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# **Apontamentos** | ||
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# Resoluções dos apontamentos | ||
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## 1. Probablidades | ||
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### **1.a** | ||
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Companhia de seguros | ||
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- Tendo o Venus: | ||
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 | ||
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Riscos de haver pelo menos 1 acidente: | ||
- baixo → 1% | ||
- medio → 10% | ||
- alto → 25% | ||
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Probablidade de ter um cliente de ter a classificacao: | ||
- baixa → 0.1 | ||
- media → 0.6 | ||
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agora considerando os acontecimentos: | ||
- A : "ter pelo menos 1 acidente" | ||
- B : ser um cliente classificado na categoria de baixo risco | ||
- M : ser um cliente classificado na categoria de medio risco | ||
- E : ser um cliente classificado na categoria de elevado risco | ||
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Logo as probablidades sao: | ||
- $P(A/B)=0.01$ | ||
- $P(A/M)=0.1$ | ||
- $P(A/E)=0.25$ | ||
- $P(B)=0.1$ | ||
- $P(M)=0.6$ | ||
- Problema: $P(A) = ??$ | ||
- $P(E) = 0.3$ | ||
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Se: $P(\Omega) = P(B) + P(M) + P(E)$ | ||
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Então: | ||
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$P(E)=P(\Omega)-( P(B) + P(M))\Leftrightarrow \\ | ||
\Leftrightarrow P(E) = 1 - (0.1+0.6)\\ | ||
\Leftrightarrow P(E) = 1 - (0.7) =0.3$ | ||
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- Considerando que: | ||
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Então podemos concluir que: | ||
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$ | ||
P(A) = P(A \cap M )+P(A \cap B )+P(A \cap E ) \\ | ||
P(A) = 0.06 + 0.001 + 0.075 \\ | ||
P(A) = 0.136 | ||
$ | ||
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- $P(A \cap M ) = 0.06$ | ||
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Pois: | ||
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$P(A/M) = \frac{P(A \cap M )}{P(M)}\Leftrightarrow$ | ||
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$P(A \cap M )=P(A/M)·P(M)\Leftrightarrow$ | ||
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$\Leftrightarrow P(A \cap M )=0.1·0.6=0.06$ | ||
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- $P(A \cap B ) = 0.001$ | ||
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Pois: | ||
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$P(A/B) = \frac{P(A \cap B )}{P(B)}\Leftrightarrow$ | ||
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$P(A \cap B )=P(A/B)·P(B)\Leftrightarrow$ | ||
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$\Leftrightarrow P(A \cap B )=0.01·0.1=0.001$ | ||
- $P(A \cap E ) = 0.075$ | ||
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Pois: | ||
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$P(A/E) = \frac{P(A \cap E )}{P(E)}\Leftrightarrow\\ | ||
\Leftrightarrow P(A \cap E )=P(A/E)·P(E)\\ | ||
\Leftrightarrow P(A \cap E )=0.25·0.3=0.075$ | ||
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**Resposta: $P(A) =0.2126$** | ||
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### **1.b** | ||
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**Problema: $P(E/A) = ??$** | ||
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$ | ||
P(E/A) = \frac{P(E \cap A )}{P(A)}\Leftrightarrow\\ | ||
P(E/A) = \frac{0.075}{0.136} = 0.5514705882352941 \approx 0.5515 | ||
$ | ||
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### **1.c** | ||
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Problema: $P(E/ \overline{A})$ | ||
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$ | ||
P(M/ \overline{A}) = P(M \cap \overline{A} )·P(\overline{A})\\ | ||
P(M/ \overline{A}) = (P(M)- P(M\cap A)) · (1- P(A))\\ | ||
P(M/ \overline{A}) = (0.6 - 0.06) · (1 - 0.136)\\ | ||
P(M/ \overline{A}) = 0.54 · 0.864\\ | ||
P(M/ \overline{A}) = 0.46656 | ||
$ | ||
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$ | ||
P(E/ \overline{A}) = P(E \cap \overline{A} )/P(\overline{A})\\ | ||
P(E/ \overline{A}) = (P(\overline{A} / E )·P(E))/P(\overline{A})\\ | ||
P(E/ \overline{A}) = ((1- P(A/E)) ·P(E))/(1-P(A))\\ | ||
P(E/ \overline{A}) = ((1-0.25) ·0.3)/(1-0.2126)\\ | ||
P(E/ \overline{A}) = 0.177165 | ||
$ |
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<!DOCTYPE html> | ||
<html> | ||
<head> | ||
<meta charset="UTF-8"> | ||
<title>**Pratica ME**</title> | ||
<style> | ||
</style> | ||
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<link rel="stylesheet" href="https://cdn.jsdelivr.net/gh/Microsoft/vscode/extensions/markdown-language-features/media/markdown.css"> | ||
<link rel="stylesheet" href="https://cdn.jsdelivr.net/gh/Microsoft/vscode/extensions/markdown-language-features/media/highlight.css"> | ||
<style> | ||
body { | ||
font-family: -apple-system, BlinkMacSystemFont, 'Segoe WPC', 'Segoe UI', system-ui, 'Ubuntu', 'Droid Sans', sans-serif; | ||
font-size: 14px; | ||
line-height: 1.6; | ||
} | ||
</style> | ||
<style> | ||
.task-list-item { list-style-type: none; } .task-list-item-checkbox { margin-left: -20px; vertical-align: middle; } | ||
</style> | ||
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</head> | ||
<body> | ||
<h1 id="pratica-me"><strong>Pratica ME</strong></h1> | ||
<h1 id="ficha">Ficha</h1> | ||
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</body> | ||
</html> |
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# **Pratica ME** | ||
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# Ficha | ||
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[voltar](../notas.html) | ||
# Teorica de ME | ||
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## Probablidades - Teoremas e Axiomas | ||
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body { | ||
background-color: black; | ||
color: white; | ||
} | ||
@media screen and (prefers-color-scheme: light) { | ||
body { | ||
background-color: white; | ||
color: black; | ||
} | ||
} | ||
background-color: #1E1E1E; | ||
color: #f7f9ff | ||
} |