problem on random variable in probability A game consists of first rolling an ordinary 6-sided die once and then tossing a fair coin once. The score, which consist of adding the number of spots showing on the die to the number of heads showing on the coin, is a random variable called X. 
a) Give the probability function for this random variable 
b) Give the CDF for this random variable 
c) Find P(X>3) 
d) Find the probability that the score is an odd integer
I'm confused what my supp(X) is? If anyone can help with this problem that would be awesome, thanks!
 A: In the discrete case, the support of the random variable $X$ is the set of values of $x$ such that $\Pr(X=x)\ne 0$. In our case, the possible values of $X$ range from $1$ to $7$, so the support of $X$ is $\{1,2,3,4,5,6,7\}$. 
Added: We find the distribution of $X$, by specifying $\Pr(X=x)$ for all values $x$ in the support of $X$. 
In order for $X$ to be $1$, we need to roll a $1$ and toss a tail. The probability of this is $\frac {1}{6}\cdot \frac{1}{2}$. Thus $\Pr(X=1)=\frac {1}{12}$.
The random variable $X$ can be $2$ in two ways: (i) we get a $2$ on the die, and roll a tail or (ii) we roll a $1$ on the die, and toss a head. The probability of (i) is $\frac{1}{6}\cdot\frac{1}{2}$. The probability of (ii) is the same. It follows that $\Pr(X=2)=\frac{1}{6}$.
You can handle the probabilities that $X=3$, $X=4$, and so on to $7$. 
For the cdf $F_X(x)$, recall that $F_X(x)$ is the probability that $X\le x$, and is defined for all real $x$.
If $x\lt 1$, the $\Pr(X\le x)=0$, so $F_X(x)=0$.
If $1\le x\lt 2$, then $\Pr(X\le x)=\frac{1}{12}$, so in this interval $F_X(x)=\frac{1}{12}$.
If $2\le x\lt 3$, then $\Pr(X\le x)=\frac{1}{12}+\frac{1}{6}$. Thus $F_X(x)=\frac{3}{12}$ in this interval.
Continue. Don't forget about $F_X(x)=1$ if $x\ge 7$. 
The remaining questions will probably not cause any difficulty.
