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I am completely stuck on this probability question involving dice. Any hints or help in the right direction would be great.

You are given $6$ numbers: $2, 3, 4, 7, 8, 5: n_1, n_2,\dots , n_6$. Assume we change a six-sided die so that the probability of getting $k$ (where $k ∈ \{1, 2, 3, 4, 5, 6\}$ as usual) is equal to $x \times n_k$ for some fixed number $x$. What is the probability of getting a $6$?

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  • $\begingroup$ Is it decided that $n_6 = 5$, or could it be any of those six numbers? $\endgroup$ – Arthur Aug 13 '15 at 12:43
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We have: $$1=\sum_{k=1}^6 P(k) = x\left(\sum_{1}^6 n_k\right) = 29x.$$ So $x=\frac{1}{29}$. So what is $P(6)?$

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  • $\begingroup$ Thanks so much for the help, I see what I have to do now, would that make the probability $\frac{5}{29}$ then? $\endgroup$ – Ralph Aug 13 '15 at 13:57
  • $\begingroup$ Correct. Just follow the formula $P(6)=n_6x$. $\endgroup$ – Thomas Andrews Aug 13 '15 at 14:22
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$$P(k) = x n_k$$ Then as $k = 1, 2, \ldots, 6$ $$n_1x + n_2x+ \ldots +n_6 = 1 \Rightarrow x = \frac{1}{\sum_{k = 1}^6 n_k}$$

Therefore, $P(k = 6) = n_6 \frac{1}{\sum_{k = 1}^6 n_k} $

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