Ch 4_Black

Ch 4_Black - Business Statistics Fifth Edition Ken Black...

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Unformatted text preview: Business Statistics Fifth Edition Ken Black Chapter 4: Probability Not to be reproduced without the written permission of F .8, Alt 4—1 4.1 Introduction to Probability (page 98) o Synonyms for probability: , 0 Probability is the language of statistics. 0 “An experiment is a process that produces outcomes.” (page 100) 0 We will only consider random experiments (page 100) 0 Characteristics of random experiments: I Can specify all possible outcomes; and, I An outcome cannot be predicted with certainty. Examgle: Today’s closing price of a security relative to yesterday. P . OSSIbIC Which one will occur? outcomes 0 Probability measures the uncertainty of the outcomes. 0 Chapters 4—6 look at ways of dealing with and modelling uncertainty. Not to be reproduced without the written permission of PB. Alt 4-2 4.2 Methods of Assigning Probabilities o The classical method (page 99) o N is the total number of possible mutually exclusive equally likely outcomes. 0 The probability of an event E equals the ratio of the number of outcomes (n5) pertaining to E to the total number of outcomes (N): P(E) = nE [N Examgle: Roll a fair die one time and observe the up face. P (rolling a 6) = Roll a pair of fair dice and observe the up faces P(both up faces are 6’s) = = { Why? Examgle: Each season, Anand’s new fashion designs are shown to a panel of experts - ten R.H.Smith undergraduates — and each secretly votes thumbs up or down. Here are historical data on the panel’s rankings of 1500 designs and the subsequent market success, measured by sales: Number of Positive Panel Votes Sales 7-10 4—6 0-3 Total Very successful(V) 380 140 80 600 Modest successful(M) 180 120 100 Disa ointment 40 40 420 Total 600 300 600 1500 What is the probability that a randomly chosen design will be given 7~10 thumbs up? Not to be reproduced without the wn'tten permission of F .8. Alt 4~3 o The Relative Frequency of Occurrence (page 99) o “. . .the probability of an event is equal to the number of times the event has occurred in the past divided by the total number of opportunities for the event to have occurred.” (page 99) 0 Assume an experiment can be repeated indefinitely under identical conditions. The probability of an event E is the long—run relative frequency with which the event E occurs. P(E) = limit (relative frequency) as number of trials 9 00 Example: Flipping a coin to find the P(Head) Number of heads Relative Frequency 3 out of 5 0.5 in each of 5 tosses 3 out of 5 2 out of 5, Etc. Not to be reproduced without the written permission of F.B. Alt 4-4 0 Subjective Probability 0 “Based on the feelings or insights of the person determining the probability.” 0 A way of assigning subjective probabilities Examgle: What is the probability that the Colts will play in the Super Bowl in 2013? Suppose you say Event (Colts Will Play in Super Bowl) Event (Pick red bead) P(Colts Play in Super Bowl) = P(Pick red bead) = DEEDS DEEDS Are you indiflerent between wagering on either event? If so, then P(Colts will play in Super Bowl in 2013) = If not, then P(Colts will play in Super Bowl in 2013) i o Regardless of how you assign probabilities, OsP(E)sl for any event Not to be reproduced without the written permission of FB. Alt 4-5 o 4.5 Addition Laws 0 Special Law of Addition (page 112) I Let X and Y denote any events in a random experiment. I If X and Y are mutually exclusive events (cannot occur simultaneously), P(X or Y) = P(X) + P(Y) S Venn diagram Example: Number of Positive Panel Votes Sales 7-10 4—6 0-3 Total Very successful(V) 380 140 80 Modest successful(M) 180 120 Disa ointment D 40 40 Total 600 300 600 1500 600 What is the probability that a randomly chosen design will be given “7—10” or “4-6” thumbs up? 13(“7—10” or “4-6”) = Not to be reproduced without the written permission of RB Alt 4—6 0 General Law of Addition (page 107) P(X or Y) = P(X) + P(Y) — P(X and Y) X and Y has been counted twice so subtract once. Examgle: Number of Positive Panel Votes Sales 7-10 4-6 0-3 Total Very successful(V) 380 140 80 Modest successful(M) 180 120 Disa ointment D 40 40 Total 600 300 600 1500 600 What is the probability that a randomly chosen design will be given 7-10 thumbs up or be “Very Successful”? P(“Very Successful” or “7-10” ) = Not to be reproduced without the written permission of RB. Alt 4-7 0 Complementary Events (page 103) 0 Not A is called the complement of A. 0 Notation : A' 0 Law: P(not A) = 1 — P(A), or P(A) = 1 — P(not A) Example (Birthday Problem): There are 50 people in a room. What is the probability that at least two of them have the same birthday? Assumptions: N0 birthdays fall on Feb 29; All 365 days are equally likely for each person; and Birthdays are independent A = {at least two people have same birthday} A, = { people have the same birthday} P(A ')= = forn=50 H P (A) = _I_l_ P(A) 23 51 3O 71 50 Not to be reproduced without the written permission ofFB. Alt 4~8 4.7 Conditional Probability (page 120) o The conditional probability of event X occurring, given that event Y has occurred, is denoted by P(X|Y) where: P(X|Y) = P(X and Y) / P(Y), provided P(Y) > 0. X and Y Example: Number of Positive Panel Votes Sales 7-10 4-6 0—3 Total Very successful(V) 380 140 80 Modest successful(M) 180 120 100 Disa ointmentD 40 40 420 Total 600 300 600 1500 What is the probability that a randomly chosen design will be “Very Successful” given that it was given “7—10” thumbs up? P(Very Successful I 7—10): P(VSl7—10)= W = Note: P (v3) = .40 while P(VSI 7-10) = Not to be reproduced without the written permission of F.B. Alt 4-9 4.6 Multiplication Laws 0 From the definition of conditional probability, P(X and Y) = P(Y) P(XIY) Obviously, P(X and Y) = P(X) P(Y IX) Examgle: Randomly pick (without replacement) 2 cards from a standard deck. Find probability of 2 hearts. X = {ISt card is a heart}, Y = {2nd card is a heart} P (X and Y) = P(X) P(YIX) = o Multiplication law useful in Probability Trees (page 129) Not to be reproduced without the written permission of F .3 Alt 4-10 0 Probability Trees (page 129) o In a probability tree, the probability for a Specific path is found by using the multiplication rule. Example: A purchasing dept finds that 75% of its special orders are received on time. Of those orders that are on time, 80% meet specification completely; of those orders that are late, 60% meet them. T = {Order is on time} M = {Meets specifications} P (T) = .75 P(MlT) = .80 P (M! T’) = .60 PM”) 2 '80 P(T and M) = .60 P(T) = .75 P(T andM’)=.15 P(Ml T’) =.60 P(T'and M): .15 P(T'andM')= .10 a. Find the probability that an order is on time and meets specifications. P(T and M) = b. Find the probability that an order meets specifications. P(M) = PL_~_) + Pt_____) II + H Not to be reproduced without the written permission of F.B. Alt 4-! l 0 Independent Events (page 123) o X and Y are independent events if and only if P(X|Y) = P(X) Example: Are events “Very Successful” and “7 — 10” independent? P(VSl7—10)= 380/600 = .63 (from page 4-9) P(VS) = 600/1500 = .40 Conclusion: The events “VS” and “7-10” (are, are not) independent because P(VSl7-10) ( = , 75) P(VS). o Multiplication Law for Independent Events (page 117) If X and Y are independent, then P(X and D = P(X) P(Y) Reason: From Multiplication Law, P(X and Y) = P(XIY) P(Y). From Independence, P(XlY) = P(X). By substitution, P(X and Y) = P(X) P(Y). Example: Roll 3 pair of fair dice and observe the up faces. P(both up faces are 6’s) = { Why? Refer to page 4—3. Not to be reproduced without the written permission of PB. Alt 4-12 0 Summary of Chapter 3 o 3 Interpretations of Probability I Classical approach (equally likely outcomes) I Relative frequency approach I Subjective approach 0 P(not A) = l — P(A) 0 Addition Law Are events A and B mutually exclusive? If yes: P(X or Y) = P(X) + P(Y) If no: P(X or Y) = P(X) + P(Y) — P(X and Y) o Conditional probability P(X | Y) = P(X and Y) / P(Y) o Multiplication Law P(X and Y) = P(X | Y) P(Y) {Useful in probability trees. 0 Determining Independence of events X and Y I If P(X | Y) = P(X), then X and Y are independent events. I If X and Y are independent events, then P(X and Y) = P(X) P(Y) Not to be reproduced without the written permission of PB. Alt 4‘13 ...
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This note was uploaded on 09/07/2011 for the course BMGT 231 taught by Professor Staff during the Fall '08 term at Maryland.

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Ch 4_Black - Business Statistics Fifth Edition Ken Black...

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