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A man and a woman decide to meet at $12:30$. If the man arrives at a time uniformly distributed between $12:15 - 12:45$, and if the woman independently arrives at a time uniformly distributed between $12:00 - 1:00$, find the probability that the first to arrive waits no longer than $5$ minutes.

Attempt: I can write $f_X(x) = 1/30, x \in\,[15,45]$ for the man and $f_Y(y) = 1/60, y \in\,[0,60]$ for the woman. Since their arriving times are independent events, I may write $f(x,y) = f_X(x)f_Y(y) = 1/1800$, where $X$ is the time that the man arrives and $Y$ is the time that the woman arrives.

So the condition we want is that $Y-X \leq 5$ or $X-Y \leq 5$ since we don't care who arrives first. This means I should compute: $ \iint_{(x,y): y-x \leq 5} f(x,y)\,dy\,dx + \iint_{(x,y): x-y \leq 5} f(x,y)\,dy\,dx.$ I think it is correct up to here, I believe where I go wrong is maybe in interpreting the limits. What I ended up computing was: $\int_{15}^{45} \int_0^{5+x} f(x,y)\,dy\,dx + \int_{15}^{45} \int_{5+x}^{60} f(x,y)\,dy\,dx$ which gives a number slightly greater than $1$. Can someone explain the limits? To get the limits I had, I drew a sketch of the line $ y = x+5 $ and considered portions either below or above this line depending on the case. Many thanks

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    @CAF You can solve this problem more easily if you note that all that those integrals are doing is computing areas of triangles or rectangles or trapezoids (times $\frac{1}{30}$, of course). Thus, simple mensuration, e.g. $\text{area of triangle} = \frac{1}{2}\times\text{base}\times\text{altitude}$ suffices.2012-12-10

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