Two balls are drawn at random from a box containing ten balls numbered $0, 1, 2, ..., 9$ Let the random variable X be the maximum of the two numbers drawn. Determine the probability function for $X$.
This is hypergeometric isn't it?
With $N = 10, n = 2$, but I can't find what $r$ (number of successes would be)?
So far I have
$$f(x) = \frac{\binom{r}{x} \cdot \binom{10 - r}{2 - x}}{\binom{10}{2}}$$
A good strategy is to go through the cumulative distribution function. $$ P(X\leq x)=P(\text{both balls are $\leq x$})=\frac{x+1\choose 2}{10\choose2} $$ And now $$ f(x)=P(X=x)=P(X\leq x)-P(X\leq x-1)=\frac{{x+1\choose 2}-{x\choose 2}}{10\choose2}=\frac{x}{45}. $$ Of course $f(0)=0$.
You can make a table. The first column and the first row are the possible outcomes of the two balls. For each combination the maximum value has been inserted.
$$ \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} \hline \text{ b1/b2 } & 0& 1 &2 &3 &4 &5 &6 &7&8&9 \\ \hline \hline \hline 0&-&1 &2 &3 &4 &5 &6&7&8&9 \\ \hline 1&1&- &2 &3 &4 &5 &6&7&8&9 \\ \hline 2& 2& 2&- & 3 &4 &5 & 6 &7&8&9 \\ \hline 3& 3 &3&3 &- &4 &5&6 &7&8&9\\ \hline 4 &4 &4 &4&4&-&5&6 &7&8&9\\ \hline 5 &5&5 &5&5&5&-&6&7&8&9 \\ \hline 6&6&6&6&6&6&6&-&7&8&9 \\ \hline 7&7&7&7&7&7&7&7&-&8&9 \\ \hline 8&8&8&8&8&8&8&8&8&-&9 \\ \hline 9&9&9&9&9&9&9&9&9&9&- \\ \hline \end{array}$$
There are $90 (=10\cdot 10-10)$ possible outcomes.
For $x=1$ we have 2 favorable outcomes. For $x=2$ There are 4 favorable outcomes. And so on. It is always the double of $x$. Thus the probability function is
$$f(x)=\begin{cases}{}\frac{2x}{90}, \text{if} \ x=0,1,\ldots 9 \\ 0, \ \text{elsewhere} \end{cases} $$