Lecture 1 Introduction Properties of Probability Methods of Enumeration Asrat Temesgen Stockholm University


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1 Lecture 1 Introduction Properties of Probability Methods of Enumeration Asrat Temesgen Stockholm University 1
2 Chapter 1 Probability 1.1 Basic Concepts In the study of statistics, we consider experiments for which the outcome can t be predicted with certainty. Such experiments are called random experiments. The collection of all possible outcomes is called the outcome space, the sample space, or simply the Space S. Example 1 Two dices are cast, and the total numbers of spots on the sides that are up are counted. Then S = {2, 3, 4, 12} Example 2 A fair coin is flipped successfully at random until the first head is observed. If we let x denote the number of flips of the coin that are required, then S={X: X = 1,2,3 }, which consists of an infinite, but countable, number of outcomes. Example 3 A fair coin is flipped successfully at random until heads is observed on two successive flips. If we let y denote the number of flips of the coin that are required, then S={Y: Y=2, 3, 4, }. In general, measurements on the outcomes associated with random experiments are called random variables, and these are usually denoted by some capital letter, such as X, Y, Z. Exercise Define the following terms: Population Sample Statistical Inference Discrete data Frequency table Histogram Mode Relative frequency histogram 2
3 1.2 Properties of probability Given a sample space S, let A be a part of the collection of outcomes is S; that is is called an event.. Then A Remark: In studying probability, the words set and event are interchangeable. Some Terminologies: denotes the null or empty set; A is a subset of B ; is the union of A and B; is the intersection of A and B; is the complement of A (i.e. all elements is S that are not in A) (See the Venn diagram in Fig of your textbook) Special Terminologies: 1. are mutually exclusive events mean that 2. are exhaustive events means that So if are mutually exclusive and exhaustive events, we know that and. The number P assigned to event A is called the probability of event A and is denoted by P (A). Definition 1.2.1: Probability is a real valued set function P that assigns to each event A in the sample space S, a number P (A), called the probability of the event A, Such that the following properties are satisfied: a. b. 3
4 c. = for each positive integer k, and = for an infinite, but countable, number of events Theorem 1.21: For each event A, Proof: We have and Thus, from properties (b) and (c), it follows that Hence. Example: A fair coin is flipped successively until the same face is observed on successive flips. Let that is, A is the event that it will take three or more flips of the coin to observe the same face on two consecutive flips. To find, we first find, where }. In two flips of a coin, the possible outcomes are and we assume that each of these four points has the same chance of being observed. Thus,. It follows from the theorem that =. Theorem 1.22: Proof: In theorem1.21, take so that.then Theorem1.23: If events A and B are such that, then Proof: We have and. Hence, from property (C),,since form property (a), Theorem : For each event A,. Proof: Since, we have, by theorem and property (b), property (a), along with theorem 1.24, shows that, for each event A, 4
5 Theorem 1.25: If A and B are any two events, then. Proof: Hence, by property (c), (1.21) However, Thus, equation 1.21, we obtain:. Substituting in, which is the desired result. Example: A faculty leader was meeting two students in Paris, one arriving by train from Amsterdam and the other arriving by train from Brussels at approximately the same time. Let A and B be the events that the respective trains are on time. Suppose we know from past experience that, and.then is the probability that at least one train is on time. Theorem1.26: If A, B and C are any three events, then Proof: Left as an exercise. Hint: Write and apply theorem Example: A survey was taken of a group s viewing habits of sporting events on TV during the last year. Let A= {watched football}, B= {watched basketball}, C= {watched baseball}. From past results,,,,,,, and. Then is the probability that a person selected from the group surveyed watched at least one of these sports. 5
6 Note: Let.Then if each of these outcomes has the same probability of occurring, we say that the outcomes are equally likely. i.e.. Under this assumption of equally likely outcomes, we have, where 1.3 Methods of Enumeration: Multiplication Principle: Suppose that an experiment (or procedure) has outcomes and, for each of these possible outcomes, an experiment (Procedure) has possible outcomes. Then the composite experiment (Procedure) that consists of performing first and then has possible outcomes. Example: A certain food service gives the following choices for dinner =soup or tomato juice; = Steak or shrimp; = French fried potatoes, mashed potatoes, or a baked potato; Corn or peas; Jello, tossed salad, cottage cheese, or coleslaw; cake, cookies, Pudding, brownie, Vanilla icecream, Chocolate icecream, or orange sherbet; coffee, tea, milk, or punch. How many different selections are possible if one of the listed choices is made for each of and? Then by the multiplication principle, there are different combinations. Note: Suppose that n positions are to be filled with n different objects. There are n choices for filing the first position, for the second,, and 1 choice for the last postion. So, by the multiplication principle, there are possible arrangements. The symbol is read. We define ; that is we say that zero postions can be filled with zero objects in one way. Definition 1.31: Each of the arrangements (in a row) of different objects is called a permutation of objects. 6
7 Example: The number of permutation of the four letters is clearly. However, the number of possible fourletter code words using the four letters letters may be repeated is if Definition 1.32: Each of the time. arrangements is called a permutation of n objects taken r at a Example: The number of possible fourletter code words, selecting from the 26 letters in the alphabet, in which all four letters are different is : Example: The number of ways of selecting a president, a vice president, a secretary, and a treasurer in a club consisting of 10 persons is: Definition 1.33: If r objects are selected from a set of objects, and if the order of selection is noted, then the selected set of objects is called an ordered sample of size Definition 1.34: Sampling with replacement By the multiplication principle, the number of possible ordered samples of size set of objects is when sampling with replacement. taken from a Example: A die is rolled five times. The number of possible ordered samples is. Note that rolling a die is equivalent to sampling with replacement. Example: An urn contains 10 balls numbered If 4 balls are selected one at a time and with replacement, then the number of possible ordered samples is =10,000. Note that this is the number of fourdigit integers between 0000 and 9999, inclusive. 7
8 Definition1.35: Sampling without replacement By the multiplication principle, the number of possible ordered samples of size r taken from a set of n objects without replacement is : n(n1) (nr+1) = n! / (nr)! Which is equivalent to. Example: The number of ordered samples of 5 cards that can be drawn without replacement from a standard deck of 52 playing cards is: (52)(51)(50)(49)(48)=52! /47! = 311,875,200. Definition : Each of the unordered subjects is called a combination of n objects taken r at a time, where =( )= Example: The number of possible 5card hands ( in 5card poker) drawn from a deck of 52 playing cards is : 52C5= ( ) = = 2,598,960 Definition : Each of the permutations of n objects, r of one type and (nr) of another type, is called a distinguishable permutation. Example: A coin is flipped 10 times and the sequence of heads and tails is observed. The number of possible 10 tuplets that result in four heads and six tails is : ( ) = = = ( ) = 210 Note: Suppose that in a set of n objects, are similar, are similar,, are similar, where =. Then the number of distinguishable permutations of the objects is =. Example: Among nine orchids for a line of orchids along one wall, three are white, four lavender, and two yellow. The number of different color displays is then ( ) = 1,260 8
9 Lecture 2 Conditional Probability Independent Events Bayes s Theorem Asrat Temesgen Stockholm University 9
10 1.4. Conditional Probability Definition 1.41: The conditional probability of an event, given that event B has occurred, is defined by: P (A/B) =, Provided that P(B) >0. Example: Suppose that P(A) = 0.7, P(B) = 0.3, and P (A ) = 0.2. Given that the outcome of the experiment belongs to B, what then is the probability of A? Solution: P(A/B) = = It is interesting to note that conditional probability satisfies the axioms for a probability function, namely, with P(B) >0, A. P(A/B) 0 B. P(B/B) = 1 C. If,,, are mutually exclusive events, then =, for each positive integer K, and = for an infinite, but countable, number of events. Proof: properties (a) and (b) are evident because P(A/B) = 0, Since P (A ) 0, P(B)>0, and P(B/B) = = 1. Property (c) holds because, for the second part of (c), = = But ( ),, are also mutually exclusive events ; so = 10
11 = + + = The first part of property (c) is proved in a similar manner. Definition 1.42: The probability that two events, A and B, both occur is given by the multiplication rule, P (A P (A P(A) P(B/A), or P(A) P(A/B) Example: A bowl contains seven blue chips and three red chips. Two chips are to be drawn successively at random and without replacement (WOR). Compute the probability a) that the first draw results in a red chip (A) and the second draw results in a blue chip(b). b) of obtaining red on the first draw and blue on the second draw. c) of drawing a red chip on each of the two draws. Solution a) P (A P(A). P(B/A) =. = b) c). = which is the same as = Example : From an ordinary deck of playing cards, cards are to be drawn successively at random and WOR. The probability that the third spade appears on the sixth draw is completed as follows. Let A be the event of two spades in the first five cards drawn, and let B be the event of a spade on the sixth draw. Thus, the probability that we wish to 11
12 compute is P (A P (A) = = and P(B/A)= Then, P (A = (0.274) (0.234) = Example: Four cards are to be dealt successively at random and WOR from an ordinary deck of playing cards. The probability of receiving, in order, a spade, a heart, a diamond, and a club is..., a result that follows from the extension of the multiplication rule and reasonable assignments to the probabilities involved. Example: A device has two components, and, but it will continue to operate with only one active component for a one year period. The probability that each component will fail when both are in operation is 0.01 in that one year period. However, when one component fails, the probability of the other failing is 0.03 in that period, due to added strain. Thus, the probability that the device fails in one year is: P ( fails/ fails) + P ( fails)p( / fails) = (0.01)(0.03)+(0.01)(0.03)= Independent Events For certain pairs of events, the occurrence one of them may or may not change the probability of the occurrence of the other. In the latter case, they are said to be independent events. Definition 1.51: Events A and B are independent if and only if P (A = P(A)P(B). Otherwise A and B are called dependent events. Note : events that are independent are sometimes called statistically independent, stochastically independent, or independent in a probabilistic sense. 12
13 Example: A red die and a white die are rolled. Let event A={4 on the red die} and event B={ sum of dice is odd}. Of the 36 equally likely outcomes, 6 are favorable to A, 18 are favorable to B, and 3 are favorable to A P(A)P(B) =. = =P (A Hence, A and B are independent by def Exercise : Let C= independent? and event D= {sum of dice is 11}. Are C and D (Ans. NO! since P(C)P(D)=. = = P(C Theorem 1.51: If A and B are independent events, then the following pairs of events are also independent: (a) A and B (b) A and B (c) A and B Proof: (a) We know that conditional probability satisfies the axioms for a probability function. Hence if P(A) 0, then P(B /A) = 1 P(B/A). Thus, P(A P(A) [1 P(B/A)] = P(A)[1P(B)], since P(B/A)= P(B) by hypothesis = P(A) P(B ) Hence, A and B are independent events. The proofs for parts (b) and (c) are left as exercises. Example: An urn contains 4 balls numbered 1, 2, 3 and 4. One ball is to be drawn at random from the urn. Let the events A, B, and C be defined by A= {1,2 }, B= {1,3}, and C={1,4}. Then P(A) = P(B)= P(C)= ½. Furthermore, P(A = ¼ = P(A)P(B), P(A 13
14 P(B which implies that A, B, and c are independent in pairs (called pairwise independence). However, since A {1}, we have P(A = P(A)P(B)P8C). That is, something seems to be lacking for the complete independence of A, B, and C. Definition 1.52: Events A, B, and C are mutually independent if and only if the following two conditions hold: a. A,B, and C are pairwise independent ; that is, P(A P(A)P(B), P(A = P(A)P(C), and P(B = P(B)P(C). b. P(A Note:(1) Definition can be extended to the mutual independence of 4 or more events. In such an extension, each pair, triple, quartet, and so on, must satisfy this type of multiplication rule. (2) If A,B,and C are mutually independent events, then the following events are also independent. (a) A and (B (b) A and (B ; (c) A and (B In addition A, B and C are mutually independent. Example: A fair sixsided die is rolled 6 independent times. Let Ai be the event that side i is observed on the ith roll, called a match on the ith trial, i= 1,2,,6. Thus, P( ) 1/6 and P( = 11/6= 5/6. If we let B denote the event that at least one match occurs, then B is the event that no matches occur. Hence, P(B)= 1P(B )= 1P( )= 15/65/65/65/65/65/6 =1 is the probability of B Bayes s Theorem Let, constitute a partition of the sample space S. That is S= and =Ø, i Of course, the events, are mutually exclusive and exhaustive (since the union of the disjoint sets equals the sample space S). Furthermore, suppose the prior 14
15 probability of the event is positive; that is P ( 0, i=1, m. If A is an event, then A is the union of mutually exclusive events, namely, A= ( Thus, P(A) = P( ) (1.61) If P(A) then P( /A) =, K= 1,2,, m (1.62) Using equation and replacing P(A) in equation 1.62, we have Bayes s theorem : P( =, k= 1,2,.., m The conditional probability P( /A) is often called the posterior probability of. Example: In a certain factory, machines I, II, and III are all producing springs of the same length of their production, machines I, II and III, respectively produce 2%, 1%, and 3% defective springs of the total production of springs in the factory, machine I produces 35%, machine II produces 25%, and machine III produces 40%. If one spring is selected at random from the total springs produced in a day, the probability that it is defective, in an obvious notation, equals. P(D)= P(I)P(D/I)+ P(II)P(D/II) + P(III)P(D/III) = ( ) + ( ) ( ) + ( )( ) = If the selected spring is defective, the conditional probability that it was produced by machine III is, by Bayes s formula, P(III/D) = = = 15
16 Note how the posterior probability of III increased from the prior probability of III after the defective spring was observed, because III produces a larger percentage of defectives than do I and II. 16
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