A Survey of Probability Concepts Chapter 5 Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Learning Objectives LO5-1 定義機率 probability, 實驗experiment, 事件event, and 結果outcome. LO5-2 Assign probabilities using a 古典classical, 實證empirical, or 主觀subjective approach. LO5-3 用加法法則 rules of addition來計算機率 LO5-4 用乘法法則rules of Multiplication 來計算機率 LO5-5 用條件次數/交叉/列聯表contingency table來計算機率 LO5-6用貝氏定理Bayes’ theorem來計算機率 LO5-7 Determine the number of outcomes using principles of counting. 5-*

漫談機率～ 面對眾多選擇，不知道該怎麼辦的時候… 骰子之前，人們用什麼來決定機遇？ 動物的骨頭！骰子 骰子何時出現在歷史上？ 公元前兩千年之前就已經存在。 機率論的起源？ 跟賭博有關！ 16世紀Girolamo Cardano就已出版”機率論”(Liber de Ludo Aleae)給賭徒許多建議：《誰，在什麼時候，應該賭博？》、《為什麼亞里斯多德譴責賭博？》、《那些教別人賭博的人是否也擅長賭博呢？》等。 “我該怎麼賭”這個問題，讓17世紀的法國賭徒”Blaise Pascal”(名作家)熱烈的請教數學家Pierre de Fermat，之間的書信後來集結成書。

漫談機率～ 面對眾多選擇，不知道該怎麼辦的時候… 骰子之前，人們用什麼來決定機遇？ Dice were originally made from the talus of hoofed animals, colloquially known as "knucklebones". These are approximately tetrahedral, leading to the nickname "bones" for dice. 骰子何時出現在歷史上？ The oldest known dice were excavated as part of a 5000-year-old backgammon set at the Shahr-e Sukhteh (Burnt City) , an archeological site in south-eastern Iran 機率論的起源？ 跟賭博有關！ 骰子的歷史與人類歷史一樣的久，常以不同形式出現在賭博中，古埃及與蘇美人的墓中也都發現過手工制骰子游戲，而在古希臘時期末與羅馬時代，擲骰子也是一種流行的游戲。 西方學者於 17世紀開始對機率理論產生興趣，其理論背景最初只是為了處理如擲骰子、輪盤、撲克牌等遊戲的賭金分配問題。

Talus, knucklebones

The oldest known dice were excavated as part of a 5000-year-old backgammon set at the Shahr-e Sukhteh (Burnt City) , an archeological site in south-eastern Iran

LO5-1 Define the terms probability, experiment, event, and outcome. PROBABILITY A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur. i.e. 0≦P(E) ≦1 5-*

Experiment, Outcome, and Event LO5-1 Experiment, Outcome, and Event An 實驗(experiment) is a process that leads to the occurrence of one and only one of several possible results. 隨機實驗乃一種過程，是不能確定預知會發生何種結果的實驗方式。 An 結果/出象(outcome) is the particular result of an experiment. 隨機實驗的每個可能結果都是elementary outcome，又稱為樣本點(sample point)。 一個隨機實驗中，所有可能outcomes的集合稱為：樣本空間(sample space)。 An 事件(event) is the collection of one or more outcomes of an experiment. 樣本空間的部分集合稱為事件。 事件中僅包含1個outcome或樣本點者，稱為簡單事件(simple event)。 事件中包含2個outcomes或2個樣本點以上者，稱為複合事件(composite event)。 5-*

Ways to Assign Probabilities LO5-2 Assign probabilities using a classical, empirical, or subjective approach. Ways to Assign Probabilities 我們根據什麼來求得事件發生的機率？3種可能方法: 客觀： 1. 古典機率論 CLASSICAL PROBABILITY Based on the assumption that the outcomes of an experiment are equally likely. 如：擲公正的銅板得正面的機率 2. 實證機率論 EMPIRICAL PROBABILITY The probability of an event happening is the fraction of the time similar events happened in the past. 主觀： 3. 主觀機率論 SUBJECTIVE PROBABILITY The likelihood (probability) of a particular event happening that is assigned by an individual based on whatever information is available. 5-*

古典機率論 Classical Probability LO5-2 古典機率論 Classical Probability 不需要做實驗，我們就應該知道擲一個公正的骰子，出現每一面的機率都應該相等，那麼若問：「雙數點出現的機率＝？」 The possible outcomes are: There are three “favorable” outcomes (a two, a four, and a six) in the collection of six equally likely possible outcomes. 5-*

有沒有人真的去做過實驗證明呢？ 擲銅板時的正面比率隨著我們擲的次數增加，比率會接近0.5 .也就是正面的機率。 (1)18 世紀的法國自然主義者布豐伯爵把一個銅板擲了？次 4040次：2048正面 正面機率：2048/4040＝0.5069 (2)英國數學家John Kerrich 在二次大戰被德國人關在牢中的時候，投擲一個銅板了?次 10,000次：5067次正面 正面機率：0.5067 (3)約1900年，英國統計學家Karl Pearson很偉大的把一個銅板擲了？次 24,000次：12,012正面 正面機率：0.5005 擲銅板時的正面比率隨著我們擲的次數增加，比率會接近0.5 .也就是正面的機率。

古典機率論 又稱先驗的機率理論 (prior probability) 它假設各個互斥的可能outcome發生的機率相同。（注意：必須真的相同，不能不可能相同的也假定相同） 用於：公平骰子、公平銅板、或公平的等面體的投擲實驗上。

互斥、耗盡條件 互斥(mutually exclusive)：一事件發生，其他事件就不能同時發生，如：擲一骰子，一次只能出現一種點數，不可能又出2點、同時又出5點。 互斥的集合：兩集合無共同元素。i.e. A∩B=Φ 耗盡(collectively exhaustive)：只要是可能發生的事件，每項實驗至少會出現其中之一。如：擲一骰子，不是出現雙數，就是出現單數。

實證機率論 Empirical Probability LO5-2 實證機率論 Empirical Probability Empirical approach to probability is based on what is called the Law of Large Numbers. 大數法則 The key to establishing probabilities empirically: a larger number of observations provides a more accurate estimate of the probability. 觀察值（次數）越多（亦即：實驗次數越多），機率的估算越準確。見本次ppt的p.12 5-*

擲一公正銅板n次的實驗結果： 大數法則

實證機率論 Empirical Probability 如果無法用古典機率得出事件的機率時，可採用此方法。只要可實驗/收集資料，都可算機率。 就是採用相對次數法(relative frequency approach) 用過去資料估算來求得機率。 例如：老師當人的機率，給A的機率。 林書豪上籃的機率，有林書豪的比賽勝算有多少。 保險精算：用歷史資料來估算是否接收某人的保單，該收多少保費。用死亡率、未來死亡風險機率等等計算。（風險機率：血液算命學）

為什麼心臟病總是突然發作 - 血液算命學 新光醫院心臟內科主任 洪惠風 「血液算命學」(Framingham Point Scores)：源自Framingham的研究，提供一套風險預估公式，分別教導男性、女性如何評估在未來十年內罹患冠狀動脈心臟病的機率，也可以作為了解心臟年齡的參考值 加總各危險因子分數： (年齡分數) + (總膽固醇值分數) + ( HDL ) + (血壓值分數) + (是否有糖病) + (是否抽菸) =總分 然後看表：分男性、女性各有不同風險機率 見：血液算命學.pdf

Empirical Probability - Example LO5-2 Empirical Probability - Example On February 1, 2003, the Space Shuttle Columbia exploded. This was the second disaster in 123 space missions for NASA. On the basis of this information, what is the probability that a future mission is successfully completed? 5-*

主觀機率論Subjective Probability - Example LO5-2 主觀機率論Subjective Probability - Example If there is little or no data or information to calculate a probability, it may be arrived at subjectively. Illustrations of subjective probability are: Estimating the likelihood the New England Patriots will play in the Super Bowl next year. Estimating the likelihood a person will be married before the age of 30. Estimating the likelihood the U.S. budget deficit will be reduced by half in the next 10 years. 5-*

主觀機率論 Subjective Probability 若無法實驗，無歷史資料，無法客觀估算，僅能主觀估算者（依據某專家的看法、意見估算）。 此機率之決定，取決於人們對事件發生的可能的主觀看法/信賴程度。 如：健保未來10年內會上漲50％的機率？ 如：經濟系女籃代表隊明年打進前4強的機率？ 如：老張未來一年內發生車禍的機率？

三種機率理論比較：例子 (1) 從一副52張樸克牌中隨機抽一張。請問：抽到老K的機率是多少？你是用的哪一種機率理論回答這問題？ 古典機率理論 (2) 幼兒看護中心中有539位幼童，他們的父母婚姻狀況如下：333位的父母已婚，182位的父母離婚，24位的父母單親。隨機抽樣一位幼童他的父母是離婚的機率是多少？ 182/539＝0.34; 實證機率論/相對機率理論 (3) 小美在40歲之前登上玉山的機率是多少？ 主觀機率…

Summarizing Probability LO5-2 Summarizing Probability 5-*

加法定理 Rules of Addition for Computing Probabilities LO5-3 Calculate probabilities using rules of addition. 加法定理 Rules of Addition for Computing Probabilities Special Rule of Addition - If two events A and B are mutually exclusive, the probability of one or the other event occurring equals the sum of their probabilities. 互斥 P(A or B) = P(A) + P(B) Events are mutually exclusive if the occurrence of any one event means that none of the others can occur at the same time. 5-*

Special Rule of Addition- Example Mutually Exclusive Events LO5-3 Special Rule of Addition- Example Mutually Exclusive Events A machine fills 蔬菜包plastic bags with a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size of the beans and other vegetables, a package might be underweight or overweight. A check of 4,000 packages filled in the past month revealed: What is the probability that a particular package will be either underweight or overweight? P(A or C) = P(A) + P(C) = .025 + .075 = .10 5-*

Basic Probability: The Complement Rule 餘集（餘事件）定理 LO5-1 Basic Probability: The Complement Rule 餘集（餘事件）定理 The complement rule is used to determine the probability of an event occurring by subtracting the probability of the event not occurring from 1. P(A) + P(~A) = 1 or P(A) = 1 - P(~A). 5-*

The Complement Rule - Example LO5-1 The Complement Rule - Example An experiment has two 互斥 mutually exclusive outcomes. Based on the rules of probability, the sum of the probabilities must be one. If the probability of the first outcome is .61, then logically, AND by the complement rule, the probability of the other outcome is (1.0 - .61) = .39. P(B) = 1 - P(~B) = 1 – .61 = .39 5-*

餘集合定理 – Example (p.155) 一家冷凍公司的產品，抽樣4000包。發現大部分的包裝符合要求重量，但是有些過輕，有些過重。詳情如下： 這些事件能否運用補集合定理? Use the complement rule to show the probability of a satisfactory bag is .900 運用補集合定理証明: 假設事件B是符合標準重量，P(B)=0.9 P(~B)=? =P(A or C) =0.025+0.075=0.1 P(B) = 1 - P(~B) = 1 – P(A or C) = 1 – [P(A) + P(C)] = 1 – [.025 + .075] = 1 - .10 = .90 28

Complement Rule- Example 互斥 Mutually Exclusive Events LO5-3 Complement Rule- Example 互斥 Mutually Exclusive Events The complement rule can also be used: Note that P(A or C) = P(~B), so P(~B) = 1 – P(B) = 1 - .900 = .10 5-*

P( A and B) is called a 聯合機率 joint probability. LO5-3 Rules of Addition for Computing Probabilities The General Rule of Addition - If A and B are two events that are 非互斥 not mutually exclusive, then P(A or B) is given by the following formula: P(A or B) = P(A) + P(B) - P(A and B) P( A and B) is called a 聯合機率 joint probability. 5-*

The General Rule of Addition LO5-3 The General Rule of Addition The Venn Diagram shows the results of a survey of 200 tourists who visited Florida during the year. The results revealed that 120 went to Disney World, 100 went to Busch Gardens, and 60 visited both. What is the probability a selected person visited either Disney World or Busch Gardens? (求聯合機率) P(Disney or Busch) = P(Disney) + P(Busch) - P(both Disney and Busch) = 120/200 + 100/200 – 60/200 = .60 + .50 – .80 5-*

General Rule of Addition– Example LO5-3 General Rule of Addition– Example What is the probability that a card chosen at random from a standard deck of cards will be either a king（老K） or a heart（紅心）? 聯集 P(A or B) = P(A) + P(B) - P(A and B) = 4/52 + 13/52 - 1/52 = 16/52, or .3077 5-*

Special Rule of Multiplication LO5-4 Calculate probabilities using the rules of multiplication. Special Rule of Multiplication The special rule of multiplication calculates the joint probability of two events A and B that are 獨立 independent. Two events A and B are independent if the occurrence of one has no effect on the probability of the occurrence of the other. This rule is written: P(A and B) = P(A)P(B) . 5-*

Special Rule of Multiplication-Example LO5-4 Special Rule of Multiplication-Example A survey by the American Automobile Association (AAA) revealed 60 percent of its members made airline reservations last year. Two members are selected at random. Since the number of AAA members is very large, we can assume that R1 and R2 are independent. What is the probability both made airline reservations last year? Solution: The probability the first member made an airline reservation last year is .60, written as P(R1) = .60 The probability that the second member selected made a reservation is also .60, so P(R2) = .60. P(R1 and R2) = P(R1)P(R2) = (.60)(.60) = .36 5-*

General Rule of Multiplication LO5-4 General Rule of Multiplication The general rule of multiplication is used to find the joint probability that two events will occur when they are 互不獨立not independent. It states that for two events, A and B, the joint probability that both events will happen is found by multiplying the probability that event A will happen by the 條件機率conditional probability of event B occurring given that A has occurred. 5-*

條件機率 Conditional Probability LO5-4 條件機率 Conditional Probability A conditional probability is the probability of a particular event occurring, given that another event has occurred. The probability of the event A given that the event B has occurred is written P(A|B).：表示事件B發生後，再發生事件A的機率。 5-*

條件機率 Conditional Probability A事件的條件機率:B事件發生後再發生A的機率，可寫為： P(A|B)＝P(A and B) P(B) P(B) ≠0 37

General Rule of Multiplication – Example (p. 148) LO5-4 General Rule of Multiplication – Example (p. 148) A golfer has 12 golf shirts in his closet. Suppose 9 of these shirts are white and the others blue. He gets dressed in the dark, so he just grabs a shirt and puts it on. He plays golf two days in a row and does not do laundry. ∵不洗衣服 ∴穿過的衣服的不會再選 What is the likelihood both shirts selected are white? 5-*

General Rule of Multiplication - Example W1：第一天挑到白球衣的事件 W2：第二天挑到白球衣的事件 這兩個事件是否互相獨立？ 若獨立，則兩天都挑到白球衣的機率用p.146 的 [5-5]式求 若不獨立，則用p.148 的 [5-6]式求

General Rule of Multiplication - Example LO5-4 General Rule of Multiplication - Example The probability that the first shirt selected is white is P(W1) = 9/12. The probability of selecting a second white shirt (W2 ) is dependent on the first selection. So, the conditional probability is the probability the second shirt selected is white, given that the first shirt selected is also white: P(W2 | W1) = 8/11. 第二件衣服只能從剩下的乾淨衣服裡選 Apply the General Multiplication Rule[5-6]: P(A and B) = P(A) P(B|A) The joint probability of selecting 2 white shirts is: P(W1 and W2) = P(W1)P(W2 |W1) = (9/12)(8/11) = 0.55 5-*

若題目有所更動… (p.149) 如果球賽是三天的話，其他條件不變，那三天都穿到白色球衣的機率為何？ P(W1 and W2 and W3 ) = P(W1) P(W2 |W1) P(W3 |W1 and W2 ) =9/12 x 8/11 x 7/10 = 0.38

若題目有所更動… 若題目註明「選手會將第一次穿過的衣服吊回衣櫃，衣服不是只穿一天」。那兩天球衣都是白色的機率為何？ 第一件選的球衣是白色的機率 W1 P(W1) = 9/12 第二件選的球衣是白色的機率 W2，跟第一次選球衣是白色的機率有關嗎？ 無關，兩事件互相獨立。 所以P(W2) = 9/12 (W1 and W2) = P(W1) x P(W2) = 9/12 x 9/12 =0.5325

聯合機率 joint probability 如果兩事件A、B獨立，請問:P(A and B)=? P(A and B)=P(A∩B)=P(A)P(B) 如果兩事件A、B互斥，請問:P(A and B)=? P(A and B) =P(A∩B)=0 若A,B互斥A∩B=，而P() = 0

Quiz: 當兩事件獨立，P(B｜A)=? P(A｜B)=? P(B), P(A) P(B|A) = P(A∩B)/P(A)=P(A)P(B)/P(A)=P(B) 2. 當三事件不互斥時，P(A∪B∪C) =? P(A)+P(B)+P(C)-P(A∩B)-P(A∩C)-P(C∩B)+P(A∩B∩C)

列聯表 Contingency Tables p.149

列聯表 Contingency Tables p.149

列聯表 Contingency Tables LO5-5 Calculate probabilities using a contingency table. 列聯表 Contingency Tables A contingency table is used to classify sample observations according to two or more identifiable characteristics measured. For example, 150 adults are surveyed about their attendance of movies during the last 12 months. Each respondent is classified according to two criteria—the number of movies attended and gender. 5-*

Contingency Table - Example LO5-5 Contingency Table - Example Based on the survey, what is the probability that a person attended zero movies? Based on the empirical information, P(zero movies) = 60/150 = 0.4 5-*

Contingency Table - Example LO5-5 Contingency Table - Example Based on the survey, what is the probability that a person attended zero movies or is male? (聯集，互斥否？) Applying the General Rule of Addition: P( zero movies or male) = P( zero movies) + P(male) – P(zero movies and male) P(zero movies or male) = 60/150 + 70/150 – 20/150 = 0.733 5-*

Contingency Table - Example LO5-5 Contingency Table - Example Based on the survey, what is the probability that a person attended zero movies if a person is male? (條件) Applying the concept of conditional probability: P( zero movies | male) = 20/70 = 0.286 5-*

Contingency Table - Example LO5-5 Contingency Table - Example Based on the survey, what is the probability that a person is male and attended zero movies? (求聯合機率，獨立否？) Applying the General Rule of Multiplication P( male and zero movies) = P(male)P(zero movies|male) = (70/150)(20/70) = 0.133 5-*

p.150上個月戲院經理工會隨機抽樣調查500個人，問：年齡、每個月看幾次電影，資料如下： 3. P(≧ 6次 or ≧ 60歲) = P(≧ 6次) + P(≧ 60歲) - P(≧ 6次 and ≧ 60歲) = 50/500 + 175/500 – 30/500 = 0.39

5. P(≧ 6次 and ≧ 60歲) = P(≧ 6次) P(≧ 60歲 |≧ 6次) = 50/500 * 30/50 = 0.06

年齡、次數是否互相獨立？要看 P(≧ 6次 | ≧ 60歲) = P(≧ 6次) ？或 不相等，故不獨立！

p.150 年齡、次數是否互相獨立？也可看下式是否成立 P(≧ 6次 and ≧ 60歲) = P(≧ 6次) P(≧ 60歲) ？ 不相等，故不獨立！

樹形圖 Tree Diagrams A tree diagram is: LO5-5 樹形圖 Tree Diagrams A tree diagram is: 1. 用來描繪條件機率、聯合機率非常好用。 Useful for portraying conditional and joint probabilities. 2. 最常被用來分析多階段的企業決策。 Particularly useful for analyzing business decisions involving several stages. 3. 多階段決策中用來計算機率非常好用。 A graph that is helpful in organizing calculations that involve several stages. 每一層討論一個階段的問題，樹枝上的機率為該可能結果的權數。 Each segment in the tree is one stage of the problem. The branches of a tree diagram are weighted by probabilities. 5-*

p.150 上個月戲院經理工會隨機抽樣調查500個人，問：年齡、每個月看幾次電影，資料如下： 根據此 contingency table可做出樹形圖

p.153

LO5-5 Tree Diagram- Example 抽取200名經理來調查他們對公司的忠誠度，問卷問題：「若其他公司給你相等或更好一點的職缺，你會留在原來的公司，還是接受其他公司的聘僱？」，問卷結果列於下表：（包含他們年資、與他們的回答） A sample of executives were surveyed about their loyalty to their company. One of the questions was, “If you were given an offer by another company equal to or slightly better than your present position, would you remain with the company or take the other position?” The responses of the 200 executives in the survey were cross-classified with their length of service with the company. 5-*

LO5-5 Tree Diagram - Example 5-*

61

貝氏定理 Bayes’ Theorem 若有新的情資，貝氏定理提供一個修正(更新)機率的方法 LO5-6 Calculate probabilities using Bayes’ theorem. 貝氏定理 Bayes’ Theorem 若有新的情資，貝氏定理提供一個修正(更新)機率的方法 Bayes’ theorem is a method for revising a probability given additional information. It is computed using the following formula: 5-*

事件B是新資訊(additional information) P(Ai)事前機率(prior probability) 根據現有資訊所得的最初機率 (the initial probability based on the present level of information) P(BIAi)條件機率 事件B的條件機率 P(AiIB)事後機率(posterior probability) 根據新資訊來修正事前機率之後所得的機率 (a revised probability based on additional information)

貝氏定理 Bayes’ Theorem 事後機率 條件機率 X 事前機率

貝氏定理 Bayes’ Theorem 貝氏定理乃描述一有先後發生順序的是件，如同樹形機率圖一般，當有新資訊時，貝氏定理便可用來更新先前事件發生的機率。更新後的機率為事後機率。 Bayes' theorem can be understood better by visualizing the events as sequential as depicted in the probability tree. When additional information is obtained about a subsequent event; it is used to revise the probability of the initial event. The revised probability is called posterior. 換句話說，我們原先有的因果關係模型：可用來預測B發生的可能性（在Ai已經發生的情況下） In other words, we initially have a cause-effect model where we want to predict whether event B will occur or not, given that event Ai has occurred. 當我們被告知B已經發生，這時，我們便進入推論模式，而我們的目標是Ai究竟是否發生，其機率為何？ We then move to the inference model where we are told that event B has occurred and our goal is to infer whether event Ai has occurred or not 總之，貝氏定理提供了一種簡單的方法，可將每個可能肇因（A）的不同影響（結果：B）的機率（ i.e. P(B|Ai) ），轉換成在該影響（結果：B）下，其肇因(A) 可能發生的機率 （i.e. P(Ai|B) ）。 In summary, Bayes' Theorem provides us a simple technique to turn information about the probability of different effects (outcomes) from each possible cause, into information about the probable cause given the effect (outcome).

貝氏定理 Bayes’ Theorem p.157 若原先的因果關係模型如下：在Ai已經發生的情況下，且已知事前機率P(Ai) ，要預測B發生的條件機率：P(B|Ai) 然後，我們被告知B已經發生（多了一些資訊），這時，我們可用Bayes' theorem來推論（到推回去）Ai發生的機率：事後機率 P(Ai|B)

範例： Umen是第三世界的國家之一。其5％的人口感染當地特殊的風土病。已知A1(得病)的機率： P(A1)=0.05  事前機率 現在有新的醫療檢測，能夠檢測當地人是否有感染到此疾病。但是不夠準確：如果已感染這疾病，檢測到（陽性反應）(B) 的機率是0.9;若沒有感染疾病，這醫療方法會誤診人得此病（檢驗呈陽性反應）的機率是0.15。已知 A1(得病)而B(偵測到)的機率： P(B|A1)=0.9 A2(未得病)而B(偵測到)的機率： P(B|A2)=0.15 如果我們隨機抽樣一人，醫療方法指出這個人有得此病。請問：此人真的有得病的機率是多少？

先定義：事前(最初事件) A1：感染疾病 A2: 沒有感染 事前機率 P(A1)=0.05, P(A2)=0.95 再定義： (新事件)醫療測試成效 B：醫療測試指出有感染疾病（如：陽性反應） 新資訊： 在病人有染病的前提下，醫療測試「陽性反應」的機率? 條件機率P(B|A1)=0.9 在病人沒有染病的前提下，醫療測試「陽性反應」的機率? 條件機率P(B|A2)=0.15 Q:測試結果是「陽性反應」的前提下，有多少人是真的有染病？換言之，我們想知道事後機率（已知有「陽性反應」但推論「是否真染病？」）：P(A1|B)=?

=0.045/0.1875=0.24 這告訴我們什麼？ (1)若從人口中隨機抽樣一人，他得到此疾病的機率是0.05：P(A1) (2)但若是此人透過醫療檢驗並且檢驗結果顯示有得病，那這人真的有得病的機率是0.24: P(A1|B) 檢測後的事後機率P(A1|B)>P(A1):若檢測得病而真正得病的機率是24％，比不減測確實準確些（不檢測，得病機率僅5％）（即使檢驗效果不夠準確） 但檢測陽性反應僅24％可能真的得病，故「陽性反應」結果僅能做參考而已

Bayes’ Theorem – Example p. 157 LO5-6 Bayes’ Theorem – Example p. 157 5-*

Bayes’ Theorem – Example p.157 事前機率 LO5-6 Bayes’ Theorem – Example p.157 事前機率 5-*

Bayes’ Theorem – Example p.157 條件機率 LO5-6 Bayes’ Theorem – Example p.157 條件機率 5-*

Bayes’ Theorem – Example p.157 條件機率joint probability P(A1∩B) = P(A1) P(B|A1)=0.3X0.03=0.009 P(A2∩B) = P(A2) P(B|A2)=0.2X0.05=0.01 P(A3∩B) = P(A3) P(B|A3)=0.5X0.04=0.02 ΣP(Ai ∩B) =0.039

Bayes’ Theorem – Example 事後機率 LO5-6 Bayes’ Theorem – Example 事後機率 5-*

也可以用下表來計算事後機率 p.159

Bayes Theorem – Example (P.159) 77

計數法則：p.160~ 如果可能的outcomes數目很大，有三種方法可以計算outcomes的數目： (1)Multiplication formula乘數法則 (2)Permutation formula 排列法則 (3)Combination formula 組合法則

Counting Rules – Multiplication 乘數法則 LO5-7 Determine the number of outcomes using principles of counting. Counting Rules – Multiplication 乘數法則 The multiplication formula indicates that if there are m ways of doing one thing and n ways of doing another thing, there are m x n ways of doing both. Example: Dr. Delong has 10 shirts and 8 ties. How many shirt and tie outfits does he have? (10)(8) = 80 Official：一實驗包含k次試驗(E1,E2,…Ek)，若每一試驗Ei有ni種結果(i=1,2,3…k)，則該隨機實驗有n1xn2xn3….xnk種可能結果 5-*

Counting Rules – Multiplication Example LO5-7 Counting Rules – Multiplication Example An automobile dealer wants to advertise that for $29,999 you can buy a convertible, a two-door sedan, or a four-door model with your choice of either wire wheel covers or solid wheel covers. How many different arrangements of models and wheel covers can the dealer offer? 5-*

Counting Rules – Permutation 排列法則 LO5-7 Counting Rules – Permutation 排列法則 A permutation is any arrangement of r objects selected from n possible objects. The order of arrangement is important in permutations. 排列順序很重要 5-*

排列法則 Permutation formula(p.169) 排列法則： 從一組含有n個元素的集合中，一次抽取r個元素(或每抽取一個，抽出不放回，連續抽r個)，則共有nPr 種不同排列。元素排列的順序很重要。 A permutation is any arrangement of r objects selected from n possible objects. The order of arrangement is important in permutations. n! n factorial = n(n-1)x(n-2)x(n-3)x….(1) 0!=1 82

範例 一般提款卡都會要求四位數密碼。那麼現在假設陳小姐剛開了個帳戶、拿了新的提款卡， (1)只是想設定任意四位密碼，請問有幾種可能性？ 根據乘數公式，四位數字中，每一位都有10種可能性，所以：10*10*10*10＝10,000 (2)想設定四位不相同數字的密碼，而且這四個數字順序不同。請問會有幾種可能性？ n=10, r=4 nPr=10!/(10-4)!=10!/6!=10x9x8x7=5,040

補充：乘數法則 乘數法則下，有抽出不放回的概念。 剛提到：想設定四位不相同的數字的密碼，而且這四個數字順序不同。請問會有幾種可能性？ 第一位數：有10種可能數字 第二位數：有9種可能數字 第三位數：有8種可能數字 第四位數：有7種可能數字 所以，10*9*8*7＝5040 這四位不同數字在不同順序的前提下，會有5040種的排列

範例： p.162 組裝3個零件，可隨便安裝排列，有幾種安裝（排列）方式？ n=3, r=3, 排列方式 = 3!/0! = 6

範例： p.163 有8種螺絲釘拴，但只有3個空位安放，有幾種安裝（排列）方式？ n=8, r=3, 排列方式 = 8!/(8-3)! = 8 X 7 X 6=336

Counting Rules – Combination 組合法則 LO5-7 Counting Rules – Combination 組合法則 A combination is the number of ways to choose r objects from a group of n objects without regard to order. 5-*

組合法則 Combination (p.170) 組合法則：從一組含有n個元素的集合中，一次抽取r個元素(或每抽取一個，抽出不放回)，這r個元素會有nCr 種不同組合可能性。這些元素排列的順序不重要。 A combination is the number of ways to choose r objects from a group of n objects without regard to order. 88

範例： p.163 Grand 16戲院的小賣部每晚得3人照顧生意，戲院有7個員工可以輪班，問：能派出幾種組合來輪班？ 7C3 = 7!/(3!4!) = 7*6*5 / 1*2*3 = 35

範例 最近某公司考慮跟別加公司合併。如果主管Adrian, Baker, Charles 被選到參與委員會商討合併的話（三選三） 請問：有幾種可能的組合？ n=3, r=3 nCr ＝3!/3!0!=1 (2) 若此委員會按責任分三種職位，請問：這三人擔任這些職位的可能性有幾種？ nPr ＝3!/0!=6

例子 自五位科長中，隨機抽取2位，則總共有多少樣本點？又若5位科長中有3位男性，2位女性，則一位男科長、一位女科長被選為副理的機率是多少？ (1)樣本空間共有樣本點數為： nCr= 5C2 =5!/2!3!=10 (2)一位男科長與一位女科長被選為副理的樣本點數為： 3C1x 2C1=(3!/2!1!)x(2!/1!1!)=3x2=6 所以機率為：6/10＝0.6

Combination and Permutation Examples LO5-7 Combination and Permutation Examples COMBINATION EXAMPLE There are 12 players on the Carolina Forest High School basketball team. Coach Thompson must pick 5 players among the 12 on the team to comprise the starting lineup. How many different groups are possible? PERMUTATION EXAMPLE Suppose that in addition to selecting the group, he must also rank each of the players in that starting lineup according to their ability. How many different rankings are possible for five players selected from the 12? 5-*