The time is now $t = 0$. Person $A$ arrives somewhere between times $0$ and $45$ uniformly at random. They stay for $30$ minutes and then leave. Person $B$ arrives somewhere between $30$ and $60$ uniformly at random. They stay for $15$ minutes and then leave. What's the probability that the two people see each other?
I've tried many things like drawing a picture geometrically (can't get it right) and trying to take a double integral. The pdf for $A$ and $B$ are $f_{A}(x) = 1/45$ and $f_{B}(x) = 1/30$ so by independence, the joint pdf is $1/1350$. Then, my bounds were
$$\int_{30}^{45} \int_{y - 15}^{y + 15} \frac{1}{1350} \mathop{dy dx}$$
but it seems like I'm going out of the desired region. Can someone help me please?
EDIT (another attempt): I think the answer is
$$\int_{30}^{60} \int_{y - 15}^{45} \frac{1}{1350} \mathop{dydx} = \frac{1}{3},$$
is it right?
Notice that if A arrives after time $30$, they must see each other (if B arrives first, B stays until at least $45$, which is the latest A can arrive, and if A arrives first, A stays until at least $60$).
If A arrives at time $t<30$, then they see each other if and only if B arrives between $30$ and $t+30$, which is an interval of size $t$. So the probability of meeting is $$\int_0^{30}\frac{t}{30\times 45}\,\mathrm dt+\int_{30}^{45}\frac{1}{45}\,\mathrm dt.$$