# Compute the probability mass function of X and Y

Let the random variable $$X$$ be the minimum and $$Y$$ be the maximum of three digits picked at random without replacement from the set $$\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$$ .

(a) Compute the probability mass function of $$X$$ .

(b) Compute the joint probability mass function of $$X$$ and $$Y$$ .

I do not know how to do the first one let alone the second one. Any help is appreciated. I'm stuck because I have't seen many examples and explanations from my lecture notes so if anyone has examples feel free to add them.

The number of ways you can choose $$\ 3\$$ integers from the set, all equally likely, is $$\ 10 \choose\ \ 3\$$. Thus the probability of any particular triple of integers being chosen is $$\ \frac{1}{10 \choose\ 3}\$$. The minimum of the three integers will be $$\ m\$$ if $$\ m\$$ is in the set of three chosen, and the other two chosen are both from the set $$\ \left\{m+1, m+2,\dots, 9\right\}\$$. In how many ways can such a set be chosen? I've hidden the answers below the fold, I'd suggest you try and work them out yourself before taking a peek.
There are exactly $$\ {9-m\choose 2}\$$ ways in which the set can be chosen if $$\ 1\le m\le 7\$$ (and $$\ 0\$$ ways if $$\ m > 7\$$). Thus $$\ \mathrm{Prob}\left(X=m\right) = \frac{9-m\choose 2}{10 \choose\ \ 3}\$$ for $$\ 1\le m\le 7\$$. The probability mass function of $$\ Y\$$ can be found similarly. For the joint distribution, $$\ m\$$ will be the mimimum and $$\ M\$$ the maximum of the three numbers chosen if both those numbers are in the set chosen, and the third lies between the two of them. There are $$\ M-m-1\$$ ways such a set can be chosen if $$\ M\ge m+1\$$ (and zero ways otherwise), so $$\ \mathrm{Prob}\left(\,X=m\ \&\ Y=M\,\right)= \frac{M-m-1}{10 \choose\ \ 3}\$$ for $$\ 1\le m\le M-2\le 7\$$.