Passing Probability Problem

The events " at least one student passes" and "no student passes" are complementary, so P[at least one student passes] = 1 - P[no student passes]. Then

$\begin{aligned} \text{P[at least one student passes]} > 99.9\% &\implies \text{1 - P[no student passes]} > 99.9\% \\ &\implies \text{P[no student passes]} < 0.1\% = \dfrac{1}{1000} \end{aligned}$

We're given that the $k$ th student has a probability of $\dfrac{1}{k + 1}$ of passing, i.e. a probability of $\dfrac{k}{k + 1}$ of failing. So

$\begin{aligned} \text{P[no student passes]} < \dfrac{1}{1000} &\implies \text{P[student 1 fails AND student 2 fails AND student 3 fails AND .... AND student } n \text{ fails]} < \dfrac{1}{1000} \\ &\implies \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{3}{4} \cdot \dfrac{4}{5} \cdot .... \cdot \dfrac{n}{n + 1} < \dfrac{1}{1000} \\ &\implies \dfrac{1}{n + 1} < \dfrac{1}{1000} \\ \\ &\implies n + 1 > 1000 \end{aligned}$

so minimum number of students needed to make sure $P > 99.9\%$ is $\boxed{1000}$

A beautiful problem indeed :) .Inclusion-exclusion principle + factoring in a way that it has very nice similarity with euler's totient function form