Suppose that the time between calls from your aunt Debie has an exponential distribution with a mean time of 3 days. What is the probability that you will get two calls in less than 3 days?
So trying to solve this I tried to calculate the probability $P(X_1+X_2\lt3)$ where $X_1 =$ days to first call and $X_2 = $ days to second call. My attempt to solve...
$$X_1+X_2 \lt 3$$ $$X_1 \lt 3-X_2$$ $$\frac{1}{9}\int_{0}^3\int_{0}^{3-X_2} e^\frac{-x_1+x_2}{3} dx_1dx_2 = 1$$
The probability isn't likely 100% so where did I make my mistake?
For clarity, what you need to integrate is:
$$\mathsf P(X_1+X_2\leq 3) = \frac 1 9 \int\limits_0^3 e^{-y/3}\left(\int\limits_0^{3-y} e^{-x/3}\operatorname d x\right)\operatorname d y$$
Just take it one step at a time.