Since you want at least two sixes, figure the probability of 0 or 1 six, then take the complement:
P(0) = (5/6)^14 * (1/6)^0 * 14!/0!/14!
P(1) = (5/6)^13 * (1/6)^1 * 14!/1!13!
calculate that and subtract it from 1. That's your answer.
A 6-sided die is rolled 14 times. What is the probability of seeing a "6" at least 2 times?
I am sooo clueless please help!
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