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Two balls are drawn at random from a box containing ten balls numbered 0, 1, ..., 9. Let the random variable X be the maximum of the two numbers drawn. Find the probability function $f(x)$.

This is hypergeometric isn't it?

So we have $N = 10, n = 2$, and the success size will be $r = x + 1$, since we can only choose from $\{0, 1, \ldots, x\}$

How do I write it in the form?

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    Hypergeometric would be useful if you wanted to know the probability of getting only balls numbered 3 or lower: ${4\choose 2}{6 \choose 0}/{10 \choose 2},$ but that is _not exactly_ what you want to know for $p(x).$ That would be $P(max \le 3).$ – 2017-02-10

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This is hypergeometric isn't it?

Somewhat similar, but no.   What you seek is the probability for selecting one of the two balls being exactly $x$ and the other from the numbers in $\{0,..,x-1\}$, when you are selecting any two balls from those in the set $\{0, .., 9\}$, without bias but no repetition.

$$\mathsf P(\max\{X_1,X_2\}=x) ~=~\dfrac{\dbinom 1 1\dbinom {?}{1}}{\dbinom {?}2}\mathbf 1_{x\in\{1,..,9\}}$$

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    Is the numerator question mark $x $? And the denominator $x+1$ as well? – 2017-02-10
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    The numerator is, the denominator is not. – 2017-02-10
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    So the numerator is $x$, but the denominator is not $x+1$? – 2017-02-10
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Here is a start. If you are drawing 2 balls without replacement from among 10, then there are ${10 \choose 2} = 45$ possible unordered outcomes. The maximum number $X$ can take values 1 through 9.

$P(X = 1) = 1/45,$ because the only possible pair with that maximum is 0 and 1.

$P(X = 2) = 2/45,$ because 02 or 12 will do.

Keep going ...

$P(X = 9) = 9/45,$ because why?