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**iknowone** 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.