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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) that it should be considered for inclusion on the list of "Generalizations of commonly asked questions" –  Dilip Sarwate Dec 10 '12 at 13:06
    
The limit on one should be $x-5 $ to $60$ I think and this gives $3/5$, but that is the incorrect answer still. –  CAF Dec 10 '12 at 13:08
    
The condition is not $Y-X\le 5$ or $X-Y\le 5$ but rather $Y-X\le 5$ and $X-Y\le 5$. –  David Mitra Dec 10 '12 at 13:31
    
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! –  CAF Dec 10 '12 at 13:31
    
I edited my previous comment to point out where your solution first went awry. –  David Mitra Dec 10 '12 at 13:39
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