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Created on: October 24, 2025

Answered using O3 by Chat01

Question

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Реши

Thought

Reading the content

I see that you're working with a classification problem for spam email. We’re provided with metrics like accuracy (0.8) and precision (0.5), and need to solve for the number of spam emails (P) and the range for true positives (TP). I'll break this down to investigate the possible outcomes for P and TP.

Determining bounds for TP

We’re solving for the range of TP (true positives) and P (total spam emails). I know accuracy = 0.8 implies TP + TN = 800, and precision = 0.5 means FP = TP, because the ratio gives this condition. So, now I need to figure out bounds for TP given these setup constraints.

Solving for P and constraints

So, I’ve narrowed down the problem: 2TP + FN + TN = 1000, and from the known relationships, TP + FN = 200, which means P = 200. So it looks like P is uniquely determined. However, I just need to double-check that all variables remain non-negative integer values while ensuring TP ranges from 0 to P.

Final bounds for TP

We know P is 200, and TP must satisfy constraints: TP >= 0, TP <= 200, and all values must be non-negative integers. I’ve checked that FP = TP, TN = 800 - TP, and FN = 200 - TP all hold up. So, TP can range from 0 to 200, and 0 is allowed, as it satisfies all conditions.

Thought for 19s

Answer

Шаг 1. Записываем условия

N=1000,Accuracy=0,8=TP+TNN    ⟹    TP+TN=0,8⋅1000=800,Precision=0,5=TPTP+FP    ⟹    TP=FP.\begin{aligned} N &= 1000,\\ \text{Accuracy} &= 0{,}8=\frac{TP+TN}{N}\;\;\Longrightarrow\;\;TP+TN=0{,}8\cdot1000=800,\\[4pt] \text{Precision} &= 0{,}5=\frac{TP}{TP+FP}\;\;\Longrightarrow\;\;TP=FP. \end{aligned}

Пусть PP — общее число писем-спама:

P=TP+FN.P = TP + FN.

Шаг 2. Используем равенство для суммы всех писем

TP+FP+FN+TN=N.TP + FP + FN + TN = N.

Подставляем FP=TPFP=TP и TN=800−TPTN = 800-TP:

TP+TP+FN+(800−TP)=1000    ⟹    TP+FN+800=1000    ⟹    TP+FN=200.TP + TP + FN + (800-TP)=1000 \;\;\Longrightarrow\;\; TP + FN + 800 = 1000 \;\;\Longrightarrow\;\; TP + FN = 200.

Но TP+FN=PTP+FN = P, следовательно

P=200.\boxed{P = 200}.

Шаг 3. Находим возможный диапазон TPTP

  • TN=800−TP≥0  ⟹  TP≤800TN = 800-TP \ge 0 \;\Longrightarrow\; TP \le 800 — условие невязкое, так как дальше будет сильнее.
  • FN=P−TP=200−TP≥0  ⟹  TP≤200.FN = P-TP = 200-TP \ge 0 \;\Longrightarrow\; TP \le 200.
  • Из Precision =0,5=0{,}5 требуется TP+FP>0  ⟹  TP>0.TP+FP>0 \;\Longrightarrow\; TP>0.

Итак, единственные ограничения: 1≤TP≤2001 \le TP \le 200 (целые значения).

TP∈[1,  200].\boxed{TP \in [1,\;200]}.

Ответ

  • P=200P = 200;
  • TP∈[1; 200]TP \in [1;\,200] (любое целое число из этого интервала).

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