Finding the joint probability of max and min functions and marginal probability mass functions So my homework question reads:
Let X1 and X2 be the numbers on two balls drawn randomly from a bag of billiard balls numbered 1 through 15. Find the joint probability mass function of Y1 = min{X1,X2} and Y2=max{X1,X2} and the marginal probability mass function of Y1 when sampling is without replacement, and then when it is with replacement
I am very confused on this problem, the answer for the first part is 1/105. I see that the 105 comes from 15 choose 2, but I have no idea where the one comes from.. any help would be greatly appreciated on all parts of this problem
 A: The joint probability mass function, $P_{Y_1,Y_2}(m,n)$, is $~\mathsf P(\min\{X_1,X_2\}=m,\max\{X_1,X_2\}=n)$; that is the probability for selecting two balls numbered $m,n$, when selecting any two balls from 15 without replacement. (Where $~1\leq m<n\leq 15~$.)
There are two ways to select those particular balls.  $\{X_1=m, X_2=n\}$ and $\{X_1=n, X_2=m\}$.  Either way, the minimum is the $m$ and the maximum is $n$.
There are $\binom{15}2\cdot2!$ outcomes that are a selection and arrangement for two balls from fifteen.   That is ${}^{15}\mathrm P_2$, the permutation count for two balls selected from 15.
There is no bias in selection of these outcomes, so the probability is $\require{cancel}\tfrac {\cancelto 1 2}{\tbinom {15}2\cdot \cancelto 1 2}$ , or $\tfrac 1{105}$.

The marginal probability mass $P_{Y_1}(m)$ will be the count of outcomes for selecting balls number greater than $m$ when selecting any ball not numbered $m$.
$P_{Y_1}(m)=\dfrac{15-m}{14}$

That is for "without replacement".   The "with replacement" senario is slightly more complicated as you have to consider the outcomes where $n=m$.   The same general proceedure is followed.  
Carry on.
