Exponential distribution: waiting time at post office 
  
*
  
*Consider a post office with two clerks. Three people, A, B and C, enter simultaneously. A and B go directly to the clerks, and C wait
  until either A or B leaves before he begins service. What is the
  probability that A is still in the post ofice after the other two have
  left when the service times are exponential with mean $\frac1{\mu}$?
  
*If $X$ is exponential with rate $\lambda$, show that $Y=[X]+1$ is geometric with parameter $p=1-e^{-1}$, where $[x]$is the largest
  integer less than or equal to $x$.

I am learning Poisson process. These are some basic exercise problems giving in the book but I am not able to get the answer.
For question (1), I try the following:
$$P(B+C<A)=\frac{2\mu}{\mu + 2\mu}=\frac23$$
I have made use of the fact that the sum of exponential r.v. is again exponential, while for exponential r.v. $X_1, X_2$, we have $$P(X_1<X_2)=\frac{\lambda_1}{\lambda_1 + \lambda_2}.$$
But the answer provided by the book is $\frac14$.
For question 2, I don't know how to start with. I write $P(Y<x)=P([X]+1<x)$. Then what can I do next?
Thanks in advance.
 A: For 1. Due to symmetry the following is obviously true $$P(X>Y)=\frac{1}{2}$$ for $X,Y \in \{A,B,C\}$, where $A, B, C$ denote the waiting times of clients $A, B, C$ respectively. Now due to the memoryless property (m.p.) of the exponential distribution we have that $$P(A>B+C)=\underbrace{P(A>B+C|A>B)}_{=P(A>C) \text{ due to m.p.}}P(A>B)=P(A>C)P(A>B)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$$
The explanation is the following:


*

*There is probability $1/2$ that $A$ finishes after $B$ (due to symmetry, there is no reason that it is something else).

*When $B$ finishes service, the service time of $C$ begins. Now the key is that the remaining time of $A$'s service is still exponential (with the same paremeter) due to the memorylessness of the exponential distribution. Thus the probability that the remaining time of $A$'s service will be longer than the time of $C$'s service is equal to the probability that simply the service time of $A$ is longer than the service time of $C$ which is (as argued before) equal to $1/2$. 


This argument is only true to the memoryless property of the exponential distribution.
For 2. Start by calculating the probability that $Y=k$ for an arbitrary $k \in \mathbb N$, i.e. $$\begin{align*}P(Y=k)&=P([X]+1=k)\\&=P([X]=k-1)=P(k-1\le X<k)=F_X(k)-F_X(k-1)\\&=1-e^{k-1}-(1-e^{-k})\\&=e^{-k}(1-e^{-1})=\left(e^{-1}\right)^k(1-e^{-1})\\&=:p^k(1-p)\end{align*}$$ with $p=e^{-1}<1$.
