# $\ X_i$ is discrete random variable, Compute $\ \sum X_i = 97$

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 ?

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

• 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. – Matti P. Jan 3 at 9:54
• 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? – bm1125 Jan 3 at 9:58

## 1 Answer

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 $$\frac{10!}{8!1!1!}$$.

• So if I understand correctly, because I have two distinct numbers ($\ 8,9$) I had to multiply $\ {10 \choose 2}$ by $\ 2!$ ? – bm1125 Jan 3 at 10:15
• 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. – drhab Jan 3 at 10:18