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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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    Solutions to this type of problem have been described in detail so many times on this site (see, for example, [here](http://math.stackexchange.com/q/250164/15941)) that it should be considered for inclusion on the list of "Generalizations of commonly asked questions"2012-12-10
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    The limit on one should be $x-5 $ to $60$ I think and this gives $3/5$, but that is the incorrect answer still.2012-12-10
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    The condition is not $Y-X\le 5$ or $X-Y\le 5$ but rather $Y-X\le 5$ *and* $X-Y\le 5$.2012-12-10
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    I understand that we use two lines, but I don't understand why we use them both at the same time. For the first case ($y-x \leq 5 => y \leq 5 +x$), so $y$ from $0$ to $5+x$ and $x$ anywhere in $[15,45]$. Similarly, for the second integral, $ x - y \leq 5 => y \geq x - 5$ so $y$ from $x-5$ to $60$ and $x$ in $[15,45]$. What is wrong with this? Thanks!2012-12-10
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    I edited my previous comment to point out where your solution first went awry.2012-12-10
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    Ok, thanks. I have the correct answer now. But I still don't understand why that works: Does making the 'or' into an 'and' not mean that both the man and the woman are the first to arrive, which does not make sense because only one of them can be the first to arrive. What did I miss? Thanks2012-12-10
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    You want $|X-Y|\le 5$. But $|z|\le a \iff z\le a\ \text{and}\ z\ge -a$. So you have $X-Y\le5$ and $X-Y\ge -5$. Note that $Y-X\le 5$ does not tell you who arrived first; for instance you could have $Y=0$ and $X=40$ here, or you could have $Y=40$ and $X=38$.2012-12-10
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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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