Let $X$ be the minimum and $Y$ be the maximum of $3$ digits picked at random without replacement from $\{0, 1, 2, \cdots, 9\}$. Find the joint distribution of $X$ and $Y$.
Without using Conditional Distribution:
Since $3$ digits are picked without replacement, it follows the multivariate hypergeometric distribution so...
$\binom{1}{1}$ ways to pick the minimum $X$.
$\binom{y-x-1}{1}$ ways to pick the middle digit.
$\binom{1}{1}$ ways to pick the maximum $Y$
$\binom{10}{3}$ ways to pick any $3$ digit combination.
I have $$P(X = x, Y= y) = \frac{\binom{1}{1} \binom{y-x-1}{1} \binom{1}{1}}{\binom{10}{3}} = \frac{y-x-1}{120}$$
Using Conditional Distribution:
$$P(X = x, Y= y) = P(X = x)P(Y = y | X = x) = \frac{\binom{1}{1} \binom{9-x}{2}}{\binom{10}{3}} \cdot \frac{\binom{y-x-1}{1} \binom{1}{1}}{\binom{10}{3}}$$ ?
Reasonings:
$\binom{1}{1}$ ways to pick the minimum$X$.
$\binom{9-x}{2}$ ways to pick the remaining $2$ digits in $P(X = x)$.
$\binom{10}{3}$ ways to pick any $3$ digit combination.
Since we are given $X = x$ in $P(Y = y | X = x)$, we only need to pick $2$ digits.
$\binom{y-x-1}{1}$ ways to pick the middle digit in $P(Y = y | X = x)$.
$\binom{1}{1}$ ways to pick the maximum $Y$.
$\binom{10}{3}$ ways to pick any $3$ digit combination.
But those two answers aren't equal.
The conditional probability $P(Y = y | X = x)$ is the ratio of number of ways to chose 3 digits such that $Y=y$ among all ways with $X=x$. The number of ways when $X=x$ is $\binom{9-x}{2}\binom{1}{1}$, therefore $$P(Y = y | X = x)=\frac{\binom{1}{1}\binom{y-x-1}{1}\binom{1}{1}}{\binom{9-x}{2}\binom{1}{1}}.$$
And the answers coincide.