# Continuous random variables – probability of a kid arriving on time for school

A father leaves his home to take his son to school between 6:15 AM and 6:45 AM and it usually takes between 30 and 40 minutes to arrive there. Let $$XX$$ and $$YY$$ be independent and random uniformly distributed continuous variables that stand for, respectively, the time of departure from home and the time spent on the way to school. What is the probability that the kid will arrive before class begins, at 7 AM?

(I’m sorry if something is misspelled or in an unusual order. I’m not a native speaker so I’m not sure if the order on “random uniformly distributed continuous variables” is correct)

I frankly have very little idea regarding how to proceed here. The first thing I’d thought about doing is converting the times to only minutes, in such a way as to make $$XX$$ and $$YY$$ comparable (in a sense). After that, imagining 7 AM as 420 minutes, we’d need to have $$X+Y<420X+Y < 420$$. After this, I'm stuck (I'm not even sure the reasoning prior to this is solid).

Any help is highly appreciated!

As you suggested, $$XX$$ and $$YY$$ can be described as two independent uniform random variables $$X∼U(375,405)X \sim \mathcal{U(375, 405)}$$, $$Y∼U(30,40)Y \sim \mathcal{U(30, 40)}$$. We are interesting in finding $$P[X+Y≤420]\mathbb{P}[X + Y \leq 420]$$. This problem can be handled with a straightforward geometric approach.
$$P[X+Y≤420]=grey areatotal area=Δy×d1+12Δy×d2Δx×Δy=5+12⋅1030=13,\mathbb{P}[X + Y \leq 420] = \frac{\text{grey area}}{\text{total area}} = \frac{ \Delta y \times d_1+ \frac{1}{2}\Delta y \times d_2}{\Delta x \times \Delta y} = \frac{5 + \frac{1}{2}\cdot10}{30} = \frac{1}{3},$$
where $$Δx=405−375=30\Delta x = 405 - 375 = 30$$, $$Δy=40−30=10\Delta y = 40 - 30 = 10$$, $$d1=(420−40)−375d_1 = (420-40)-375$$ and $$d2=390−(420−40)d_2 = 390 - (420-40)$$