Sampling 3 out of 20 numbered balls from a box, what is distribution of max number drawn? The tittle may be a little misleading as I couldn't sum up the problem in one sentence.
There are 20 numbered balls in a box and we pick 3 without replacement and without caring about the order that we picked them.
Let Random Variable $X$ express the largest number out of the three.

$-$ Find $p_{X}(x) = P(X=x),$ for all the possible $x$'s.
$-$  Then, find the cumulative distribution function: $F_{X}(x) = P(X \leq\ x) $

The set of all possible outcomes consists of ${20\choose3}=1140$ elements.
All the possible $x$ values are $:  {3,4,5,...,19,20}$.
There are too many possible $x$'s to calculate the density of $X$ for each one individually, so I am trying to find a general expression for that. Also, I think the $2nd$ question will become easier if the first one is solved.
Any help would be appreciated. Thanks in advance!
 A: Let's look at a specific value, say $P(X=12)$.  For the highest ball to be $12$, you have to pick one ball from the set $\{12\}$ and two balls from the set $\{1,2,3,\dots,11\}$.
So there should be a total of $1\cdot {11\choose 2}$ ways to do this.
You can generalize this idea to continue the problem.
A: Let $X_1$, $X_2$, $X_3$ denote balls drawn from the urn without replacement. You seek to compute $$\Pr\left(\max\left(X_1,X_2,X_3\right) \leqslant x \right) = \Pr\left( X_1 \leqslant x ,X_2  \leqslant x ,X_3 \leqslant x \right)$$ 
The latter probability is the probability of choosing 3 marked balls (those whose numbers are $\leqslant x$) out of total $N = 20$ balls, see hypergeometric distribution:
$$
  \Pr\left( X_1 \leqslant x ,X_2  \leqslant x ,X_3 \leqslant x \right) = \frac{ \binom{x}{3}}{\binom{N}{3}} [ 3 \leqslant x \leqslant N ] = \frac{x (x-1)(x-2)}{N (N-1)(N-2)} [ 3 \leqslant x \leqslant N ] = \frac{x (x-1)(x-2)}{6840} [ 3 \leqslant x \leqslant 20 ] 
$$
Then 
$$
  \Pr\left(\max(X_1,X_2,X_3)=x\right) = \Pr\left(\max(X_1,X_2,X_3) \leqslant x\right) - \Pr\left(\max(X_1,X_2,X_3) \leqslant x-1\right) = \frac{(x-1)(x-2)}{2280}  [ 3 \leqslant x \leqslant 20 ] 
$$
Here is a quick verification with simulation in Mathematica:

A: Otherwise, you can use a simple EXCEL macro to find the values:
Sub Macro1734572()
'
' Macro1734572 Macro
'
CONT = 1

For I = 1 To 18
For J = I + 1 To 19
For K = J + 1 To 20

Cells(CONT, 1) = I
Cells(CONT, 2) = J
Cells(CONT, 3) = K
Cells(CONT, 4) = WorksheetFunction.Max(Cells(CONT, 1), Cells(CONT, 2), Cells(CONT, 3))
CONT = CONT + 1

Next K
Next J
Next I

CONT1 = 1
For L = 1 To 20
Cells(CONT1, 7) = L
Cells(CONT1, 8) = WorksheetFunction.CountIfs(Range(Cells(1, 4), Cells(1140,  4)), L)
CONT1 = CONT1 + 1
Next L

'
End Sub

This macro can be upgraded and this new version Works with numbers between 3 and 50 numbered balls. It was added an INPUTBOX Command. 
Sub Macro1734572b()
'
' Macro1734572b Macro
'
Columns("A:H").Select
Range("H1").Activate
Selection.Delete Shift:=xlToLeft
Range("A1").Select

10 Q = INPUTBOX("ENTER Q VALUE (Q MIN = 3 - Q MAX = 50)", "Q")

   If (Q > 50 Or Q < 3) Then
   MsgBox "False"
   GoTo 10
   End If

CONT = 1

For I = 1 To Q - 2
For J = I + 1 To Q - 1
For K = J + 1 To Q

Cells(CONT, 1) = I
Cells(CONT, 2) = J
Cells(CONT, 3) = K
Cells(CONT, 4) = WorksheetFunction.Max(Cells(CONT, 1), Cells(CONT, 2), Cells(CONT, 3))
CONT = CONT + 1

Next K
Next J
Next I

CONT1 = 1
For L = 1 To Q
Cells(CONT1, 7) = L
Cells(CONT1, 8) = WorksheetFunction.CountIfs(Range(Cells(1, 4), Cells(19600, 4)), L)
CONT1 = CONT1 + 1
Next L
'
End Sub

