Is the probability of getting AT LEAST one 6 1/36?
There are three disjoint possibilities:
EVENT A: Dice-1 gets a 6 and Dice-2 doesn't,
EVENT B: Dice-2 gets a 6 and dice 1 doesn't,
EVENT C: Both dice get a 6.
So the probability of getting AT LEAST one 6 is the sum of the probabilities of these three events. P(A) + P(B) + P(C) = 1/6*5/6 + 5/6*1/6 + 1/6*1/6 = 11/36.
You can get this same number by considering the only other possibility which is:
EVENT D: Neither dice rolls a 6.
This probability is P(D) = 5/6*5/6 = 25/36, and of course we want 1-P(D) = 11/36.