Suppose that for some $p \in \left(0, 1\right)$ and some $n \in \mathbb{N}$, we have $n$ independent Bernoulli random variables, $X_{1}, X_{2}, \dots, X_{n}$, each with mean $p$. We shall call $X_{1}, X_{2}, \dots, X_{n}$ the "items". We define a test to be a subset of $\left\{1, 2, \dots, n\right\}$. The result of a test, t, will be $1$ if there exists some $j \in t$ with $X_{j} = 1$ and will otherwise be $0$. Given some $\epsilon \in \left(0, 1\right)$, the objective is to choose a set of tests, T, and a function that will take the results of those tests and output a vector $Y \in \left\{0, 1\right\}^{n}$ such that $T$ is as small as possible while ensuring that $P\left(Y \neq \left(X_{1}, X_{2}, \dots, X_{n}\right)\right) \leq\epsilon$.
Of course, we could test each item individually, i.e. let $T = \left\{\left\{1\right\}, \left\{2\right\}, \dots, \left\{n\right\}\right\}$, and be guaranteed to find $X = \left(X_{1}, X_{2}, \dots, X_{n}\right)$ with $n$ tests but my hope is that it's possible to be more efficient than that, particularly for small $p$. It was shown by Chan et al (2011) that if the number of items equal to $1$ is $d$ then you can succeed with $\mathcal{O}\left(d\log\left(n\right)\right)$ as $n \to \infty$, which is better than individual testing if $d = \mathcal{o}\left(\frac{n}{\log\left(n\right)}\right)$, as $n \to \infty$, but will presumably be worse than individual testing in our case, where $\mathbb{E}\left(d\right) = pn$.
Is it possible to do better than individual testing for certain values of $p$ in the problem described here, even if the number of tests still grows linearly in $n$ as $n \to \infty$? For example, is it possible to have $\lvert T \rvert \sim \alpha n$ as $n \to \infty$ for some $\alpha \in \left(0, 1\right)$? What if we add experimental noise, for example there is some probability $q \in \left(0, 1\right)$ that the test gives the wrong result?