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so there are 3 ways when we can get 2 dies to have the sum of 4 (1,3),(2,2),(3,1) and there are 36 possible outcomes so we would expect the dies to have a sum of 4 1/12 times
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i'm pretty sure that's not the answer.
the anseweris 83 my bad E(x)=N.p binomial distribution
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pretty sure
Okay thanks.
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