You have a list of chores to do at home, but are expecting family to arrive shortly. The amount of time until their arrival (measured in hours) can be modeled as an Exp(2) random variable. Your list of chores contain 4 activities, each of which take 15 minutes to complete. Let X be the (whole) number of chores you complete before your family arrives. Find the CDF of X.
So I set Y~Exp(2) = time until family's arrival and X as the number of chores done. I'm not really sure how to proceed from there?
Outline: The notation $\text{Exp}(2)$ is somewhat ambiguous. The exponential with density function $\lambda e^{-\lambda y}$, for $y\gt 0$, is sometimes described as being $\text{Exp}(\lambda)$ and sometimes described as being $\text{Exp}(1/\lambda)$. (Note that $1/\lambda$ is the mean.) I do not know what notation your course/book uses. So I will use $\lambda$, which could either be $2$ or $1/2$.
The tasks each take $1/4$ of an hour. The probability that $X=0$ is the probability the arrival time is $\lt \frac{1}{4}$. This is $1-e^{-\lambda/4}$.
The probability that $X=1$ is the probability that the arrival time is between $1/4$ and $1/2$. This is $e^{-\lambda/4}-e^{-2\lambda/4}$.
The probabilities that $X=2$, $X=3$, and $X=4$ are calculated in a similar way.
Now that you have the probabilities, the cumulative distribution can be found.