I need some help to solve this problem:
There are two shops and each can serve at most one client per period. The number of clients that arrive in each period to each shop is
\begin{cases} 0 & \text{with probability} & p\\ 1 & \text{with probability} & 1-2p\\ 2 & \text{with probability} & p \end{cases}
where $p \in [0, 1/2]$. Clients arrivals to the two shops are independent.
Question 1: What is the expected number of clients the two shops combined serve per period?
Assume now the two shops can cooperate: they still can serve at most one client per period but if one of them is faced with 2 clients and the other has none, the overburdened shop can divert one of his clients to the other shop.
Question 2: Again, what is the expected number of clients they are able to serve per period?
What I come up with is the following:
Answer of question 1: Each shop serves $1$ client with probability $1-2p$ (when one comes) and $1$ client with probability $p$ (since when two comes, it can only serve one). This means each shop serves an expected number of clients equal to $1-2p + p = 1-p$. Since the two shops face the same arrival process, they combined serve on average $2(1-p)$ clients. Does it seem correct to you? If yes, do you have an alternative solution?
Answer of question 2: here I need your help! I have some ideas but I would like to see your opinion on how this can be solved!
Thanks!
Let $X_i$, $i=0,1$ be the number of customers that arrive at shop $i$ during some given period. Let $C_i$, $i=0,1$ be the number of customers served by shop $i$ during that period. For question 1, it is clear that $C_i = \min\{X_i, 1\}$, and hence \begin{align} \mathbb P(C_i=0) &= \mathbb P(X_i=0) = p\\ \mathbb P(C_i=1) &= \mathbb P(X_i>0) = 1-p. \end{align} It follows that $$ \mathbb E[C_i] = 0\cdot p + 1(1-p) = 1-p, $$ and so the expected number of customers served is $$ \mathbb E[C_1+C_2] = \mathbb E[C_1] + \mathbb E[C_2] = 2(1-p). $$ For question 2, we have $$ C_i = \min\{X_i,1\} + \mathsf 1_{\{X_i=0,X_{1-i}=2\}}, $$ and hence \begin{align} \mathbb P(C_i=0) &= p(1-p)\\ \mathbb P(C_i=1) &= 1-p(1-p). \end{align} So the expected number of customers served is $$ \mathbb E[C_1] + \mathbb E[C_2] = 2(1-p(1-p)). $$