# Finding range, partiton and finding the dependency of a random variable

How to do this question part (a)

range(X) =(2,3,4,5,....12) right the range is usually from 2 to 12

range(y) =(2,3,4,5,6) the range is usually from two to 6 right?

range(z) = (0,1,2,3,4,5)

range(w)?? I dont think im putting this in a proper notation can anyone please help

and i dont know how to do the rest can anyone help me out

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(a) $1$ can also be present in Range$(Y)$. What is $\mathcal A_X$, and, is $f_X$ the density function or what? –  Berci Mar 12 '13 at 13:33
what is a density function? –  jyuserersh Mar 12 '13 at 13:36
My question was earlier.. ;) –  Berci Mar 12 '13 at 13:46

I think your best bet is to make a $6\times6$ table of the $6^2=36$ possible outcomes when two dice are rolled. It helps to imagine that one die is red and the other green so that they are easily distinguishable. In making your table, use the format $(4,2)$ to represent a 4 on the red die and a 2 on the green die.
Then, for each of the four random variables, make a separate $6\times6$ table in which the value of the random variable is recorded for each of the $36$ outcomes. This is tedious, but illuminating, and you will soon spot regularities that will speed up the process. The entries in the table for $W$ are obtained by multiplying the corresponding entries in the tables for $X$ and $Y.$ A hint for this one: the range of $W$ will include a large number of different values; no value occurs more than twice.
To find the partition $\mathcal{A}_X,$ group the outcomes in the original table according to the value of $X$. For example, the set corresponding to $X=4$ is $\{(1,3),(2,2),(3,1)\}$.
Assuming that $f_X(x)$ means $\Pr[X=x],$ you will use the fact that for fair dice the $36$ outcomes are all equally likely. Computing $\Pr[X=4],$ for example, then becomes a matter of counting the number of outcomes such that $X=4$ and dividing by $36.$ We saw above that there are three such outcomes, which makes $f_X(4)$ equal to $1/12.$
Two events $E$ and $F$ are independent if $\Pr[E\cap F]=\Pr[E]\Pr[F].$ You can compute $\Pr[X=7\text{ and }Z=1]$ by counting the outcomes for which those values are attained. Then compare with the product of $\Pr[X=7]$ and $\Pr[Z=1]$ to determine whether the events are independent.