The Statement of the Problem:
Suppose that $X$ and $Y$ are independent random variables with $F_X(x) = 2e^{-2x}, x \gt 0,$ and $F_Y(y) = 3e^{-3y}, y \gt 0.$
Find $ \mathbb P \{ 2X + 3Y \gt 2 \}$.
SOLUTION:
The solution states that $2X \sim 2\text{Expo}(2) \sim \text{Expo}(1)$ and $2Y \sim 3\text{Expo}(3) \sim \text{Expo}(1)$ and proceeds from there by identifying the sum of the two with a gamma-distributed random variable. This, of course, implies that $X \sim \text{Expo}(2)$ and $Y \sim \text{Expo}(3)$. However, what's given above in the statement of the problem I take to be the CDFs of these random variables (because of the capital "$F$"s), not the PDFs. And so, differentiating the above (to obtain the PDFs), we have
$$f_X(x) = -4e^{-2x} \text{ and } f_Y(y) = -9e^{-3y} $$
which are obviously not the PDFs of two exponentially distributed random variables with parameters $2$ and $3$, respectively. Is this mistake, or is this there some "trick" I'm missing here?
Obviously, the given functions "$F_X$" and "$F_Y$" must be density functions. If they were CDF's, they must be non decreasing. However, both of them are decreasing. It must be a mistake!