Example: (Ross, p.338 #48(a)). Consider an $n$ -server parallel queueing system where customers arrive according to a Poisson process with rate $\lambda,$ where the service times are exponential random variables with rate $\mu,$ and where any arrival finding all servers busy immediately departs without receiving any service. If an arrival finds all servers busy,
Question: find the expected number of busy servers found by the next arrival.
It is easy to show that if we had a single server, and at $t=0$, it was busy, than the probability that the customers find the server busy is $\frac{\lambda}{\mu + \lambda}$. Hence $$T_1 := \text{# of busy servers} = 1 * \frac{\lambda}{\mu + \lambda} + 0*(1-\frac{\lambda}{\mu + \lambda}) = \frac{\lambda}{\mu + \lambda}.$$
Given that we have $n$ out of $n$ busy servers at $t=0$ which are independent from each other, the expected number of busy servers should be $$n* \frac{\lambda}{\mu + \lambda},$$ but in the lecture notes where I got this example, the author first finds $T_1$, then considers a $k$-server system with all servers busy and separates the cases according to whether first a customers arrives or at least one of servers finishes it's job. Then it tries to form a recursive relation between $T_i's$, and obtains quite a complex answers.
However, I cannot understand what is wrong with my answers. After all, we throw $n$ independent coins, each of which having probability $p$ to get heads, then we would expect to see $p*n$ heads. But, in this case, for some reason the auhors thinks that this logic does not work. Why?
I just made small simulation of the above system, with $n = 10$ and #of simulations = 100000. As far as numbers tell, the expected number of busy servers found by the customers is quite close to
$$\sum_{i=1}^n \frac{u_i}{u_i+\lambda}.$$ and this is the answer the author gives
Your approach and result are correct. You mention independence twice. The linearity of expectation does not require independence; your approach and result would be correct even without it. Since you didn’t give any details about the author’s more complicated solution, there’s not much more to say.