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​​Discussion 8A

​​Conditional Probability

​​The probability of an event ​​$A$​ occurring given that an event ​​$B$​ occurs is ​​$\Pr[A \mid B] = \frac{\Pr[A \cap B]}{\Pr[B]}$​.

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​​Since ​​$\Pr[B \mid A] = \frac{\Pr[A \cap B]}{\Pr[A]}$​, we know ​​$\Pr[A \cap B] = \Pr[A] \Pr[B \mid A] = \Pr[B] \Pr[A \mid B]$​.

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​​Rearranging gives us Bayes’ Rule: ​​$\Pr[A \mid B] = \frac{\Pr[A] \Pr[B \mid A]}{\Pr[B]}$​.

​​Chain Rule

​​We saw that ​​$\Pr[A \cap B] = \Pr[A] \Pr[B \mid A]$​. This extends to any number of events:

​​​​$\Pr[\bigcap_{i}^n A_i] = \Pr[A_1] \times \Pr[A_2 \mid A_1] \times \Pr[A_3 \mid A_1 \cap A_2] \times \cdots \times \Pr[A_n \mid \bigcap_i^{n-1} A_i]$​

​​Total Probability

​​There are two ways for an event ​​$B$​ to occur: with event ​​$A$​ occurring and without event ​​$A$​ occurring: ​​$\Pr[B] = \Pr[A \cap B] + \Pr[\overline{A} \cap {B}]$​.

​​Using conditional probability, we obtain ​​$\Pr[B] = \Pr[A] \Pr[B \mid A] + \Pr[\overline{A}] \Pr[B \mid \overline{A}]$​.

​​Applying this to the denominator of Bayes’ rule leads to a common pattern: ​​$\Pr[A \mid B] = \frac{\Pr[A] \Pr[B \mid A]}{\Pr[A] \Pr[B \mid A] + \Pr[\overline{A}] \Pr[B \mid \overline{A}]}$​

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​​This extends to any partitioning of ​​$\Omega$​, not just ​​$A$​ and ​​$\overline{A}$​, so ​​$\Pr[B] = \sum_{A_i} \Pr[B \cap A_i] = \sum_{A_i} \Pr[A_i] \Pr[B \mid A_i]$​.

​​Disjointness

​​Events ​​$A$​ and ​​$B$​ are disjoint if ​​$\Pr[A \cap B] = 0$​ (in the discrete case).

​​Independence

​​Events ​​$A$​ and ​​$B$​ are independent if

• ​​​​$\Pr[A \cap B] = \Pr[A] \Pr[B]$​

• ​​​​$\Pr[A] = \Pr[A \mid B]$​

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​​Special case to consider: if either event has a probability of ​​$0$​ of occurring, then are they independent? are they disjoint? What if either event has a probability of ​​$1$​ of occurring?

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