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Let $\ X_1, X_2, , \dots , X_{10} $ be a discrete random variable with uniform distribution between $\ 0 $ to $\ 10 $. Compute $\ P\{ \sum_{i=1}^{10} \ X_i = 97 \} $, the variables are independent.

My attempt:

So I can either get 97 by having nine 10's and a 7, eight 10's and 9 & 8 or seven 10's and three 9's ?

$$\ P\{ \sum_{i=1}^{10} X_i = 97 \} = {10 \choose 1 } \cdot \frac{1}{11}^{10} + {10 \choose 2} \cdot \frac{1}{11}^{10} + {10 \choose 3} \cdot \frac{1}{11}^{10} $$

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  • $\begingroup$ Well, you could consider the number of integer solutions to $$ x_1 + x_2 + \ldots + x_{10} = 97 $$ You can do it with the stars and bars method. $\endgroup$ – Matti P. Jan 3 at 9:54
  • $\begingroup$ Yes at first I tried to use the formula of k objects into n bins but I have a limitation since I can only have up to 10 objects in each bin. Could you maybe guide me how to apply this limitation? $\endgroup$ – bm1125 Jan 3 at 9:58
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Your solution is almost correct.

The correct solution is:$$11^{-10}\left(\frac{10!}{9!1!}+\frac{10!}{8!1!1!}+\frac{10!}{7!3!}\right)$$

Observe that $8$ and $9$ are distinct numbers.

So e.g. $(10,10,9,10,10,10,10,10,8,10)$ and $(10,10,8,10,10,10,10,10,9,10)$ are distinct possibilities.

That is why $\binom{10}{2}=\frac{10!}{8!2!}$ must be replaced by $\frac{10!}{8!1!1!}$.

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  • $\begingroup$ So if I understand correctly, because I have two distinct numbers ($\ 8,9 $) I had to multiply $\ {10 \choose 2} $ by $\ 2! $ ? $\endgroup$ – bm1125 Jan 3 at 10:15
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    $\begingroup$ That is a way to put it. Personally I would rather say that there are $\frac{10!}{8!1!1!}$ (multinomial coefficient) ways to split up $10$ objects in $3$ separate groups consisting of $8$, $1$ and $1$ objects respectively. $\endgroup$ – drhab Jan 3 at 10:18

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