Sample Space and Probability
Bertrand's paradox: 一个很有意思的悖论, 4 个不同的答案都看上去很合理.
- Bertrand's Paradox (with 3blue1brown) - Numberphile
- More on Bertrand's Paradox (with 3blue1brown) - Numberphile
Total probability theorem and Bayes' rule¶
Total Probability Theorem
Let \(A_1 ,\dots , A_n\) be disjoint events that form a partition of the sample space (each possible outcome is included in exactly one of the events \(A_1, . . . , A_n\)) and assume that \(\Pr(A_i) > 0\), for all \(i\). Then, for any event \(B\), we have
\[\begin{aligned} \Pr(B) &= \Pr(A_{1} \cap B)+ \cdots + \Pr(A_{n} \cap B) \\ &= \Pr(A_{1})\Pr(B \mid A_{1}) + \cdots + \Pr(A_{n})\Pr(B | A_{n}) \end{aligned}\]
Bayes' Rule
Let \(A_1 , A_2,\dots , A_n\) be disjoint events that form a partition of the sample space, and assume that \(\Pr(A_i) > 0\), for all \(i\). Then, for any event \(B\) such that \(\Pr(B) > 0\), we have
\[\begin{aligned}\Pr(A_{i} \mid B) &= \frac{\Pr(A_{i})\Pr(B \mid A_{i})}{\Pr(B)}\\ &= \frac{\Pr(A_{i})\Pr(B \mid A_{i})}{\Pr(A_{1})\Pr(B\mid A_{1})+\cdots+ \Pr(A_{n})\Pr(B \mid A_{n})} \end{aligned}\]