Uniform Probability Distribution CDF and Probability Suppose a value $x$ is chosen at random in the interval $[0,10]$. In other words, $x$ is an observed value of a random variable $X \sim \mathrm{UNIF}(0,10)$. The value $x$ divides the interval $[0,10]$ into two sub-intervals. 
a.) Find the CDF of the length of the shorter sub-interval.
b.) What is the probability that the ration of lengths of the shorter to the longer sub-interval is less than $\frac{1}{4}$
I have no idea to start with this problem. The book says the answer to a.) is $\mathrm{UNIF}(0,5)$ but I am not sure how they found this as a CDF. 
 A: When the cut falls in $[0;5]$, the shorter-length will be uniformly distributed over $[0;5]$ and when the cut falls in $[5;10]$, the shorter-length be uniformly distributed over $[10-10;10-5]$... so...
Alternatively:
Let $X$ the cut-point and $S$ be the shorter-length.   Now for any length $s: 0\leq s\leq 5$ then $S=s$ means either $X=s$ or $X=10-s$.   So the probability density of $S$ is:
$$\begin{align}f_{S}(s) & = (\;f_X(s)+f_X(10-s)\;)\quad \big[0\leq s\leq 5\big] \\[2ex] & = \tfrac 1 5\quad \big[0\leq s\leq 5\big]\end{align}$$
Which is the probability density function of $\mathcal U[0;5]$
Hence: $\Pr(0\leq S\leq s) = \frac s 5 \quad \big[0\leq s\leq 5\big]$

Now you wish to find $\Pr(0\leq \frac{S}{10-S}\leq \frac 1 4)$
A: Note that the length of the shorter interval ranges from $0$ to $5$. If it crosses $5$, then the other interval becomes the 'shorter' interval. Since the longer interval follows a uniform distribution (let us denote it by $X$), then $10 - X$ is also a uniform distribution which ranges from $0$ to $5$.
Therefore, $CDF = \frac{x}{2}, x \in [0,10] = UNIF(0,5)$
