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The passage around this formula

Suppose an evaluation runs n independent trials of a consequential task and observes zero failures above the severity threshold $\tau$ . It is tempting to read that as evidence the risk is negligible. It is not, and the size of the gap has a name: the rule of three. If the true failure probability is $q_\tau$ , the probability of observing zero events in n independent trials is (1-$q_\tau$)^n ; setting that probability equal to 0.05 and solving gives n $\ln(1-q_\tau)$ $\approx$ -3 , and for small $q_\tau$ , where $\ln(1-q_\tau)$ $\approx$ -$q_\tau$ , this reduces to the boundary

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