Fraction of sampled positives being really infected or not

Putting all together, our rough expectation is that our sample of $m$ individuals will contain $m_1\approx p\cdot m$ infected, although we shall write it within this section as an equality (` $m_1 = p\cdot m$'), and ditto for other related numbers. Out of these $m_1$ infected, $\pi_1 \cdot m_1$ will be tagged as positive and $(1-\pi_1)\cdot m_1$ as negative. Of the remaining $m_2= (1-p)\cdot m$, not infected, $\pi_2\cdot m_2$ will be tagged as positive and $(1-\pi_2)\cdot m_2$ as negative. In sum, the expected numbers of positive and negative will be

$$n_P = \pi_1\cdot m_1 + \pi_2\cdot m_2$$ (1)

$$n_N = (1-\pi_1)\cdot m_1 + (1-\pi_2)\cdot m_2\,,$$ (2)

which we can rewrite as

$$n_P = \pi_1\cdot p\cdot m + \pi_2\cdot (1-p)\cdot m$$ (3)

$$n_N = (1-\pi_1)\cdot p\cdot m + (1-\pi_2)\cdot (1-p)\cdot m\,,$$ (4)

So, just to fix the ideas with a numerical example and sticking to $\pi_1=0.98$ and $\pi_2=0.12$ of Ref. [16], in the case we sample 10000 individuals we get, assuming 10% infected ($p=0.10$),

• number of infected in the sample: 1000 (and hence 9000 not infected);
• infected tagged as positive: 980;
• infected tagged as negative: 20;
• not infected tagged as positive: 1080;
• not infected tagged as negative: 7920;
• total number of positive: 2060;
• total number of negative: 7940;
• fraction of the positives really infected: $980/2060 = 47.6\,\%$.