post office probability Assume that each person that goes to post office, has three choices A, B and C to post anything. The probability of choosing each type is : p(A)=1/5, p(B)=3/5 and p(C)=1/5. How many people must go to the post office until each type would be selected for at least once?
+I mean the expectation of at least once.
what I think is that this problem can be solved by Geometric dis:


*

*X1: The number of people that must go to the post office to choose a new type of posting

*X2: The number of new people that must go to the post office to choose a new type of posting (While one type has been selected before)

*X3: The number of new people that must go to the post office to choose a new type of posting (While two types have been selected before)
 A: Let $N_A$ be the number of people that go to the post office until A is chosen. Likewise, $N_B$ and $N_C$ be the number of people that go to the post office until B and C are chosen respectively. Therefore, $N =\max(N_A, N_B, N_C)$ is the number of people that go to the post office until each type is selected at least once. We want to find $\mathbb{E}(N)$. By inclusion-exclusion principle,
$N = \max(N_A, N_B, N_C) = N_A +  N_B +  N_C - \min(N_A, N_B) -  \min(N_A, N_C) - \min(N_B, N_C) + \min(N_A, N_B, N_C)$
Therefore, by linearity of expectation
$\mathbb{E}(N) = \mathbb{E}(\max(N_A, N_B, N_C)) = \mathbb{E}(N_A) +  \mathbb{E}(N_B) +  \mathbb{E}(N_C) - \mathbb{E}(\min(N_A, N_B)) -  \mathbb{E}(\min(N_A, N_C)) - \mathbb{E}(\min(N_B, N_C)) + \mathbb{E}(\min(N_A, N_B, N_C))$
Using the fact that
$N_A \sim \text{Geom} \left(\frac{1}{5}\right)$, $N_B \sim \text{Geom} \left(\frac{3}{5}\right)$, $N_C \sim \text{Geom} \left(\frac{1}{5}\right),\min(N_A, N_B) \sim \text{Geom} \left(\frac{4}{5}\right)$, $\min(N_A, N_C) \sim \text{Geom} \left(\frac{2}{5}\right)$, $\min(N_B,N_C) \sim \text{Geom} \left(\frac{4}{5}\right)$ and $\min(N_A, N_B, N_C) \sim \text{Geom} \left(1\right)$
we get
$\mathbb{E}(N) = \mathbb{E}(\max(N_A, N_B, N_C)) = 5 + \frac{5}{3} +  5 - \frac{5}{4} -  \frac{5}{2} - \frac{5}{4} + 1 = \frac{23}{3}$.
A: Since the probabilities $p_A$, $p_B$, $p_C$ are unequal there are $2^3=8$ states to consider, corresponding to the eight different subsets $X$ of $\{A,B,C\}$. For each subset $X$ denote by $E_X$ the expected number of additional customers until the postmaster has seen all three types of customers, assuming that he has already  seen the types contained in $X$. We then have the eight unknowns
$$E_\emptyset,\quad E_A,\quad E_B,\quad E_C,\quad, E_{AB},\quad E_{AC},\quad E_{BC},\quad E_{ABC}=0\ .$$
Among these variables we have equations of the following kind:
$$E_{BC}=1+p_A E_{ABC}+p_B E_{BC}+p_C E_{BC}\ ,$$
and six more like this. Solve the resulting system, and $E_\emptyset$ will be your final answer. (I obtained $E_\emptyset={23\over3}$.)
A: Do you mean, expectation of at least one?
Because if not, there is no finite amount of people which can guarantee that each post will be posted. 
