New to stats: Bivariate Discrete Random Variables: how does this work? The problem: 
A fair coin is tossed twice. Let X be the number of heads and let Y be the
indicator function of the event X = 1. Find P(X = x,Y = y) for all appropriate values
of x and y.
I've been treating $S$, the sample space for the experiment, as
$S=\{ \{T,H\}, \{H,T\}, \{H,H\}, \{T,T\} \}$ and defining the DRV $X$ as:
$$ 
             X(\omega) = \begin{cases}
                             0,      & \omega = \{T,T\}\\
                             1,      & \omega = \{H,T\}, \{T,H\}\\
                             2,      & \omega = \{H,H\} 
                                \end{cases}
$$
and the DRV $Y$ as:
$$
Y(\omega) = \begin{cases}
                             1,      & \omega = \{H,T\}, \{T,H\}\\
                             0,      & \omega = \{H,H\}, \{T,T\}
                                \end{cases}
$$
Is this in the right direction? I don't have a solution to verify with so if y'all don't mind, maybe have a hidden one for me once I get it? 
I'm truly not sure if I"m on the right track. 
Thanks people!
 A: Yes, you're doing great! Now put in your different combinations of x and y and count! Hint: it all should sum up to 1 ;)
Good luck!
A: I don't know if I am exactly hitting the point of your question, but I hope that I am anyway helping you in some sort of way.
I would address this problem using the definition and the concept of probability measure. Referring to Kolmogorov's axioms of probability, a probability measure $\textit{P}$ on $\Omega$ with sigma algebra $A$ is a function $\textit{P} : A \to [0,1]$ such that:
1) $\textit{P}(\emptyset) = 0$ (probability measure of the empty set);
2) $\textit{P}(\bigcup_{i=1}^\infty E_{i}) = \sum_{i=1}^\infty \textit{P}(E_{i})$ 
for any pairwise disjoint sets $E_{1}, E_{2}, ..., \in A$ (countable additivity);
3) $\textit{P}(\Omega) = 1$ (certain event).
In your case, the set $ \Omega $ is the set of all possible outcomes of your experiment, so $ \Omega :=  \{ \{T,H\},\{H,T\},\{H,H\},\{T,T\} \} $, and $A$ is the powerset on $\Omega$.
Then define the following probability measure:
$ P(\omega) = 1/\text{Card}(\omega) = 1/4 $ for each $ \omega \in \Omega $. This is the uniform distribution on the set, and we can use it because of the physical characteristics of the experiment (as pointed out in whuber♦'s answer).
Note that we haven't define $P$ on all of the sets of $A$, which we would have to do in order to have a measure, but defining $P$ on these single-element-sets is sufficient to induce a measure in the whole space.
So now you can claim that there is a unique proability measure on $ A $ that is consistent with definition.
The last step is the following: considering the set $ \{ \{H,T \}, \{T,H\} \} $;  $ P( \{ \{H,T \}, \{T,H\} \} ) $ is the disjoint union of its building blocks, i.e. $ P( \{H,T \} ) $ and $ P( \{T,H\} ) $, and from bullet 2) above this is the sum $ P( \{H,T \} ) + P( \{T,H\} ) $, which is equal to $1/2$. You have now a measure of each set you are interested to.
