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| Sep 29, 2017 | » | Conditional Probability
1 min; updated Feb 12, 2023
Definition Where \( \mathbb{P}(A|B) \) is the probability of event \(A\) given that event \(B\) occurs: $$ \mathbb{P}(A|B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} $$ Notice that if \(P(B) = 0\) then it’s meaningless to condition \(A\) on \(B\). Satisfying the Axioms of Probability Axiom 1: Normalization $$ \mathbb{P}(B|B) = \frac{\mathbb{P}(B \cap B)}{\mathbb{P}(B)} = 1 $$ Axiom 2: Non-negativity Because \(A \cap B \subseteq B \), then \( 0 \le \mathbb{P}(A|B) \le 1 \).... |
| Sep 29, 2017 | » | The Bayes Formula
1 min; updated Mar 14, 2021
The Formula By definition… $$ \mathbb{P}(A) = \mathbb{P}(A \cap B) + \mathbb{P}(A \cap B^{c}) $$ From conditional probability … $$ \mathbb{P}(A) = \mathbb{P}(A|B) \ \mathbb{P}(B) + \mathbb{P}(A|B^c) \ \mathbb{P}(B^c) $$ Therefore $$ \mathbb{P}(B|A) = \frac{ \mathbb{P}(B \cap A) }{ \mathbb{P}(A) } $$ $$ = \frac{ \mathbb{P}(A|B) \mathbb{P}(B) }{ \mathbb{P}(A|B) \ \mathbb{P}(B) + \mathbb{P}(A|B^c) \ \mathbb{P}(B^c) } $$ Switching the roles of the events is convenient because in many problems, one of the conditional probabilities is easier to calculate.... |