# Lecture_8
Author: Zheyuan Wu
Date: 2023-02-03

#### Zheyuan Wu

#### 2023-02-03

## Conditional Probability

### Independent events

Independent events: if the occurrence of one event does not change the probability of the other event occurring.

\[ P(A|B)=P(A \cap B)/P(B)=P(A)\\ P(A\cap B)=P(A)\times P(B) \]

Properties of Independence

- If A is independent of B, B is also independent of A

- If A is independent of B, A is also independent of \(B^c\)

\[ Proof: P(A|B)={P(A\cap B^c)\over{P(B^c)}}\\ ={P(A)-P(A\cap B)\over{1-P(B)}}\\ ={P(A)-P(A)\times P(B)\over{1-P(B)}}\\ =P(A){(1-P(B))\over{1-P(B)}}\\ =P(A) \]

- If A is independent of B, \(A^c\)is also independent of B

- If A is independent of B, \(A^c\)is also independent of \(B^c\)

- Any event is independent of the empty event

### Mutually Exclusive vs Independence

For two mutually exclusive event \(P(A\cap B)\) is zero. To be independent \(P(A|B)=P(A)\) But \(P(A|B)\) is zero because \(P(A\cap B)\) is zero.

Example: Let A, B and C be three disjoint event with P(A)=0.2, P(B)=-0.3, P(C)=0.5. Find \(P({A\cap B^c \over{B\cup C}})\)

\[ P({A\cap B^c \over{B\cup C}})={P((A\cup B^c)\cap(B\cup C))\over{P(B\cup C)}}\\ ={P(\phi \cup C)\over P(B\cup C)}\\ ={0.5\over 0.8}\\ ={5\over 8} \]

## Rules of probabilites

### Multiplication Rule

When tow events were independent we could write

\[ P(A\cap B)=P(A|B)\times P(B)\\ =P(B|A)\times P(A)\\ =P(A)\times P(B) \]

This was because \(P(A)=P(A|B)\), which is not generally true.

### Addition Rule

For two mutually exclusive events,

\[ P(A\cup B)=P(A)+P(B) \]

if they are not mutually exclusive.

\[ P(A\cup B)=P(A)+P(B)-P(A\cap B) \]

### Complement Rule

For any envent

\[ P(A)=S-P(A^c)=1-P(A^c) \]