I was looking at a very recent [question](https://math.stackexchange.com/questions/2698475/probability-that-the-sum-of-n-positive-numbers-less-than-2-is-less-than-2#2698475), and I wanted to generalize it as 
such:

> 
> Suppose that $\{x\_1, x\_2, \cdots x\_n\}$ are n independently uniformly
>  distributed real numbers from 0 to k, then what is the probablity that
>  their sum is less k?
> 
> 
> 

I want to check if this type of question is solvable using probability and Bayes Theorem as I have attempted below. Thus we require to find:
$$Pr(x\_1+x\_2+\cdots+x\_n\leq k | x\_1, x\_2, \cdots x\_n \in [0, k])$$
$$=\dfrac{Pr(x\_1, x\_2, \cdots x\_n \in [0, k] | x\_1+x\_2+\cdots+x\_n\leq k)Pr(x\_1+x\_2+\cdots+x\_n\leq k)}{Pr(x\_1, x\_2, \cdots x\_n \in [0, k])}$$

Now: 

$$Pr(x\_1, x\_2, \cdots x\_n \in [0, k] | x\_1+x\_2+\cdots+x\_n\leq k) = 1$$

and 

$$Pr(x\_1, x\_2, \cdots x\_n \in [0, k])$$
$$=Pr(x\_1\in [0, k])\cdots Pr(x\_n\in [0, k]) $$
$$=\dfrac{x\_1\cdots x\_n}{k^n}$$
The above is true since the $x\_i$'s are independent and uniformly distributed between 0 and k.

Now how do you compute: $Pr(x\_1+x\_2+\cdots+x\_n\leq k)$ to find the last term? Is my use of Bayes Theorem correct?