lecture_08
Let [E_i =] prob. that [i]th hand [( 1 \le i \le 4)] gets exactly one ace.
[P( E_1) = \frac{ {4 \choose 1} {48 \choose 12}}{ {52 \choose 13}]
[P( E_2|E_1) = \frac{ {3 \choose 1} {36 \choose 12}}{ {52 \choose 13}}, \dots]
[E = E F \cup E F^c]
[EF \cap EF^c = EFEF^c = \emptyset]
[P( E) = P( EF) + P(EF^c)]
[P( E) = P( E|F) P(F) + P( E|F^c)P(F^c) = P( E|F) P( F) + P( E|F^c)(1 - P( F))]
Multiple choice exam, each question has [m] answer choices.
[p = ] prob student understans the material.
[1 -p = ] prob the student does not understand the material.
[C:] event "correct answer"
[K:] understans the answwer and the issue.
[P( K|C) = \frac{ P( K C)}{ P( C)} = \frac{ P( K)}{ P( C)} = \frac{ P}{ P( C|K) P( K) + P( C|K^c) P( K^c)}] [= \frac{ P}{ 1 \cdot p + \frac{ 1}{ m} (1 - p)} = \frac{ m p}{ 1- p + m p} = \frac{ m p}{ 1 + (m - 1) p}]
- [S = \bigcup{ F_i}, F_i \cap F_j \notequal \emptyset]
- Then [E = E \cap (\bigcup{ F_i}) = \bigcup{ E F_i} ]
- [P( E) = \sum{ P( E F_i) = \sum\limits_{i}{ P( E|F_i)P( F_i)]
Theorem [P( F_j|E) = \frac{ P( E F_j)}{ P( E)}]
[P( F_j|E) = \frac{ P( E|E_j)P( F_j)}{ \sum\limits_{ i = 1}^{n}{ P( E|F_i) P( F_i)}}]
- Bayes' formula
