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if an octahedral die is rolled 9 times, what is the probability that only the last 2 rolls will be a multiple of 3?
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If there are 2 multiples of three out of 8, then it's 2/8 x 2/8 = 1/16.
hmmm
But then you need to guarantee that the others are NOT multiples of three, so...
1/16 x (6/8)^7 = 2187/262144 = approximately 0.008342742919921875
Thanks for the help OakTree!
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