correlated or independent Let $(X_1, X_2)$ be a randomly chosen pair out of $\{1,2, \ldots, 20\}$ (draw without repetition).
Are both events $$E_1:=\{X_1 \geq 8\}$$ and $$E_2:=\{X_2 \geq 12\}$$ positive or negative correlated. Are they independent?
$$P(E_1\cap E_2) = \frac{9}{20}$$
and
$$P(E_1) \cdot P(E_2) = \frac{13}{20} \cdot \frac{9}{20} = \frac{117}{400}$$
on the basis of
$$P(E_1 \cap E_2) > P(E_1)\cdot P(E_2)$$
-> positiv correlated?
--- Add ---
We expect that both events $$(E_1\text{ and }E_2)$$ happen at the same time (the pair is chosen in one draw).

The notation $(X_1, X_2)$ means $X_1$ is the result of the first draw and $X_2$ the result of the second draw which occurs without replacement.  Both $X_1$ and $X_2$
are uniformly distributed on $\{1, 2, \ldots, 20\}$, but they are not 
independent random variables.
$$P(E_1) = P\{X_1 \geq 8\} = \frac{13}{20}, 
~~ P(E_2) = P\{X_2 \geq 12\} = \frac{9}{20}.$$
But $E_1$ and $E_2$ are independent events if and only if $E_1^c$ and $E_2$ are independent events, in which case we would have $P(E_2 \mid E_1^c) = P(E_2)$.  But clearly,
$$P(E_2 \mid E_1^c) = \frac{9}{19} > \frac{9}{20} = P(E_2)$$
and so $E_1$ and $E_2$ are dependent events.
Since $P(E_1^c) = 1 - P(E_1)$, the law of total probability gives
$$\begin{align*}
P(E_2) &= P(E_2\mid E_1)P(E_1) + P(E_2\mid E_1^c)P(E_1^c)\\
&= P(E_2\mid E_1^c) + (P(E_2\mid E_1) -P(E_2\mid E_1^c))P(E_1)
\end{align*}$$ 
which shows that
$$\min\{P(E_2\mid E_1), P(E_2\mid E_1^c)\} 
\leq P(E_2) \leq \max\{P(E_2\mid E_1), P(E_2\mid E_1^c)\},$$
and since $P(E_2 \mid E_1^c)  >  P(E_2)$,
we conclude that 
$$P(E_2 \mid E_1)  <  P(E_2) ~\text{and}~ 
P(E_2 \cap E_1) = P(E_2 \mid E_1)P(E_1)  <  P(E_2)P(E_1).$$
We can also calculate
$$P(E_2 \mid E_1) = \frac{P(E_2) - P(E_2\mid E_1^c)P(E_1^c)}{P(E_1)}
= \frac{\frac{9}{20}-\left(\frac{9}{19}\times\frac{7}{20}\right)}{\frac{13}{20}}
=\frac{9 \times 12}{13 \times 19}$$
and so 
$$P(E_2 \cap E_1) = P(E_2 \mid E_1)P(E_1) = \frac{9 \times 12}{13 \times 19}
\times \frac{13}{20} = \frac{9}{20}\times \frac{12}{19}$$
in contrast to the $P(E_2 \cap E_1) = \dfrac{9}{20}$ claimed by the OP.
