Probability/mass function & variance problem with a dice and a coin We have a coin and a dice, both are fair, and we declare $X$ and $Y$ so they correspondingly describe the events $X=1$ and $X=0$ wheter we get Heads or Tails and $Y=\text {outecome from dice 1, 2, 3, 4, 5, 6}$. We also declare $Z=XY$
I need to find the range of $Z$ and compute its mass. Also, i need to find its average value and variance.
I think range of $Z$ is {${0, 1, 2, 3, 4, 5, 6}$.}
I haven't really understood this chapter in probability yet. A little help for the first part of my question would be very helpful to move on to the next questions.
 A: You're right about the range.
If $X=0$ with probability $\frac{1}{2}$ then we know $Z=0$.
If $X=1$ with probability $\frac{1}{2}$ then we know 
$Z=1$ with probability $\frac{1}{6}$
$Z=2$ with probability $\frac{1}{6}$
$\vdots$
$Z=6$ with probability $\frac{1}{6}$
From here, you should be able to compute the pmf.
Then use the fact that $$E(Z)=\sum zP(Z=z)$$ 
and
$$Var(Z)=E(Z^2)-E(Z)^2$$
where
$$E(Z^2)=\sum z^2P(Z=z)$$ 
A: Indeed: $Z\in\{0,1,2,3,4,5,6\}$, as it will be $0$ if tails shows, or the coin result if heads shows.
Whech tills us the probability mass function $$\mathsf P(Z=z) = \begin{cases}\tfrac 12&:&z=0\\\mathsf P(X=1\cap Y=z)&:& z\in\{1,2,3,4,5,6\}\\0&:& \text{otherwise}\end{cases}$$
For the mean, $\mathsf E(Z) = \mathsf E(XY)$ and because of independence ...
For the variance, $\mathsf{Var}(Z)= \mathsf E(X^2Y^2)-\mathsf E(XY)^2$ and, again, because of independence...

Ps: recall $\mathsf E(X)=\tfrac 12, \mathsf E(Y)=\tfrac 72, \mathsf{Var}(X)=\tfrac 1{4}, \mathsf {Var}(Y)=\tfrac{35}{12}$
