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Let $\mathrm{X}_{\mathrm{1}}=1$ and $\mathrm{X}_{\mathrm{i}}, 1<\mathrm{i}\leq\mathrm{N}$, be $\mathrm{N}$ independent and uniformly distributed random variables over the set $\{1 / \mathrm{i}, 2 / \mathrm{i}, \ldots,(\mathrm{i}-1) / \mathrm{i}\} .$ As $\mathrm{N}$ tends towards infinity,what is the probability that $\mathrm{N} \cdot \min \left(\mathrm{X}_{1}, \mathrm{X}_{2}, \ldots, \mathrm{X}_{N}\right)$ is greater than 2018?

Tentative solution: $$\begin{align*}P\left(N\min \left(X_{1}, \ldots, X_{N}\right) > 2018\right) &= P\left(\min \left(X_{1}, \ldots, X_{N}\right)>\frac{2018}{N}\right) \\ &= P\left(X_{1}>\frac{2018}{N}, \ldots, X_{N}>\frac{2018}{N}\right) \\ &= P\left(X_{1}>\frac{2018}{N}\right) \ldots P\left(X_{N}>\frac{2018}{N}\right) \end{align*}$$

I am not sure on how to deal with $\lim_{N\to\infty} P(X_{1}>\frac{2018}{N}) \ldots P\left(X_{N}>\frac{2018}{N}\right)$

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