A fair coin is tossed once, and then a fair dice is rolled two independent times. We jot down which side of the coin, $C$, (Heads(H) or Tails(T)), appears and also the values from the dice $X$ and $Z$, ($1, 2, 3, 4, 5, 6$) that appear. Let the random variable $$Y= \begin{cases} X + Z, & C=H \\ X - Y, & C=T \end{cases}. $$
Describe the initial range $Ω$. How many objects does it include? Compute the mass and allocation of $Y$. Also compute its average value $E(Y)$ and variance $\mathrm{Var}(Y)$.
The initial range is
$$Ω =\{H11, H12, H13, \cdots, H21, H22, \cdots, H66, T11, T12, T13, \cdots T21, T22,\cdots, T66\}.$$ It includes $2!×6!×6!$ objects. Range of $Y$ maybe is: $$Y = \{-5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}.$$ Now, about the mass: For example: \begin{align*} p_y(2) &= P(X=1, Y=1)+P(X=3, Y =1)\\ &= \dfrac{1}{6}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{1}{6}. \end{align*}
and the rest goes on like this. So, I just want to know if I understand how to calculate pmf. If I do undestand I guess I can move on to calculating all the rest.
There should only be $2 \cdot 6 \cdot 6 = 72$ possibilities.
\begin{align} p_y(2)= P(H11) + P(T31)+P(T42)+P(T53)+P(T64) \end{align}