Find the Probability that duration of one run is longer than another One day a run begins at 5:31 p.m. and ends at 5:46 p.m. The following day another run begins at 5:31 p.m. and ends at 5:47 p.m. We have a watch to measure the runtime but it shows only hours and minutes (not seconds). What is the probability that the run the first day lasted longer then the second day? 
I keep getting that the run on Day1 can last from 14:01 to 15:59 minutes and Day2 can last from 15:01 to 16:59 minutes. Then:
$$
P(Day2<Day1) = [indpt]=  \int\int f_X(x)*f_Y(y) dxdy = 
$$
and assuming that the duration of the runs have a uniform distribution f = 1/(b-a) = 1/(59/30), I get:
$$
=\int\int (30/59)*(30/59) dxdy = (30/59)^2*0.95*0.95≈0.23
$$
when I integrate from 15+(1/60) to 15+(58/60), i.e. the period of time where Day2 can be smaller than Day1. This is incorrect since the answer should be 1/24 ≈ 0.0416, can anyone please show me where I'm wrong/ how to solve it?
 A: We have four times involved so define an r.v. for each. The range for each of the times is $1$ minute and it is simpler to use minutes as our units. We define the random variables as:
\begin{align}
X &= \text{Day 1 Start Time - 5:31pm (in minutes)} \\
Y &= \text{Day 2 Start Time - 5:31pm (in minutes)} \\
Z &= \text{Day 1 End Time - 5:46pm (in minutes)} \\
W &= \text{Day 2 End Time - 5:46pm (in minutes)} \\
\end{align}
So the pdfs are $f_X(x)=f_Y(y)=f_Z(z)=f_W(w)=1$ and the ranges are $X,Y,Z\in(0,1)$ and $W\in(1,2)$.
We want the probability that $Y-X\gt W-Z$. So we integrate as follows:
\begin{align}
P(Y-X\gt W-Z) &= \int_{x=0}^{1} \int_{y=x}^{1} \int_{z=1-(y-x)}^{1} \int_{w=1}^{z+(y-x)} 1\cdot 1\cdot 1\cdot 1\;dw\;dz\;dy\;dx \\
&= \int_{x=0}^{1} \int_{y=x}^{1} \int_{z=1-(y-x)}^{1} (z+y-x-1) \;dz\;dy\;dx \\
&= \int_{x=0}^{1} \int_{y=x}^{1} \left[z^2/2+yz-xz-z\right]_{z=1-y+x}^{1}  \;dy\;dx \\
&= \int_{x=0}^{1} \int_{y=x}^{1} \left(y^2/2-xy+x^2/2\right)  \;dy\;dx \\
&= \int_{x=0}^{1} \left[y^3/6-xy^2/2+x^2y/2\right]_{y=x}^{1}  \;dx \\
&= \int_{x=0}^{1} \left(1/6-x/2+x^2/2-x^3/6\right) \;dx \\
&= \left[x/6-x^2/4+x^3/6-x^4/24\right]_{x=0}^{1} \\
&= 1/24.
\end{align}
A: It seems to me the answer should be about $\frac{1}{8}$.
You're correct that the valid potential lengths are 14:01 - 15:59 and 15:01 - 16:59 for day 1 and day 2, respectively.  Just to make the numbers nicer let's recenter to 0, so that the times are 0:00 - 1:58 for day 1 and 1:00 - 2:58 for day 2.  Note the possible time in seconds for day 1's length is 118 seconds and ranges from 0 to 118 in seconds, while day 2 ranges from 60 to 178 seconds, also with total length 118 seconds.
Now let $X \sim Unif(0,118)$ and $Y \sim Unif(60,178)$ be random variables corresponding to the lengths of day 1's and day 2's runs, respectively.  Thus the random variables live on the square $[0,118] \times [60,178]$ and we want to know $P(X > Y)$.  This is simply the area of the triangle in the bottom right of this plot divided by the total area, which you can see is about $\frac{1}{8}$.
Specifically, the joint pdf is
$$
f_{X,Y}(x,y) = \frac{1}{118^2}
$$
and the probability is
\begin{align*}
P(X > Y) & = \int_{60}^{118} \int_y^{118}\frac{1}{118^2} \, dx \, dy \approx 0.1208.
\end{align*}
