Probability and distribution function Can some one please tell me how to do it?
Let $(i,j)$ denote the numbers on the top faces when a pair of fair dice is rolled. 
Let $X(i,j)=i+j-3$. 
Find the range of $X$. 
Find the probability distribution function $f_{X}(x)$ of $X$. 
Find each of the following


*

*$p(x>11)$

*$p(x<7)$

*$p(x \ge0)$

 A: The range: The values of $i$ and $j$ range from $1$ to $6$. So the smallest possible value of $i+j$ is $2$, and the largest is $12$. So $i+j-3$ ranges over the integers from $-1$ to $9$.
Thus the range of $X$ is the set of all integers $n$ such that $-1\le n\le 9$. You could write it also as $\{-1,0,1,1,3,4,5,6,7,8,9,\}$.
The probability distribution function: Recall that $f_X(x)=P(X=x)$. We need to specify $f_X(x)$ for all $x$ in the range.  
As a start, we find $f_X(-1)=P(X=-1)$. We have $X=-1$ precisely if $i+j=2$, which happens if both dice show a $1$. The probability of this is $\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$.
Next we find $f_X(0)$. We have $X=0$ precisely if $i+j=3$. This happens if the yellow die shows $1$ and the pink shows $2$, or the other way around. So $f_X(0)=\frac{2}{36}$.
Next we find $f_X(1)$. Note that $X=1$ precisely if $i+j=4$, that is, if two dice give sum $4$. We have probably long known that the sum of two fair dice is $4$ with probability $\frac{3}{36}$.
Let's get slightly more abstract. Let $Y=X+3$. Then $Y=i+j$ is just the sum of two dice, and $X=k$ iff $Y=X+k$. So $f_X(1)=f_Y(4)=\frac{3}{36}$.
Similarly, $f_X(2)=f_Y(5)=\frac{4}{36}$, $f_X(3)=f_Y(6)=\frac{5}{36}$, $f_X(4)=f_Y(7)=\frac{6}{36}$, $f_X(5)=f_Y(8)=\frac{5}{36}$, and so on.
The numbered questions:  $1$. We want $P(X\gt 11)$. Since $X=Y-3$, We have $X\gt 11$ precisely if $Y \gt 14$. Of course the sum of two dice cannot be $\gt 14$, and therefore $P(X\gt 11)=0$.
$2$. This one is more work.  The most mechanical way to do it is to observe that $X \lt 7$ precisely if $X$ takes on the values $-1$ to $6$. So add up the values of $f_X(k)$ obtained earlier, $k=-1$ to $k=6$. 
Or else note that this is the probability that $Y \lt 10$, that is, the probability that the sum $Y$ of two dice is less than $10$. It is easier to find the probability that $Y\ge 10$. This is $\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{6}{36}$. So the probability that $Y \lt 10$ is $\frac{30}{36}$.
$3$. More of the same.
