Understanding the density matrix for pure states vs. mixed states So, there's a question in my textbook that explains the following scenario: 
A physicist runs two experiments A and B to prepare quantum systems in a variety of initial states. In experiment A he uses a probabilistic machine that can
prepare a single quantum system in one of $n$ possible pure states $\{\lvert \psi_1 \rangle, \lvert \psi_2 \rangle, ..., \lvert \psi_n \rangle \}$ with corresponding probabilities $\{\lvert p_1 \rangle, \lvert p_2 \rangle, ..., \lvert p_n \rangle \}$. In experiment $B$, instead, he generates m non-interacting quantum systems, each of them prepared in its corresponding lower energy state $\{\lvert \phi_1 \rangle, \lvert \phi_2 \rangle, ..., \lvert \phi_m \rangle \}$. Let $\rho_A$ and $\rho_B$ be the density matrix operators for the quantum states prepared in experiments $A$ and $B$, respectively.
And asks the following:


*

*Write down an expression for both $\rho_A$ and $\rho_B$ and derive an expression for the expected value of $\hat{O_A}$ and $\hat{O_B}$ (the Hermitian operators describing an observable of the system A and kth system of system B, respectively). 

*Let $P_j$ and $P_k$ be the projectors associated to the states $\lvert \psi_j \rangle$ and $\lvert \psi_k \rangle$ produced by $A$. Discuss whether the product $P_jP_k$ vanishes.
For question 1, I wrote the following for system $A$:
$$\rho_A = \lvert \psi_j\rangle\langle \psi_j\rvert \tag{1}$$
$$\hat{O_A} = Tr(\hat{O_A} \rho_A) = Tr(\hat{O_A}\lvert\psi_j\rangle\langle \psi_j\rvert) \tag{2}$$
My justification for equation 1 is that system A produces pure states only and that only "a single" state is generated, so therefore the density matrix only contains one term (here j represents the jth state generated). However, the question mentions their probabilities which I haven't included in my answer because only one state is generated. Should $p_j$ be the coefficient in equation (1) as opposed to just the number 1? 
For system $B$, I have the following:
$$\rho_B = \frac{1}{m}\sum_i\lvert \phi_i\rangle\langle \phi_i\rvert \tag{3}$$
My justification for this is that experiment B describes a mixed state, which produces a total of m states and therefore each state has a probability of $\frac{1}{m} $. I am unsure, however, about how to write the expected value for experiment $B$.
For the second question, I can express the product as
$$ P_jP_k = \lvert \psi_j\rangle\langle \psi_j\rvert \psi_k\rangle\langle \psi_k\rvert \tag{4}$$
But as I'm given no information regarding the orthogonality of the states for experiment $A$, how am I supposed to deduce whether this equation vanishes or not? 
If anyone could provide any hints or insight into my answers, it would be greatly appreciated. I am not fully confident that I correctly understand the differences between pure and mixed states.
 A: Each pure State $|\psi_n\rangle$ of the quantum system is a multi-particle state describing all particles of the system. So your pure states are all possible states of the system. Since it is not prepared in a state $|\psi\rangle$ there are plenty of possibilites of states in which your systems could be. 
Let us say you prepared your system in $|\psi\rangle$, then it is in a pure state. What physicists call a pure state is essentially the density matrix / statistical operator:
$$\hat{\rho} = |\psi\rangle \langle\psi|$$
Since you are given the probabilities of the state, which are just real numbers, you can assume your system to be in a mixed state. Also, probabilites are usually not denoted as kets. For a mixed state you write:
$$\hat{\rho} = \sum_i p_i|\psi_i\rangle \langle \psi_i|$$
Also for the probabilites must hold:
$$1 = \sum_i p_i$$
For your second question, the states are not required to be orthogonal nor form a complete orthonormal system, but they are required to be normalized.
I hope that helps you. 
Edit: For question 1 you should probably also consider writing out the trace using a complete orthonormal system of states e.g. the eigenstates of the density matrix.
