Assigning probabilities to Events l Random experiment l

Assigning probabilities to Events l Random experiment l l Examples l l l 2 a random experiment is a process or course of action, whose outcome is uncertain. 擲一粒骰子，觀察其出現何種點數抽取一份統計考試成績，觀察其分數在馬路上假裝跌倒，看看多少時間後有沒有人來扶你（當然也不太鼓勵你, 做這樣的試驗, 以免自信心受傷） Jia-Ying Chen

Assigning probabilities to Events l l 3 Performing the same random experiment repeatedly, may result in different outcomes, therefore, the best we can do is consider the probability of occurrence of a certain outcome. To determine the probabilities we need to define and list the possible outcomes first. Jia-Ying Chen

Sample Space l Determining the outcomes. l l l 4 Build an exhaustive list of all possible outcomes. Make sure the listed outcomes are mutually exclusive. A list of outcomes that meets the two conditions above is called a sample space. Jia-Ying Chen

Sample Space: S = {O 1, O 2, …, Ok} O 1 O 2 Sample Space a sample space of a random experiment is a list of all possible outcomes of the experiment. The outcomes must be mutually exclusive and exhaustive. 5 Simple Events The individual outcomes are called simple Event events. An event is any collection Simple events of one or more simple eventscannot be further decomposed Our objective is to into constituent determine P(A), the outcomes. probability that event A will occur. Jia-Ying Chen

Assigning Probabilities 6 l Given a sample space S={O 1, O 2, …, Ok}, the following characteristics for the probability P(Oi) of the simple event Oi must hold: l Probability of an event: The probability P(A) of event A is the sum of the probabilities assigned to the simple events contained in A. Jia-Ying Chen

Approaches to Assigning Probabilities… l. There are three ways to assign a probability, P(Oi), to an outcome, Oi, namely: l. Classical approach: make certain assumptions (such as equally likely, independence) about situation. l. Relative frequency: assigning probabilities based on experimentation or historical data. l重複進行此一實驗許多次，並觀察該事件出現次數的比例 l有一位英國統計學家Pearson(1857 -1936)很神勇的擲一個銅板 24, 000次，結果出現 12, 012次正面，出現正面的機率為 12012/24000=0. 5005。 7 Jia-Ying Chen

Example 1 l A quiz contains multiple-choice questions with five possible answers, only one of which is correct. A student plans to guess the answers because he knows absolutely nothing about the subject. a. Produce the sample space for each question. b. Assign probabilities to the simple events in the sample space you produced. c. Which approach did you use to answer part b d. Interpret the probabilities you assigned in part b 9 Jia-Ying Chen

Solution l l 10 a. {a is correct, b is correct, c is correct, d is correct, e is correct} b. P(a is correct) = P(b is correct) = P(c is correct) = P(d is correct) = P(e is correct) =. 2 c. Classical approach d. In the long run all answers are equally likely to be correct. Jia-Ying Chen

Joint, Marginal, and Conditional Probability l We study methods to determine probabilities of events that result from combining other events in various ways. l There are several types of combinations and relationships between events: l l 13 Intersection of events Union of events Dependent and independent events Complement event Jia-Ying Chen

Intersection 14 l The intersection of event A and B is the event that occurs when both A and B occur. l The intersection of events A and B is denoted by (A and B; A∩B). l The joint probability of A and B is the probability of the intersection of A and B, which is denoted by P(A and B) Jia-Ying Chen

Intersection l Additional Example – 1 l The number of spots turning up when a six-side die is tossed is observed. Consider the following events. § § l A: The number observed is at most 2. B: The number observed is an even number. Determine the probability of the intersection event A and B. 3 1 A 2 4 15 A and B B 6 P(A and B) = P(2) = 1/6 5 Jia-Ying Chen

Marginal Probability l These probabilities are computed by adding across rows and down columns l 在兩個或兩個以上類別的樣本空間中，若僅考慮一類別個別發生的機率稱之。 Mutual fund outperforms the market (B 1) Top 20 MBA program (A 1) Not top 20 MBA program (A 2) Mutual fund doesn’t outperform the market (B 2) 0. 11 0. 29 0. 06 0. 54 Marginal Prob. P(Ai) P(A 1 and B 1)+ P(A 1 and B 2) = P(A 2 and B 1)+ P(A 2 and B 2) = P(A Marginal Probability P(Bj) 16 Jia-Ying Chen

Marginal Probability l These probabilities are computed by adding across rows and down columns Mutual fund outperforms the market (B 1) Top 20 MBA program (A 1) P(A 1 and B 1) + Not top 20 MBA program P(A 2 and B 1 = (A 2) P(B 1) Marginal Probability P(Bj) 0. 17 17 Mutual fund doesn’t outperform the market (B 2) P(A 1 and B 2) + P(A 2 and B 2 = P(B 0. 83 2) Marginal Prob. P(Ai) . 40. 60 Jia-Ying Chen

Conditional Probability l 已知A 事件發生下，另一B事件發生的機率，稱為在A發生條件下，A的條件機率。 P(A|B) = P(A and B) / P(B) l Example 6. 2 (Example 6. 1 – continued) l l l Find the conditional probability that a randomly selected fund is managed by a “Top 20 MBA Program graduate”, given that it did not outperform the market. Solution P(A 1|B 2) = P(A 1 and B 2) =. 29 =. 3949 P(B 2). 83 18 Jia-Ying Chen

Independence & Union l Independent events l Two events A and B are said to be independent if P(A|B) = P(A) or l 19 P(B|A) = P(B) That is, the probability of one event is not affected by the occurrence of the other event. l The union event of A and B is the event that occurs when either A or B or both occur. l It is denoted “A or B” ; (A∪B). Jia-Ying Chen

Example 2 l l 20 Suppose we have the following joint probabilities A 1 A 2 A 3 B 1 0. 2 0. 15 0. 1 B 2 0. 25 0. 05 Compute the marginal probabilities, Jia-Ying Chen

Solution l 21 P(A 1)=0. 2+0. 25=0. 45, P(A 2)=0. 15+0. 25=0. 4, P(A 3)=0. 1+0. 05=0. 15, P(B 1)=0. 2+0. 15+0. 1=0. 45, P(B 2)=0. 25+0. 05=0. 55 Jia-Ying Chen

Example 3 l The following tables lists the joint probabilities associated with smoking and lung disease among 60 to 65 year-old-men. He is a smoker He is a nonsmoker He has lung disease 0. 1 0. 03 He does not have lung disease 0. 21 0. 66 One 60 to 65 year old man is selected at random. What is the probability of the following events a. He is a smoker b. He does not have lung disease c. He has lung disease given that he is a smoker d. He has lung disease given that he does not smoke l 22 Jia-Ying Chen

Solution l l 23 a. P(He is a smoker)=0. 1+0. 21=0. 31 b. P(He does not have lung disease )=0. 21+0. 66=0. 87 c. P(He has lung disease｜ he is a smoker)= P(He has lung disease and he is a smoker)/P(he is a smoker)=0. 1/0. 31=0. 323 d. P(He has lung disease ｜ he does not smoke)= P(He has lung disease and he does not smoke)/P(he does not smoke)=0. 03/0. 69=0. 044 Jia-Ying Chen

Probability Rules and Trees l l l We present more methods to determine the probability of the intersection and the union of two events. Three rules assist us in determining the probability of complex events from the probability of simpler events. Complement Rule l l 24 The complement of event A (denoted by AC) is the event that occurs when event A does not occur. The probability of the complement event is calculated by A and AC consist of all the simple events in the sample space. Therefore, P(A) + P(AC) = 1 - P(A) Jia-Ying Chen

Multiplication Rule l For any two events A and B P(A and B) = P(A)P(B|A) = P(B)P(A|B) l When A and B are independent P(A and B) = P(A)P(B) 25 Jia-Ying Chen

Addition Rule For any two events A and B P(A or B) = P(A) + P(B) - P(A and B) P(A) =6/13 + A P(B) =5/13 _ P(A and B) =3/13 26 P(A or B) = 8/13 B A or B Jia-Ying Chen

Addition Rule When A and B are mutually exclusive, P(A or B) = P(A) + P(B) A 27 B B P(A and B) = 0 Jia-Ying Chen

思考!! l What is the difference between mutually exclusive and Independent ? l Mutually exclusive l l l Independent l l 28 P(A or B) = P(A) + P(B) P(A and B)=0 P(A and B) = P(A)P(B) P(A∣B)=P(A) or P(B∣A)=P(B) Jia-Ying Chen

Example 4 l a. b. c. d. 29 Approximately 10% of people are lefthanded. If two people are selected at random, what is the probability of the following events? Both are right-handed Both are left-handed One is right-handed and the other is lefthanded At least one is right-handed Jia-Ying Chen

Solution l l 30 a. P(R and R)=0. 81 b. P(L and L)=0. 01 c. P(R and L)+P(L and R)=0. 09+0. 09=0. 18 d. P(R and L)+P(L and R)+P(R and R)=0. 09+0. 81=0. 99 Jia-Ying Chen

貝氏定理（Bayes’ Theorem） l Example l l 32 Medical tests can produce false-positive or falsenegative results. A particular test is found to perform as follows: l Correctly diagnose “Positive” 94% of the time. l Correctly diagnose “Negative” 98% of the time. It is known that 4% of men in the general population suffer from the illness. What is the probability that a man is suffering from the illness, if the test result were positive? Jia-Ying Chen

貝氏定理（Bayes’ Theorem） l Solution l Define the following events l l l 33 D = Has a disease DC = Does not have the disease PT = Positive test results NT = Negative test results Build a probability tree Jia-Ying Chen

貝氏定理（Bayes’ Theorem） l Solution – Continued l The probabilities provided are: l l 34 P(D) =. 04 P(PT|D) =. 94 P(PT|DC) =. 02 P(DC) =. 96 P(NT|D)=. 06 P(NT|DC) =. 98 The probability to be determined is Jia-Ying Chen

貝氏定理（Bayes’ Theorem） Prior Likelihood probabilities |D T P ( P(D ) 4 0. = P(D C )= . 96 ) P P( N T|D 4 9. = )= C) T|D P(P P( N . 06 2 =. 0 T|D C )= 36 Posterior probabilities . 98 Jia-Ying Chen