A software‐reliability engineer executes $80$ independent test runs of a new application and observes zero failures. She wants a $95\%$ confidence interval for the true failure probability $p$. Which statement best explains why the usual normal‐approximation confidence intervals fail when $x=0$, and why the “Rule of Three” provides a valid upper bound?

When $x=0$, the sample proportion $\hat{p}=0$ gives a standard error of zero, so Wald‐type intervals collapse to $[0,0]$, understating uncertainty. The Rule of Three uses the exact binomial upper bound $p>\frac{3}{n}$, yielding a sensible one‐sided $95\%$ bound.

The Agresti–Coull interval adds pseudo‐counts and always gives a two‐sided interval even at the boundary, so no special “Rule of Three” is needed.

The Wilson score interval remains valid at $x=0$ by shifting the center, so one can always use Wilson instead of the Rule of Three.

Both the Wald and Rule of Three methods rely on the normal approximation, so they both break down when $x=0$; you must instead double the sample size.