What is the probability that the ball picked first is numbered higher than the ball picked second ... 
Probability isn't my skill - I don't really know how to do these kind of questions.  Could someone help explain to how to calculate the solution?
 A: Let our sample space $S$ be the set of all ways in which he can pick three balls in succession with replacement.  I.e. $S\simeq\{(a,b,c)\in\mathbb{Z}^3~:~0\leq a\leq 9, 0\leq b\leq 9, 0\leq c\leq 9\}$
Check: is $S$ an unbiased sample space?  I.e. are all outcomes equally likely to occur?  The answer is yes it is.  This is good as it allows us to use counting methods to calculate the probability.
What is the size of $S$?

$|S|=10^3$

Now, how can we describe the event we are interested in calculating the probability of?  $a>b>c$.  This occurs when $a,b,c$ are all distinct and when they are in decreasing order.
How many ways can we pick an outcome $(a,b,c)$ such that $a>b>c$?
Related question:  How many ways can we pick an unordered subset of three distinct numbers?

Note that there are exactly as many ways to pick $(a,b,c)$ such that $a>b>c$ as there are ways to pick $\{a,b,c\}\subset\{0,1,2,\dots,9\}$ since there is a clear bijection between the two.  There are $\binom{10}{3}$ ways to pick three numbers.

Finish the problem by using the definition of probability in an equiprobable (unbiased) sample space:
$$Pr(A)=\frac{|A|}{|S|}$$
A: Think of your sample space as the set of ordered triples $\{(x,y,z):0\leq x,y,z\leq 9\}$. Here $x$ represents the outcome of the first draw, $y$ the outcome of the second, and $z$ the outcome of the third. Each point in the sample space has probability $1/1000$.
Therefore to determine the desired probability, it is enough to find the size of the set
$$ \{(x,y,z):0\leq z<y<x\leq 9\} $$
The size of the set is
$$ \sum_{z=0}^9\sum_{y=z+1}^9\sum_{x=y+1}^91=\sum_{z=0}^9\sum_{y=z+1}^9(9-y)=\sum_{z=0}^9\Big[9(9-y)-\frac{9\cdot 10}{2}+\frac{z(z+1)}{2}\Big]$$
$$=\sum_{z=0}^9\Big[36+\frac{z^2-17z}{2}\Big]=360-\frac{17}{2}\cdot\frac{9\cdot10}{2}+\frac{9\cdot10\cdot19}{12}=120$$
using the identities
$$ \sum_{n=1}^Nn=\frac{n(n+1)}{2}$$
and
$$ \sum_{n=1}^Nn^2=\frac{n(n+1)(2n+1)}{6}$$
Therefore the probability is $\frac{120}{1000}=\frac{3}{25}$.
