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Feb 28, 2017 at 16:49 history edited Taylor CC BY-SA 3.0
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Feb 28, 2017 at 15:11 comment added Dilip Sarwate You are correct; the change from $m$ to $n$ in $\binom{n}{Y}$ occurred in Michael Hardy's edit. I have changed it back. Nonetheless, your answer still needs fixing a little (or more clarification) because $\binom{m}{k} = \binom{m}{m-k}$ and so the event $\{X=x\}$ is, in general, the union of two disjoint events $\{Y=k\}$ and $\{Y=m-k\}$, with a special case when $m$ is even and $k=\frac m2$.
Feb 28, 2017 at 14:48 comment added Taylor @DilipSarwate the $n$ wasn't there when I answered this. Perhaps it's a mistake
Feb 28, 2017 at 12:44 comment added Dilip Sarwate This answer is incorrect because it does not take into consideration the relationship between $n$ and $m$. For example, if $n < m$, then $X=0$ if $Y \in \{0, n+1, n+2, \ldots , m\}$.
Feb 28, 2017 at 8:31 vote accept noob
Feb 28, 2017 at 4:08 comment added jbowman You could simplify this a little by noting that if the inverse exists, $P(X=x) = x$, and if it doesn't, $P(X=x) = x * (\text{# of Y for which }f(Y) = x)$.
Feb 27, 2017 at 20:35 history answered Taylor CC BY-SA 3.0