Three hundred parachutes

Idea. Any failure probability larger than the bound would make “no failure in 300 tests” an event with less than 5% probability, so such values are rejected. The bound solves $(1-p{)}^{300}=0.05$.

Step 1. Solving $(1-p{)}^{300}=0.05$ gives $$p=1-0.05^{1/300}\approx \mathrm{0.00994.}$$

Step 2: the rule of three. Taking logarithms, $300\ln (1-p)=\ln 0.05\approx -3.0$, and $\ln (1-p)\approx -p$ for small $p$, so $p\approx \frac{3}{300}=0.01$. In general, after $n$ trials with no failure, the 95% upper bound is about $\frac{3}{n}$.

The lesson. Zero failures does not mean zero risk. With 300 clean tests you can only say, with 95% confidence, that the failure rate is below about 1%.