# Lecture_6
Author: Zheyuan Wu
Date: 2023-01-30

#### Zheyuan Wu

#### 2023-01-30

## Introduction to Probability

- \(P(\phi)\) is 0.

- For any finite collection \(E_1,E_2,...E_m\) of disjoint event \[ P(E_1 \cup E_2 \cup .... \cup E_m)=P(E_1)+P(E_2)+...+P(E_m) \]

- If \(A\subseteq B\), then \(P(A)\leq P(B)\)

- \(P(A^c)=1-P(A)\)

- $P(AB)=P(A)+P(B)-P(AB) $

- \(P(A\cup B \cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)\)

## Counting

The Multiplication Rule: If a task can be completed in K stages and stage i has n_i outcomes, regardless of the outcomes of the previous stages, then the task has \(n_1,n_2,..n_k\) outcomes.

Example. coin tossed 10 times.

\(2^{10}=1024\)

Select first second, and third place winder from a group of 4 final-list. (select without replacement.)

\(4\times 3\times 2=24\)

### Equal likely outcome

- The probability of an event \(E\) is the likelihood of the occurrence of E, This is denoted \(P(E)\). (\(0\leq P(E) \leq 1\))

- Let S be a sample space with N outcomes that are equally likely to occur. Then the probability of each outcome is 1/N.

- If \(N(E)\) denotes the number of outcomes in the Event \(E\), then

\[ P(E)={{N(E)}\over N} \]

Example: Suppose we roll 2 die separately, the probability of rolling 6 is.

\(E=\{(1,5),(2,4),(3,3),(4,2),(5,1)\},N=5\)

\(P(E)=5/36=0.139\)

### Permutations and Combinations

- If a distinction is made between the outcomes of the stages, we say the outcome are ordered. Otherwise we say the outcomes are unordered.

- The ordered outcomes are called permutations of k units. The number of permutations of k unites selected from a group of n units is denoted by \(P_{k,n}\).

- The unorderd outcomes are called combinations of k units. The number of combinations of k units selected from a group of n units is denoted by \((^n_k)\).

To compute the number of permutations of k units:

\[ P_{k,n}=n(n-1)(n-2)...(n-k+1)={n!\over{(n-k)!}} \] where \(m!=m(m-1)(m-2)..(2)(1)\) and \(0!=1\).

and \(P_{n,n}=n!\)

To compute the number of combinations of k units

\[ \begin{pmatrix}n\\k\end{pmatrix}={P_{k,n}\over{P_{k,k}}}={n!\over{k!(n-k)!}} \] choose k form n

Example. Select two card from a deck of 52 cards.

- When the first card to player 1 and second card to player 2

\[ P_{2,52}={52!\over{(52-2)!}}=52\times 51=2652 \] * When both card to player 1

\[ C_{2,52}=\begin{pmatrix}52\\2\end{pmatrix}={{52\times 51}\over 2!}=1326 \]