time it takes to service a car with exponential random variable with rate 1 Need help with this question here. Ill post exactly what it says then show my ideas so far.
"The time it takes to service a car is an exponential random variable with rate 1. (a) If A brings his car in at time 0 and B brings her car in at time t, what is the probability that B's car is ready before A's car? (Assume service times are independent and service begins upon arrival of the car.)
(b) If both cars are brought in at time 0, with work starting on B's car only when A's car has been completely serviced, what is the probability B's car is ready before time 2?"
I have the answers from the back of the book. (a) $(1/2)e^{(-t)}$ (b) $1-3e^{(-2)}$
I tried setting up an integral: $\int_{t}^{\infty}\int(e^{(-x)})dydx$ where Y is A's car and X is B's car but I don't know what bounds to put for X. So I tried working backward from the answer and arrived at $(1/2)\int_{t}^{\infty}e^{(-x)}dx$. Is it $\int_{t}^{\infty}\int_{x}^{\infty}e^{(-(x+y))}dydx$?
All help is appreciated.
 A: I hope you can do the first part without integration.  First A has to take until least t which has probability $e^{-t}$.  Then at that time the probability B finishes first is 1/2 by symmetry.  The probability B finishes first is the product $(1/2)e^{-t}$.  The key thing to remember about the exponential distribution which may be counterintuitive is that it is memoryless, so the fact that they have already been working on A for some time does not make it any more likely to finish before B which they are just starting on since A and B have the same exponential distributions of waiting times.  It's similar to flipping a coin until you get heads.  If you've already gotten 100 tails in a row, you are no more likely to get a head any sooner than you were when you started.  The waiting time there is a geometric distribution which is the discrete analogue of the exponential distribution.
If you do set up integrals for the first part as an exercise, keep in mind that the variable for B is the waiting time relative to when it starts, not absolute time relative to time 0 when A starts.
For the second part you can set up a double integral.  It doesn't matter whether you use A or B for the outer integral as long as it goes from 0 to 2, and then set the limits for inner one accordingly.  The amount of that 2 taken by one takes will reduce the amount that the other can take.
A: Both cars has to be finished before time 2. Thus you have to calculate $P(X+Y<2)$.
$$P(X+Y<2)=\int_0^2 \int_0^{2-y}e^{-(x+y)} \, dx \, dy$$
A has to be finished before B. Thus A has time from $0$ to $2-y$. And the finishing time of B has to always greater than the finishing time of A: $y>2-x$ 
X can vary from 0 to 2. Therefore Y can vary from 0 to 2, too.
