Probability of measuring two observables in a mixed state Lets say i have density Matrix on the usual base
$$ \rho = \left(
\begin{array}{cccc}
 \frac{3}{14} & \frac{3}{14} & 0 & 0 \\
 \frac{3}{14} & \frac{3}{14} & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & \frac{4}{7} \\
\end{array}
\right) $$
from this two states
$$|v_1\rangle=\frac{1}{\sqrt{2}}\left(
\begin{array}{c}
 1\\
 1  \\
 0  \\
 0  \\
\end{array}
\right);|v_2\rangle=\left(
\begin{array}{c}
 0\\
 0  \\
 0  \\
 1  \\
\end{array}
\right)$$ with weights $\frac{3}{7}$ and $\frac{4}{7}$ respectively
And two observables A and B
$$A=\left(
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 2 & 0 \\
 0 & 0 & 0 & 2 \\
\end{array}
\right) $$
$$B=\left(
\begin{array}{cccc}
 3 & 0 & 0 & 0 \\
 0 & 4 & 0 & 0 \\
 0 & 0 & 3 & 0 \\
 0 & 0 & 0 & 4 \\
\end{array}
\right)
$$
If you measure $A$ and $B$ at the same time, what you are doing is measuring $A$ with some state and measuring $B$ with not necessarily the same state but it may be the same.
So if I am trying to measure probability of measuring 2 and 4, that is $p(2)\times p(4)$ ?
Thats $p(2) = \frac{4}{7} $ and $p(4) = \frac{11}{14}$ then obtaining the two at the same time is  $ \frac{4}{7}\times\frac{11}{14} = \frac{22}{49}$
What confuses me is, why is lower than $\frac{4}{7}$, since $|v_2\rangle$ has that chance of being measured, the other state just adds chances of measuring 4 with B.
Whats really going on here?
 A: If you are measuring both observables at the same time (which is possible, since the two observables commute), then you have to do one measurement which measures both quantities. Therefore the measurement must result in one of the common eigenstates of $A$ and $B$. Now it turns out that $A$ and $B$ have an unique common set of eigenstates, which are just the basis states. Therefore the probabilities are just the diagonal elements of the density matrix, that is (naming $a$ the measurement result of $A$ and $b$ the measurement result of $B$):
$$\begin{aligned}
p(a=1 \land b=3) &= \frac{3}{14}\\
p(a=1 \land b=4) &= \frac{3}{14}\\
p(a=2 \land b=3) &= 0\\
p(a=2 \land b=4) &= \frac{4}{7}
\end{aligned}$$
Note that $p(a=2 \land b=4) \ne p(a=2)\cdot p(b=4)$ since the events are not independent of each other. But that's not specifically quantum; if you know $a=2$, you already know that $b$ must be $4$, while from learning $a=1$ you don't learn anything about the measurement result $b$. Already classical probability theory tells you that in that case, the product formula for the probabilities doesn't hold.
Indeed, given that there is a single combined measurement, you could split the measurement into two steps; first doing the measurement (which produces the values for $a$ and $b$), and then, reading the measurement results (which is a completely classical process, described by ordinary probability theory).
