# Find the probability that the woman has to wait longer than 7 minutes for the arrival of the man.

A man and a woman agree to meet at a restaurant between 6:00 and 6:15. Assuming their arrival times are uniformly distributed over the interval (0,20) and that their arrivals are independent, find the probability that the woman has to wait longer than 7 minutes for the arrival of the man.

My main struggle is how to set up the limits of integration. I set up the bounds to be "0 < y < x+7 < 20".

So to integrate set up the integral for the y to be 0 to x+7, and I set up the integral for the x to be 0 to 13. Doing this double integral I got the probability to be 0.325. Now I think that this is correct, but I'm confused on how to know that I set up the integral limits correct.

For instance I tried setting up the bounds like this " 0 < y-7 < x < 20",

and then I set up the integral for the x to be y-7 to 20, and the integral for the y to be 0 to 27. Doing the double integral for that I got 0.91125, which is different from what I got from my previous solution. So I'm confused as to why it's different and how to set up the limits of integration for the double integral properly.

Thanks so much for your help.

• The integral of $y$ should still only be between $0$ and $20$, since $y$ is $0$ outside that range. – Nitin Nov 7 '16 at 21:22

Your first integral is correct. For the second, the bounds on $y$ should be $7$ to $20$. This is because the inequality $0<y-7<20$ becomes $7<y<27$, but $y$ must also satisfy the inequality $0<y<20$. Satisfying both of these is equivalent to saying $7<y<20$.