Calculate $P(1 \le X \le 2)$ Let $X$ be a random variable with distribution function
$$f(X)= \left\{\begin{array}{ll}
0, &  X < 0, \\
X/8, & 0 \le X < 1, \\
1/4 + X/8, & 1 \le X < 2, \\
3/4 + X/12, & 2 \le X < 3\end{array}\right.
$$
Calculate $P(1 \le X \le 2).$
 A: Hint: $P(a \leq X \leq b) = F(b) - lim_{n\rightarrow\infty}F(a-\frac{1}{n})$.
A: Let us analyze the cumulative distribution function at the interesting points $x=1$ and $x=2$. For reasons of tradition, we will call the cdf by the name $F(x)$.
As $x$ approaches $1$ from the left, $F(x)$ approaches $\frac{1}{8}$.
But note that from the definition given, we have $F(1)=\frac{1}{4}$. So $\Pr(X\le 1)=\frac{1}{4}$. There is a dramatic jump in probability when we hit $x=1$.
This means there must be a point mass of $\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$ at $x=1$. 
A similar analysis shows that there is a point mass of size $\frac{1}{4}$ at $x=2$. 
The weight up to and including $x=2$ is $\frac{3}{4}$. The weight $\le 1$ is $\frac{1}{4}$. 
Thus $\Pr(1\lt X\le 2)=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$.  But this is not quite what we want, we want $\Pr(1\le X\le 2)$. So we must add $\Pr(X=1)$ to $\Pr(1\lt X\le 2)$. The result is $\frac{1}{2}+\frac{1}{8}$.
Remark: I visualize the situation as follows. Our distribution is a hybrid between a discrete distribution and a continuous distribution.  From $x=0$ to $x=1$, we have wire of density $\frac{1}{8}$. At $x=1$, we have a point mass of $\frac{1}{8}$. From $1$ to $2$, we have wire of density $\frac{1}{8}$. At $2$ we have a point mass of $\frac{1}{4}$. From $2$ to $3$ we have wire of density $\frac{1}{12}$. There is nothing before $0$ or after $3$. 
