Let X be uniform on $[0, 10]$. Let $Y$ be exponential with $E(Y ) = 5$. Find $P(X < Y )$ From the given information above. I'm able to derive that
$f(x) = \cfrac{1}{10}$ from $0\leq x \leq 10$
and
$f(y) = \cfrac{1}{5}e^{\cfrac{-1}{5}y}$ for $0 < y$
I think since they're asking for P(X < Y) it's safe to assume that X and Y are independent (is this a fair assumption?) so $f(x,y) = \cfrac{1}{50}e^{\cfrac{-1}{5}y}$ for $0 \leq x \leq 10$ and $0 < y$
so $P(X < Y) = \int_0^{\infty}\int_0^y\cfrac{1}{50}e^{\cfrac{-1}{5}y}dxdy = \cfrac{1}{2}$
Have I understood and done this problem correctly?

$$P(X<Y)=\int_{[0,10]\times[0,\infty)}1_{x\le y}f_{X,Y}(x,y)d(x,y)$$
Now, since $X$ and $Y$ are independent, the joint density function is the product $f_{X,Y}(x,y)=f_X(x)f_Y(y)$. Using partial integration, one finds
$$\begin{align*}
\int_{[0,10]\times[0,\infty)}1_{x\le y}f_{X,Y}(x,y)d(x,y) &=\int_0^\infty \int_0^{10} 1_{x\le y}\frac{1}{10}\frac{1}{5}e^{-\frac{1}{5}y}dxdy\\
&=\int_0^\infty\int_0^{min(10,y)} \frac{1}{10}\frac{1}{5}e^{-\frac{1}{5}y}dxdy\\
&=\int_0^{10}\frac{y}{10}\frac{1}{5}e^{-\frac{1}{5}y}dy+\int_{10}^\infty\frac{10}{10}\frac{1}{5}e^{-\frac{1}{5}y}dy\\
&=\frac{1}{50}\Big[-5ye^{-\frac{1}{5}y}\Big]_{y=0}^{y=10}-\frac{1}{50}\int_0^{10}-5e^{-\frac{1}{5}y}dy+e^{-2}\\
&=-e^{-2}+-\frac{1}{2}(e^{-2}-1)+e^{-2}\\
&=\frac{1}{2}(1-e^{-2})
\end{align*}$$
