I have an unlimited supply of standard 6-sided dice. What's the fewest number of dice that I have to simultaneously roll to be a
t least 90% likely to roll at least one 6? You may use a calculator to help you with the computations if you like -- in fact you'll almost certainly want to -- but your final answer should be a positive integer, and you should explain how you got it.
P( rolling 6) = 1/6 P(NOT rolling 6) = 5/6 P(AT least one 6) = 1 - 5/6 P(After x number of trials to obtain at least one 6) = 0.9 Then 0.9 = 1 - (5/6)ˣ