# From Accuracy to Precision: Solution

*English translation by SOTA – AI Community of the Russian original. Organisers who would like this translation removed can email sota.ai.community@gmail.com.*

*Task 5 of the school stage of the All-Russian School Olympiad (VsOSh) 2025/26 in artificial intelligence (region group III), grades 9–11. Answer and official solution.*

**Answer:** $\left[\dfrac{2}{3}, 1\right]$

**Solution.**

From $\text{Accuracy} = \dfrac{TP + TN}{1000} = 0.8$ we obtain

$$TP + TN = 800 \quad \Longrightarrow \quad FP + FN = 200.$$

Since $FN = P - TP = 400 - TP$, we have

$$FP = 200 - (400 - TP) = TP - 200.$$

Then

$$\text{Precision} = \frac{TP}{TP + FP} = \frac{TP}{TP + (TP - 200)} = \frac{TP}{2TP - 200}.$$

Constraints on the meaningful values:

$$FN \ge 0 \Rightarrow TP \le 400, \qquad FP \ge 0 \Rightarrow TP \ge 200,$$

that is, $TP \in [200, 400]$. The function

$$f(TP) = \frac{TP}{2TP - 200} = \frac{1}{2 - \frac{200}{TP}}$$

decreases on this segment, therefore

$$TP = 200 \ \Rightarrow\ \text{Precision} = 1, \qquad TP = 400 \ \Rightarrow\ \text{Precision} = \frac{400}{600} = \frac{2}{3}.$$

Thus, $\text{Precision} \in \left[\dfrac{2}{3}, 1\right]$.