Given two random variables $X$ and $Y$ where $X \in [a, b]$ and $Y \in [c, d]$, $a < c < b < d$, what is the probability of $X$ and $Y$ being withing $Z$ units from each other?
For example: John and Jane both have a 45 minute lunch break. John eats sometime between 11 and 12.45, and Jane between 11.15 and 13.00. What is the probability of their lunch breaks overlapping?
Sorry, I'm not good with operating probabilities through densities and integrals, so this solution might be incorrect, but you'll get the idea:
I will reformulate the task as "John starts to eat between 11 and 12.45", etc.
Then we have that random variable $Y$, uniformly distributed on $[15,120]$ (in minutes) should get its value between $X$ and $X+45$, where random variable $X$ has uniform distribution on $[0,105]$.
$$P(X<Y<X+45) = P(Y<X+45) - P(Y<X)$$
To calculate these probabilities we will use convolution:
$$F_Y(y) = \frac{y-15}{120-15}, \ y \in [15,120], \ 0 \text{ if less than 15 and 1 if greater than 120}$$ $$f_X(x) = \frac{1}{105-0}, \ x \in [0;105], \ 0 \text{ otherwise}$$
Thus we calculate the second probability, restricting to the range of possible values of $Y$ and $X$ (as PDF of $X$ turns to $0$ elsewhere):
$$P(Y<X) = \int_{[15,105]}P(Y<x)f_x(x)dx = \int_{[15,105]}F_Y(x)f_X(x)dx =$$ $$= \int_{[15,105]}\frac{x-15}{120-15}\frac{1}{105-0}dx = \frac{18}{49}. $$
Then we compute the first probability that should be bigger:
$$P(Y<X+45) = \int_{[15,105]}P(Y<x+45)f_x(x)dx = \int_{[15,105]}F_Y(x+45)f_X(x)dx =$$ $$= \int_{[15,75]}\frac{x+45 -15}{120-15}\frac{1}{105-0}dx + \int_{[75,105]}1\frac{1}{105-0}dx = \frac{20}{49} + \frac{2}{7} = \frac{34}{49}. $$
So our probability should be
$$\frac{34}{49} - \frac{18}{49} = \frac{16}{49}. $$
