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Suppose that 8 dice are rolled. What is the probability that the sum of the eight dice is 9?

I would interpret this question as: What is the probability that we get exactly 7 ones and one 2. We have 8 possible indices and out of them we choose 7 for the ones. The remain index will go for the remaining two.

Therefore

$P(sum = 9) = \binom{8}{7} (\frac{1}{6})^8 (\frac{5}{6})^0$

Is this correct? Thanks in advance!

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  • $\begingroup$ What is that $(\frac{5}{6})^0$ for? $\endgroup$ – Jorge Fernández Hidalgo Feb 26 '15 at 14:24
  • $\begingroup$ I was using the general formula of $\binom{n}{k} \theta^k (1-\theta)^{n-k}$ $\endgroup$ – geomquestion Feb 26 '15 at 14:25
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There are eight possible outcomes (result sequences) that get you $7$ ones and one $2$. Each outcome has probability $\frac{1}{6^8}$ .Hence the probability is $\frac{8}{6^8}$

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  • $\begingroup$ Thanks! That's another way of looking at it, but much more straightforward. $\endgroup$ – geomquestion Feb 26 '15 at 14:25

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