If $x \in [0,3]$ and $y \in [0,4]$, what is the probability of $x < y$? Posting this separate question as per request from the answerer's comment on my previous question.
From Question 54 of the practice book of the GRE math subject test:
What if one were to choose numbers $x\in [0,3]$ and $y\in [0,4]$? Why is the probability of $x$ being less than $y$ equal to $\frac 58$? And if so, I take it that the complement, probability of $x \ge y$, is $\frac 38$.
 A: We have $x\in[0,3]$ and $y\in [0,4]$. We are looking for the probability that $x<y$. Essentially, we are looking for the probability that $(x,y)$ is not in the triangle bounded by $y=0$, $x=3$, and $x=y$, as opposed to inside. The area (i.e., measure) of the total set of possibilities is $12$, while the area of the triangle is $4.5$. This leaves an area of $7.5$ of favorable results. Thus, the answer is $$\frac{7.5}{12}=\frac{15}{24}=\frac{5}{8}.$$
A: If $y\in[3,4]$ (which happens with probability $\frac{1}{4}$), then $x<y$ almost surely. If $y\in[0,3]$ (which happens with probability $\frac{3}{4}$), then $x<y$ with probability $\frac{1}{2}$ by symmetry. Then, the desired probability is
$$
\frac{1}{4}\times 1+\frac{3}{4}\times\frac{1}{2}=\frac{5}{8}\cdot\tag{$*$}
$$

More rigorous version: (the short version is probably more useful under the GRE exam condition)
\begin{align*}
\Pr(X<Y)&=\Pr(X<Y \& Y\in[3,4])+\Pr(X<Y\& Y\in[0,3])
\end{align*}
which implies
$$
\Pr(X<Y)=\Pr(X<Y\mid Y\in[3,4])\Pr(Y\in[3,4])+\Pr(X<Y|Y\in[0,3])\Pr(Y\in[0,3]).
$$
This is in fact ($*$) because with the joint density of $(X,Y)$ being $\frac{1}{3}\frac{1}{4}=\frac{1}{12}$, we have
\begin{align*}
\Pr(X<Y\mid Y\in[3,4])&=\frac{\int_3^4\int_0^3(1/12)dx dy}{\Pr(Y\in[3,4])}=\frac{(1/4)}{(1/4)}=1
\end{align*}
and
$$
\Pr(X<Y|Y\in[0,3])=\frac{\int_0^3\int_0^y(1/12)dx dy}{\Pr(Y\in[0,3])}=\frac{3/8}{3/4}=\frac{1}{2}\cdot
$$
The "symmetry" argument earlier refers to the fact that
$$
\frac{\int_0^3\int_0^y(1/12)dx dy}{\Pr(Y\in[0,3])}=\frac{(1/12)\int_0^3\int_0^y1dx dy}{(3/4)}=\frac{\int_0^3\int_0^y1dx dy}{9}
$$
equals the ratio of half the area of a $3\times 3$ square to the area of the square itself, which of course is $\frac{1}{2}$.
