mission. While we cannot know these quantities a priori, we can calculate how many exoEarth candidates are required to place compelling constraints on these factors.

The binomial distribution function, $$P\left(x,n,p\right) = \frac{n!}{x! (n-x)!}\, p^x\, \left(1 - p \right)^{n-x},$$ expresses the probability $P$ of $x$ successes out of $n$ tries, if each try has an independent probability $p$. For the exoEarth science goal at hand, $n=N_{\rm EC}$ represents the number of exoEarth candidates, $x$ is the number of candidates that exhibit the feature for which we are searching (e.g., water), $p = \eta_x \epsilon$ is the probability of detecting the feature, $\eta_x$ is the statistical fraction of exoEarth candidates that exhibit the feature, and $\epsilon$ is our detection efficiency of the feature.

In the worst case scenario of a null result in which zero detections of the desired feature are made, we should require a scientifically compelling constraint on the quantity $\eta_x$. Setting $x=0$ and assuming $\epsilon \approx 1$, we find the simple expression $$P\left(0,N_{\rm EC},\eta_x \right) = \left(1 - \eta_x \right)^{N_{\rm EC}}.$$ Solving for $N_{\rm EC}$ gives $$N_{\rm EC} = \frac{\log{(1-C)}}{\log{(1-\eta_x)}},$$ where $C = 1 - P$ is the confidence of the constraint on $\eta_x$. Figure \[confidence\_curves\] shows the number 