Probability depending on x being the difference between two numbers An urn contains 10 balls, labeled  1 – 10 .  A set of two balls is drawn from the urn, and  let  X  record  the absolute value of the difference.  Compute  Prob( X < 2 ).  
My understanding is that this is without replacement and so the probability of x<2 is just the probability of x=1 plus x=0. Therefore, the probability of the difference being 0 is 10/100 and the probability of the difference being 1 is 18/100. So x<2 is just (10/100)+(18/100). Is this correct?

Let us draw the balls, and then look at them one at a time.
There are two possibilities: (i) The first ball is $1$ or $10$ or (ii) The first ball is $2$ to $9$.
In Case (i), the probability that the second ball is a neighbour is $\frac{1}{9}$. In Case (ii), the probability the second ball is a neighbour is $\frac{2}{9}$. 
Note that Case (i) has probability $\frac{2}{10}$, and Case (ii) has probability $\frac{8}{10}$. Thus the required probability is $\frac{2}{10}\cdot \frac{1}{9}+\frac{8}{10}\cdot \frac{2}{9}$. This simplifies to $\frac{1}{5}$.
Another way: Think of the numbers as $10$ chairs in a row. There are $\binom{10}{2}$ equally likely ways to pick a set of $2$ chairs.  There are $9$ ways to pick $2$ consecutive chairs. Thus our probability is $\frac{9}{\binom{10}{2}}$. This is $\frac{1}{5}$.  
