Jointly distributed exponential random variables Attempting to solve the following problem I am confused about what to use as the probability density function

Problem

The time that it takes to service a car is an exponential 
random variable with rate 1.

If A.J. brings his car in at time 0 and M.J.
brings her car in at time t, what is the probability 
that M. J.'s car is ready before A. J.'s car? 
(Assume that service times are independent and service 
begins upon arrival of the car.)


Exponential Random Variable

A continuous random variable whose probability density 
function is given, for some $\lambda &gt; 0$, by

$
f(x) = \left\{
\begin{array}{l l}
\lambda e^{-\lambda x} &amp;\text{if } x \ge 0\\
0 &amp;\text{if } x &lt; 0
\end{array} \right.
$

is said to be an exponential random variable (or, more simply, 
is said to be exponentially distributed) with parameter $\lambda$. 

How I'm approaching the problem

My thought is to let $\lambda = 1$, let $f( x, y) = \lambda^2 e^{-\lambda x} e^{-\lambda y}$, let $X$ denote A. J.'s car, let $Y$ denote M. J.'s car and solve for $P\{ X &lt; Y\} = \int\limits_0^\infty \int\limits_t^y f( x, y) dx dy$  but immediately I am troubled with the though that this is not the proper joint distribution function?
I would approach the problem as follows.  Assume that $t \ge 0$.  First find the probability that the service time for A.J.'s car is $\le t$.  Of course if this happens, then A.J. will beat M.J.  If the first car is not serviced by time $t$, then by the memorylessness of the exponential, the probability that A.J. finishes before M.J. is $1/2$.  Now you can put the pieces together. 

Or else more simply do the same calculation from M.J.'s point of view. For M.J. to "win," we need first of all to have A.J.'s  service time be $\ge t$. If that happens, the probability that M.J. wins is $1/2$.
The exponential distribution is memoryless. So by the time MJ arrives, AJs service starts out as a fresh new exponential. So $P=0.5$ since both have the same distribution