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(D) The total number of outcomes is $n 	imes n = n^2$.We want to find the number of pairs $(x, y)$ such that $x$ divides $y$ or $y$ divides $x$.Let $

Vedclass · Accepted Answer

$-\frac{1}{n} + \frac{2}{n^2} \sum_{k=1}^n \left[ \frac{n}{k} \right]$

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(D) The total number of outcomes is $n 	imes n = n^2$.We want to find the number of pairs $(x, y)$ such that $x$ divides $y$ or $y$ divides $x$.Let $S$ be the set of pairs $(x, y)$ such that $x|y$ or $y|x$.This is equivalent to counting pairs where $x|y$ plus pairs where $y|x$,and subtracting the pairs where both $x|y$ and $y|x$ (which happens when $x=y$).For a fixed $x=k$,the number of $y$ such that $k|y$ is $\lfloor \frac{n}{k} floor$.Thus,the number of pairs $(x, y)$ such that $x|y$ is $\sum_{k=1}^n \lfloor \frac{n}{k} floor$.Similarly,the number of pairs $(x, y)$ such that $y|x$ is $\sum_{k=1}^n \lfloor \frac{n}{k} floor$.The pairs where $x=y$ are $(1,1), (2,2), \ldots, (n,n)$,which are $n$ pairs.By the Principle of Inclusion-Exclusion,the number of favorable outcomes is $\sum_{k=1}^n \lfloor \frac{n}{k} floor + \sum_{k=1}^n \lfloor \frac{n}{k} floor - n = 2 \sum_{k=1}^n \lfloor \frac{n}{k} floor - n$.The probability is $\frac{2 \sum_{k=1}^n \lfloor \frac{n}{k} floor - n}{n^2} = \frac{2}{n^2} \sum_{k=1}^n \lfloor \frac{n}{k} floor - \frac{1}{n}$.

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