Probability Xavier and Yolanda meet for lunch Xavier and Yolanda plan to meet for lunch between noon and 1 p.m.
They arrive independently with uniform distribution on [0, 1]. Yolanda will
wait 30 min. for Xavier, but Xavier will only wait 15 min. for Yolanda.
What is the probability that they meet? 
I solved a problem similar to this by finding $P(\vert X-Y\vert\leq 30)$, but in that example they were both waiting for 30 minutes. How would I set up the bounds for the distribution function in this scenario, when they are waiting for different times?

Take $Y$ to be the time Yolanda arrives and $X$ the time Xavier arrives.
We have two cases. $X \leq Y$ or $Y \leq X$.


*

*If $X \leq Y$ in order for them to meet we need that $Y \leq X + 15$. Since $X$ and $Y$ are independent this happens with probability $P(X \leq Y) \cdot P(Y \leq X + 15)$.

*If $Y \leq X$ in order for them to meet we need that $X \leq Y + 30$. Since $X$ and $Y$ are independent this happens with probability $P(Y \leq X) \cdot P(X \leq Y + 30)$.


The final probability that they meet is the sum of the above probabilities
$$ P_{meet} = P(X \leq Y) \cdot P(Y \leq X + 15) +  P(X \leq Y) \cdot P(X \leq Y + 30)$$
