Calculate probability $P(\max(X,Y)>2)$ Let $X$ and $Y$ be a uniformly distributed random variable over the intervals $[0,4]$ and $[0,3]$, respectively. If $X$ and $Y$ are independent events, then compute the probability, $P(\max(X,Y)>2)$.
 A: Hint
$$\Pr\{\max[X,Y]>2\}=1-\Pr\{\max[X,Y]\le 2\}=1-\Pr\{X\le2,y\le2\}$$
A: $$P(\max(X,Y)>2)=P(X>2\cap X\ge Y)+P(Y>2\cap X<Y)$$
The remaining probabilities are easy to compute if you sketch a plot of the joint PDF, which is
$$P(X=x,Y=y)=\begin{cases}\frac1{12}&\text{for }(x,y)\in[0,4]\times[0,3]\\[1ex]0&\text{otherwise}\end{cases}$$
Then
$$P(X>2\cap X\ge Y)=\int_2^3\int_0^x\frac{\mathrm dy\,\mathrm dx}{12}+\int_3^4\int_0^3\frac{\mathrm dx\,\mathrm dy}{12}=\frac{11}{24}$$
$$P(Y>2\cap X<Y)=\int_2^3\int_0^y\frac{\mathrm dx\,\mathrm dy}{12}=\frac5{24}$$
which are the integrals of the joint PDF over the orange and green regions below, respectively. (The dashed line is $Y=X$. The region in blue is the rest of the support of the joint density. Elsewhere the probability is $0$.) The total probability is then $\frac23$.

A: Maybe the quickest way to do this is to find the probability that they're both less than $2,$ and then subtract that from $1.$
$$
\Pr(X<2\ \&\ Y<2) = \Pr(X<2)\Pr(Y<2) = \frac 2 4\cdot \frac 2 3 = \frac 1 3.
$$
Therefore the probability that at least one of them is more than $2$ is $2/3.$
