A problem with two uniformly distributed random variables A man and a woman agree to meet at a certain
location about $\text{12:30}$. If the man arrives at
a time uniformly distributed between $\text{12:15}$ and
$\text{12:45}$, and if the woman independently arrives at
a time uniformly distributed between $\text{12:00}$ and $\text{13:00}$, find the probability that the first to arrive
waits no longer than $5$ minutes. What is the probability
that the man arrives first?
Surely I have problems to understand the limit of integration for the second part.
(a) Let $X$ be the time the man arrives and $Y$ the time the woman arrives (both in fractions of 1 hour)
We want the probability $P(|X-Y|\le\frac{1}{12})$
The probability densities for each variable are
$$f_x(t) = \begin{cases}2  &\frac{1}{4} \le t \le \frac{3}{4}\\
0  & \text{otherwise}. \end{cases}$$
and
$$f_y(t) = \begin{cases}1  &0 \le t \le 1\\
0  & \text{otherwise}. \end{cases}$$
The probability is given by
$$P(|X-Y|\le\frac{1}{12})=P(Y-\frac{1}{12}\le X \le Y+\frac{1}{12})=
\int_{s=\frac{1}{4}}^{s=\frac{3}{4}} \int_{y-\frac{1}{12}}^{y+\frac{1}{12}} f_x(t)\cdot f_y(t) dt~ds = \int_{s=\frac{1}{4}}^{s=\frac{3}{4}} \left(\int_{y-\frac{1}{12}}^{y+\frac{1}{12}} f_x(t)~dt\right)\,ds = \frac{1}{6}$$
(b) For the second part we have to calculate $P(X<Y)$ but I don't know what to do about the limits of integration. could someone help me?
I'm finding these kind of problems very difficult. Could you suggest me a strategy ? 
 A: Are you simply asking for help with the second part?
We can think about it in the following way:
Can we compute the probability that $X$ is smaller than $t \in \mathbb{R}$? Can we compute the probability that $Y = t$? Then, we can solve the problem by conditioning.
$\mathbb{P}(X<Y) = \int_0^1 \mathbb{P}(X<Y|Y=t)f_Y(t)dt = \int_0^1 \mathbb{P}(X<t)dt$.
Now, split the last integral into the summation of integrals on sections on $[0,1/4), [1/4,3/4], (3/4,1]$. Can you solve it from here? 
----------EDIT------------
So how do we determine $\mathbb{P}(X<t)$? We know that $X$ is uniformly distributed on $[1/4,3/4]$; therefore, it can never be less than $1/4$ and must always be less than $3/4$. Hence, for $t<1/4$, $\mathbb{P}(X<t) = 0$, and for $t>3/4$, $\mathbb{P}(X<t) = 1$. Now, for $t\in [1/4,3/4]$, we have that $\mathbb{P}(X<t)$ follows the cumulative distribution function (CDF) of a uniform distribution; hence,
$\mathbb{P}(X<t) = \frac{t-1/4}{3/4-1/4} = 2t-1/2$.
If you are unsure about the CDF of a uniform distribution, try to determine it by integration and by simple geometry.
