# the probability of the sum of a exponential r.v. and uniform r.v. that are independent?

I am currently working on an old exam for practice for my final. The question I am having trouble with is this.

"Suppose that X~Exponential(mean=1) and Y~Uniform[0,3] are independent continuous random variables. Compute Pr(X+Y<=2)?"

I know how to compute the sum of two independent random variables that have the same distribution, but not if they have different distributions. Do I have to use the convolution formula? or can I just assume that because they are independent I do find the probability of each distribution when they are equal to or less than two? or can I use some manipulation of Bayes' theorem?

The joint density is $\frac{1}{3}e^{-x}$ on the part $K$ of the first quadrant that is between the $x$-axis and the line $y=3$. Draw the region.
We want the probability that $X+Y\le 2$. Draw the line $x+y=2$. We want the probability that $(X,Y)$ lands in the part $C$ of $K$ that is "below" the line $x+y=2$. This probability is $$\iint_C \frac{1}{3}e^{-x}\,dx\,dy.$$ Express this integral as an iterated integral. It is most convenient to integrate first with respect to $x$, with $x$ going from $0$ to the boundary line $x=2-y$. Then $y$ goes from $0$ to $3$.
Remark: There are other ways to proceed, which can be marginally easier. For example, we could first find the probability that $X+Y\gt 2$. Then we are integrating with respect to $x$, from $2-y$ to $\infty$, and then with respect to $y$, from $0$ to $3$.