
probability of one 9 
10*(1/12) 
(because there are ten ways to do it and each has p = 1/12) 
Add these together and then subtract from 1 and you get the probability you seek: 
prob(at least two 9's) = 0.200