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ece108: add bayes law 2

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eggy
2023-03-31 11:50:17 -04:00
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@@ -927,3 +927,8 @@ $$\text{independent}\iff Pr\{A\cup B\}=Pr\{A\}+Pr\{B\}-Pr\{A\}Pr\{B\}$$
**Bayes' theorem** provides a general formula for conditional probability:
$$Pr\{A|B\}=\frac{Pr\{B|A\}}{Pr\{B\}}$$
Formally, this can be solved without $Pr\{B\}$:
$$Pr\{A|B\}=\frac{Pr\{A\}Pr\{B|A\}}{Pr\{A\}Pr\{B|A\}+Pr\{\overline A\}Pr\{B|\overline A\}}$$