probability theory-expected vales for exponential distribution

Here's a problem that I got when I self-study probability theory with the course recorded at Harvard in stat 110.

A post office has 2 clerks. Alice enters the post office while 2 other customers,

Bob and Claire, are being served by the 2 clerks. She is next in line. Assume

that the time a clerk spends serving a customer has the Exponential(lambda) distribution.

(b) What is the expected total time that Alice needs to spend at the post

office?

the answer gives that the expected waiting time(waiting in line) is 1/(2lambda) and the expected time being served in 1/lambda, so the total time is 3/(2lambda)

I don't understand why the expected waiting time(waiting in line) is 1/(2lambda)

the solution says that the minimum of two independent

Exponentials is Exponential with rate parameter the sum of the two

individual rate parameters.

where does the rationale of the statement above come from.......

thanks..

Re: probability theory-expected vales for exponential distribution

Hey pyromania.

What you looking at comes from an area known as an order statistics. If you want to prove the result yourself calculate Min(A,B).

To start you off consider P[Min(A,B)] which is given by P(A > x and B > x) = P(C < x) = P(A > x)P(B > x) [independence] which is equal to [1 - P(A < x)][1 - P(B < x)] and P(A < x) is the CDF for A(x) and the P(B < x) is the CDF of B(x). Now differentiate both sides to get the PDF of C and compare the two.