Here is an exam problem with the work shown:
A man and a woman agree to meet at a certain location at about 12:30 pm. The man will arrive at a time uniformly distributed between 12:15 and 12:45, and the woman will arrive at a time uniformly distributed between 12:00 and 1:00. Find the probability that the first to arrive has to wait no longer than 10 minutes.
Solution: Let $X$ be the time man arrives and $Y$ the time the woman arrives. We may assume that $X$ is uniform on $(15,45)$ and $Y$ uniform on $(0,60)$. We know that $X$ and $Y$ are independent. By drawing a picture and calculating the areas of appropriate regions we can conclude that the desired probability is $\frac23$.
Alternatively, by using that the joint density of $X$ and $Y$ is $\frac13\times\frac1{60}=\frac1{1800}$, we can compute $$\begin{align}\mathbb P(|X-Y|\le10)&=\color{blue}{\mathbb P(-10\le Y-X\le10)=\int_{15}^{45}\int_{x-10}^{x+10}}\frac1{1800}\,\mathrm dy\,\mathrm dx\\&=\frac1{1800}\int_{15}^{45}20=\frac1{1800}20\times30=\frac13\end{align}$$
I am extremely confused by the step that I circled. How did it change to $Y-X$, not $X-Y$?
This is my work:
Let $x$ denote the time at which the man arrives. Let $y$ denote the time at which the woman arrives.
$$x-y\le10,y-x\le10\to|x-y|\le10$$
$$\begin{align}
\mathbb P(|X-Y|\le10)&=\mathbb P(-10\le Y-X\le10)\\
&=\int_0^{60}\int_{y-10}^{y+10}\left(\frac1{30}\right)\left(\frac1{60}\right)\,\mathrm dx\,\mathrm dy\\
&=\int_0^{60}\left(\left[\frac1{1800}x\right]_{x=(y+10)}-\left[\frac1{1800}x\right]_{x=(y-10)}\right)\,\mathrm dy\\
&=\int_0^{60}\left[\frac{y+10}{1800}-\frac{y-10}{1800}\right]\,\mathrm dy\\
&=\int_0^{60}\left[\frac{20}{1800}\right]\,\mathrm dy\\
&=\frac1{90}\int_0^{60}1\,\mathrm dy\\
&=\frac{60}{90}\\
&=\frac23
\end{align}$$
I get $\frac23$, the professor gets $\frac13$ (but also says above it is $\frac13$?)
Best Answer
Note that, because $\mathsf P(\lvert X-Y\rvert \leq 10) = \mathsf P(\rvert Y-X\lvert \leq 10)$ , therefore it could go either way.
They chose to use $\mathsf P(-10\leq Y-X\leq 10)$ since that displays how they obtained the bounds for their inner integral. Here's what they did with a few more steps for clarity.
$\begin{align} \mathsf P(-10\leq Y-X\leq 10) & = \int_{15}^{45}\mathsf P(x-10\leq Y\leq x+10\mid X=x)f_X(x)\operatorname d x \\[1ex] & = \int_{10}^{45}\int_{\max\{0,x-10\}}^{\min\{60,x+10\}} f_{X}(x)f_{Y}(y)\operatorname d y\operatorname d x & \star \\[1ex] & = \int_{10}^{45}\int_{x-10}^{x+10} \frac 1{45-10}\cdot\frac1{60-0}\operatorname d y\operatorname d x \\[1ex] & = \frac{45-10}{45-10}\cdot \frac{20}{60} \\[2ex] & = \frac 1 3 \end{align}$
Note: I included the $\max$ and $\min$ functions on the line marked $\star$ as a reminder to beware that the bounds don't exceed the support, as happens occasionally in these problems. Fortunately it is not a concern for this problem as all values of $X$ lying more than $10$ minutes inside the boundary of $Y$. But be careful.
In fact, this seems to be where you went awry.
$\begin{align} \mathsf P(-10\leq X-Y\leq 10) & = \int_{15}^{45}\mathsf P(y-10\leq X\leq y+10\mid Y=y)f_Y(y)\operatorname d y \\[1ex] & = \int_{0}^{60}\int_{\max\{15,y-10\}}^{\min\{45,y+10\}} f_{X}(x)f_{Y}(y)\operatorname d x\operatorname d y & \star \\[1ex] & = \int_0^{25}\int_{15}^{y+10}\frac 1{30}\frac 1{60} \operatorname d x\operatorname d y + \int_{25}^{30}\int_{y-10}^{y+10} \frac 1{30}\frac1{60}\operatorname d x\operatorname d y + \int_{30}^{60}\int_{y-10}^{45}\frac 1{30}\frac1{60}\operatorname d x\operatorname d y \\[2ex] & = \frac 1 3 \end{align}$