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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.

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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\ $.

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