Math  /  Data & Statistics

Question02/11/2024
Applying Bayes Rule P(AB)=P(BA)P(A)P(BA)P(A)+P(BA)P(A)P(A \mid B)=\frac{P(B \mid A) P(A)}{P(B \mid A) P(A)+P(B \mid \sim A) P(\sim A)} A=A= you have the flu, B=B= you just coughed Assume: P( flu )=0.05P( cough  flu )=0.80P( cough flu)=0.2\begin{array}{l} P(\text { flu })=0.05 \\ P(\text { cough } \mid \text { flu })=0.80 \\ P(\text { cough } \mid \sim f l u)=0.2 \end{array} what is P (flu | cough)?
Bayesian classifiers -This is where the "naïve" in "naïve Bays" comes in: if we make naïve assumptions about the generative model for each label, we can find a rough approximation of the generative model for each class, and then proceed with the Bayesian classification. - Different types of naive Bayes classifiers rest on different naïve assumptions about the data. - The naïve Bayes classification algorithm was built on the assumption of independent events, to avoid the need to compute there messy conditional probabilities. - If everything was independent, the world of probability would be a much simpler place.

Studdy Solution

STEP 1

1. P(flu)=0.05 P(\text{flu}) = 0.05
2. P(coughflu)=0.80 P(\text{cough} \mid \text{flu}) = 0.80
3. P(coughflu)=0.2 P(\text{cough} \mid \sim \text{flu}) = 0.2

STEP 2

1. حساب احتمال السعال.
2. تطبيق قاعدة بايز لإيجاد P(flucough) P(\text{flu} \mid \text{cough}) .

STEP 3

لحساب احتمال السعال P(cough) P(\text{cough}) نستخدم:
\[ P(\text{cough}) = P(\text{cough} \mid \text{flu}) \cdot P(\text{flu}) + P(\text{cough} \mid \sim \text{flu}) \cdot P(\sim \text{flu}) $

STEP 4

نحسب P(flu) P(\sim \text{flu}) :
\[ P(\sim \text{flu}) = 1 - P(\text{flu}) = 1 - 0.05 = 0.95 $

STEP 5

نحسب P(cough) P(\text{cough}) :
\[ P(\text{cough}) = (0.80 \times 0.05) + (0.2 \times 0.95) $
\[ P(\text{cough}) = 0.04 + 0.19 = 0.23 $

STEP 6

تطبيق قاعدة بايز لإيجاد P(flucough) P(\text{flu} \mid \text{cough}) :
\[ P(\text{flu} \mid \text{cough}) = \frac{P(\text{cough} \mid \text{flu}) \cdot P(\text{flu})}{P(\text{cough})} $
\[ P(\text{flu} \mid \text{cough}) = \frac{0.80 \times 0.05}{0.23} $
\[ P(\text{flu} \mid \text{cough}) = \frac{0.04}{0.23} \approx 0.1739 $
الاحتمال هو:
0.1739 \boxed{0.1739}

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