If a die is tossed twice, are the smaller and larger of the outcomes independent random variables? A balanced die  is tossed twice. Let X and Y denote the smaller and larger of the two faces, respectively. Are X and Y independent random variables?
 A: The basic idea of independence is that the knowledge of the outcome of one random variable does not affect our knowledge of the outcome of the other.  Say I conduct the experiment of rolling the die twice.  I record the minimum and maximum value, but I don't tell you what they are yet.
Before I tell you, could you say that there is a positive probability that $Y \le 2$?  Sure:  I could have rolled $(X,Y) \in \{(1,1), (1,2), (2,2)\}$, all of which would satisfy the property that the maximum is at most $2$.
But the experiment was already done.  I just haven't told you the result.  I decide to tell you the value of $X$ only:  I tell you that $X = 5$.  Now does this knowledge affect the aforementioned probability that $Y \le 2$?  Absolutely.  What was before a probability $\Pr[Y \le 2] = \frac{4}{36} = \frac{1}{9}$, is now impossible:  $\Pr[Y \le 2 \mid X = 5] = 0$.  Therefore, $X$ and $Y$ cannot be independent.
(This is not a rigorous proof, but it does furnish a basic motivation and understanding of the concept of independence, by means of a counterexample.)

By contrast, what would be an example of random variables that are related to the outcome of the two die rolls that would be independent?  Suppose $U$ is the random variable that is $0$ if the outcome of the first die roll is even, and $1$ if the outcome of the first die roll is odd.  And suppose $V$ is the random variable that equals twice the value of the second die roll.  Are $U$ and $V$ independent?  Our intuition says yes, because the die is fair and the die has no means of remembering any previous outcomes and so cannot change its result based on what has happened to it before.  If I tell you that $V = 10$, that gives you no information about what the first die roll could have been.  Similarly, if I tell you in a different experiment that $U = 0$, that says nothing about what happens in the second roll.
A: Definition Two random variables are said to be independent if
$$P(\{X \in A\}\cap \{ Y \in B\}) = P(\{X \in A\}) P(\{Y \in B\})$$
for all $A,B \subseteq \mathbb{R}$. 
If $A = \{3\}$ then $Y$ must be $\{3,4,5,6\}$ by construction. Then, let $B = \{3,4,5,6\}$. Therefore, 
$$P(\{X \in A\}\cap \{ Y \in B\}) = \frac{4}{21}$$
However, $P(\{Y \in B\}) > 0$ and $P(\{X \in A\}) = \frac{4}{21}$ so the equality cannot hold and the random variables are not independent. 
To see the probability computations, write out the sample space: 
$$\Omega = \big\{(1,1), (1,2), \ldots, (1,6), (2,2), \ldots, (6,6)\big\}$$
and observe that $|\Omega| = 21$. 
