Introduction To Probability PDF

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ImmaculateJupiter4509

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Prestige Institute of Management and Research

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probability statistics mathematics introduction to probability

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This document provides an introduction to probability, covering fundamental concepts and calculations. It explains different approaches to probability, including classical and conditional probability. The document also addresses various types of events, such as mutually exclusive, independent, and collectively exhaustive events. The material includes examples to illustrate the concepts.

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Introduction to Probability What are the chances that sales will decrease if we increase prices? How likely is that project will be finished on time? What is the chance that a new investment will be profitable? What is the likelihood a new assembly method will increase productivity ? Pro...

Introduction to Probability What are the chances that sales will decrease if we increase prices? How likely is that project will be finished on time? What is the chance that a new investment will be profitable? What is the likelihood a new assembly method will increase productivity ? Probability Probability is numerical measure of likelihood that an event will occur. It is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. Probability values are always assigned on a scale from 0 to1. A probability near zero indicates an event is unlikely to occur, probability near 1 indicates an event is almost certain to occur. Other probabilities between 0 to 1 represent degrees of likelihood that event will occur. Terminology Experiment : It is a process that generates well- defined outcomes. Each outcome is called an event. Random Experiment: In an experiment where all possible outcomes are known and in advance if the exact outcome cannot be predicted, is called a random experiment. Example: Tossing a coin throwing a dice Terminology Sample Space: Set of all possible experimental outcomes Example Tossing a coin A = {H,T} throwing a dice B = {1,2,3,4,5,6} Sample Point: Experimental outcomes Event: Sample point or collection of sample points Classical Approach P(A) = Number of favorable outcomes/Total number of possible outcomes Unions and Intersections Set notation is the use of braces to group numbers o The union of sets X, Y is denoted X Y An element is part of the union if it is in set X, set Y, or both “X or Y” o The intersection of sets X, Y is denoted X ∩ Y An element is part of the intersection if it is in set X and set Y “X and Y” Copyright ©2020 John Wiley & Sons, Inc. 7 Event Types Mutually exclusive events Collectively Exhaustive events Independent and dependent Events Compound events Equally Likely events Complementary events Mutually Exclusive Events o Events with no common outcomes o Occurrence of one event precludes the occurrence of the other event o Example: if you toss a coin and get heads, you cannot get tails Independent Events Copyright ©2020 John Wiley & Sons, Inc. 9 Independent Events o The occurrence or nonoccurrence of one event does not affect the occurrence or nonoccurrence of the other event(s) o The probability of someone wearing glasses is unlikely to affect the probability that the person likes milk o Many events are not independent The probability of carrying an umbrella changes when the weather forecast predicts rain If events are independent, then: P  X  Y   P  X  , and P Y  X   P  Y  P  X  Y  is the probability that X occurs given that Y has occurred. begin underline end underline Copyright ©2020 John Wiley & Sons, Inc. 10 Collectively Exhaustive Events and Complementary Events Collectively Exhaustive Events o Contains all possible elementary events for an experiment o The sample space for an experiment can be described as mutually exclusive and collectively exhaustive Complementary Events o The elementary events of an experiment not in X comprise its complement o Complementary events are denoted X′, which is pronounced as “not X” P  X   1 P  X  Business Statistics 10e by Ken Black, Sanjeet Singh Copyright ©2020 John Wiley & Sons, Inc. 11 Copyright  2022 Wiley India Pvt. Ltd. All rights reserved. Marginal, Union, Joint, and Conditional Probabilities Copyright ©2020 John Wiley & Sons, Inc. 12 Rules of Probability Rule of addition Rule of Multiplication Addition Laws The General Law of Addition is used to find the probability of the union of two events P  X  Y   P  X   P Y   P  X  Y  Copyright ©2020 John Wiley & Sons, Inc. 14 Addition Laws Addition Law Example: Workers were asked which changes in office design would increase productivity 70% chose noise reduction (N), 67% chose more storage space (S) 56% chose both noise reduction and more space What is the probability that a person chooses either noise reduction or more storage space? P ( N ) .70 P ( S ) .67 P ( N  S ) .56 P ( N  S ) .70 .67 .56  0.81 Copyright ©2020 John Wiley & Sons, Inc. 15 Addition Laws A joint probability table displays the intersection (joint) probabilities along with the marginal probabilities of a given problem The joint probability table for the office productivity problem o Inner cells show joint probabilities o Outer cells show marginal probabilities Increase storage space: Increase storage space: Yes Increase storage space: No Total Noise reduction: Yes 0.56 0.14 0.70 Noise reduction: No 0.11 0.19 0.30 Noise reduction: Total 0.67 0.33 1.00 P( N  S )  P(N )  P(S )  P( N  S ) .70 .67 .56 .81 Copyright ©2020 John Wiley & Sons, Inc. 16 Addition Laws The complement of a union is the probability that that the outcome is neither X nor Y P(neither X nor Y )  P (not X  not Y )  1  P ( X  Y ) For the office productivity problem, since the probability of N or S was 0.81, the probability that a worker chooses neither noise reduction nor storage space is 1 – 0.81 = 0.19 Copyright ©2020 John Wiley & Sons, Inc. 17 Addition Laws The Special Law of Addition applies to mutually exclusive events. P( X  Y )  P  X   P Y  Since there is no intersection of X and Y, there is no need to subtract the joint probability (because it is zero) For example, if workers are asked which most hinders their productivity, and 20% beginunderline endunderline cite lack of direction, 18% cite too much work, 18% cite lack of support, 8% cite inefficient process, 7% cite lack of supplies, and 29% cite other reasons, what is the probability that a worker cites too much work or inefficient process? Pwork  process  Pwork   Pprocess .18 .08 .26 Business Statistics 10e by Ken Black, Sanjeet Singh Copyright ©2020 John Wiley & Sons, Inc. 18 Copyright  2022 Wiley India Pvt. Ltd. All rights reserved. Multiplication Laws General Law of Multiplication P X  Y   P  X   PY | X   P Y   P X | Y  Used to find the joint probability Example: A company has 140 employees, of which 30 are supervisors (S). Eighty of the employees are married (M), and 20% of the married employees are supervisors. If a company employee is randomly selected, what is the probability that the employee is married and is a supervisor? 80  PM   .5714 140  P  M  S   P  M   P  S | M   .5714 .20  .1143 Copyright ©2020 John Wiley & Sons, Inc. 19 Multiplication Laws Special Law of Multiplication If X and Y are independent, P(X ∩ Y) = P(X) · P(Y) Example: A study found that 28% of Americans believe that the ATM has had a most significant impact on everyday life (A). A different study found that 71% of workers believe that working in a team reduces stress (S). These studies are unrelated, so they can be considered independent. o What is the probability that a randomly selected person believes that an ATM is significant AND is less stressed working in a team? o P(A ∩ S) = P(A). P(S) = (.28)(.71)=.1988 Business Statistics 10e by Ken Black, Sanjeet Singh Copyright ©2020 John Wiley & Sons, Inc. 20 Copyright  2022 Wiley India Pvt. Ltd. All rights reserved. Conditional Probability The Probability of an event occurring when given that another event has occurred. Conditional Probability of event A , event B has already occurred = P(A|B) = P(A∩B)/P(B) Similarly Conditional Probability of event B , event A has already occurred = P(B|A) = P(A∩B)/P(A) Bayes' Theorem Bayes' Theorem, named after 18th-century British mathematician Thomas Bayes, is a mathematical formula for determining conditional probability. Conditional probability is the likelihood of an outcome occurring, based on a previous outcome having occurred in similar circumstances. Bayes' theorem provides a way to revise existing predictions or theories (update probabilities) given new or additional evidence. The theorem is also called Bayes' Rule or Bayes' Law and is the foundation of the field of Bayesian statistics. Bays’ Theorem Let X1, X2,…, Xn be a set of events associated with the sample space S, in which all the events X1, X2,…, Xn have a non-zero probability of occurrence. All the events X1, X2,…, Xn form a partition of S. Let Y be an event from space S for which we have to find probability, then according to Bayes’ theorem: P(Xi )  P(Y | Xi ) P(Xi | Y)  P(X1)  P(Y | X1)  P(X2 )  P(Y | X2 )  P(Xn )  P(Y | Xn ) Prior and Posterior probability In Bayesian statistics, prior probability is the probability of an event before new data is collected, while posterior probability is the probability of an event after new data is collected. Bays’ Formula is used to update a given set of prior probabilities of a given event in response to arrival of new information Examples Suppose an item is manufacture by three machines X, Y, and Z. All the three machines have equal capacity and are operated at the same rate. It is known that the percentages of defective items produced by X, Y, and Z are 2, 7, and 12 per cent respectively. All the items produced by X, Y, and Z are put into one bin. From this bin, one item is drawn at random and is found to be defective. What is the probability that this item was produced on Y? Examples Let A be the defective item. We know the prior probability of defective items produced on X, Y, and Z, that is, P(X) = 1/3; P(Y) = 1/3 and P(Z) = 1/3. We also know that P(A|X) = 0.02, P(A|Y) = 0.07, P(A|Z) = 0.12 Now having known that the item drawn is defective, we want to know the probability that it was produced by Y. That is P(Y|A) = P(A|Y). P(Y) / (P(X). P(A|X)+P(Y). P(A|Y)+P(Z). P(A|Z)) = (0.07). (1/ 3)/(1/ 3) (0.02) + (1/ 3) (0.07) + (1/ 3) (0.12) = 0.33

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