probabilities of arrival time John and Mary arrive under the clock tower independently. Let X be John's arrival time and let Y be Mary's arrival time. If John arrives first and Mary is not there then he will leave. If Mary arrives first then she will wait up to one hour before leaving. John's arrival time, X, is exponentially distributed with mean of 1. Mary's arrival time has density $f(y) = {2y\over 9}$, $0≤y≤3$. Calculate the probability they will meet. 
What I've tried. 
I figured this problem should be the summation of two probabilities. The probability that Mary arrives before John + The probability that John arrives within an hour of Mary. 
the joint distribution is $f(x,y) = {2y\over 9}e^{-x}$, therefore the first probability could be written as $P[X>Y]$ of the joint distribution. 
$$\int_0^3 \int_0^y f(x,y) dxdy$$
this gives the answer of .822, which is well above the answer given in the book of .1125, without the second probability even being found. 
I'm not even sure how to set up the limits for the second probability.  
Any help would be appreciated. 

The probability that Mary arrives first, and within an hour of John can be calculated as follows. 
$$P[Y<X<Y+1] = \int_{0}^{3}\int_{0}^{y} \frac{2ye^{-x}}{9} = .822$$
