Solving Intersection in 2 Ways

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$X = \{1, 2, 3, ..., n\}$, uniformly destributed, independent from $Y$

$Y = \{1, 2, 3, ..., n\}$, uniformly destributed, independent from $X$

$P(X > Y \cap X = k) = $ ?

Method 1

$P(X > Y \cap X = k)$

$= P(X = k)P(X>Y|X=k)$

$=\frac{1}{n} \cdot \frac{k-1}{n}$

Method 2

$P(X > Y \cap X = k)$

$= P(X>Y)P(X=k|X>Y)$

$= \frac{n-y}{n} \cdot \frac{1}{n-y}$ ?

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Method 1 looks correct

Method 2 introduces an undefined $y$ which you should probably try to avoid, perhaps something like

$P(X > Y \cap X = k)$

$= P(X>Y)\,P(X=k\mid X>Y)$

$= \displaystyle \sum_{y=1}^n P(Y=y)\,P(X>Y\mid Y=y )\,P(X=k\mid X>Y, Y=y)$

$= \displaystyle \sum_{y=1}^n \frac1n \cdot \frac{n-y}{n} \cdot \frac{I[k \gt y]}{n-y} $ using an indicator function

$= \displaystyle \sum_{y=1}^{k-1} \frac{1}{n^2} $

$= \displaystyle \frac{k-1}{n^2}$