Bstat4 - Lessons in Business Statistics Prepared By P.K....

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Unformatted text preview: Lessons in Business Statistics Prepared By P.K. Viswanathan Chapter 4: Probability A Conceptual Framework Introduction Managers will have to cope with uncertainty in many decision situations. For example, a marketing manager may assume that he knows the volume of sales in the next year precisely. This is not true because he may know roughly what the next year sales will be, but cannot give the exact number. There is some uncertainty. Concepts of probability will help you measure uncertainty and perform associated analyses. This chapter provides a conceptual framework of probability and the various probability rules that are essential in making effective business decisions. 1) Probability-Meaning and Concepts Probability refers to chance or likelihood of a particular eventtaking place. For example, you may like to ask " what is the chance that the company will achieve a sale of more than $1million in the coming quarter?" An event is an outcome of an experiment. An experiment is a process that is performed to understand and observe possible outcomes. Examples include tossing a coin, drawing a card from a well- shuffled pack of cards, or rolling a pair of dice. Set of all outcomes of an experiment is called the sample space. Example In a manufacturing unit three parts from the assembly are selected. You are observing whether they are defective or non-defective. Determine a) b) The sample space. The event of getting at least two defective parts. Solution a) Let S = Sample Space. It is pictured as under: D = Defective G = Non-Defective b) Let E denote the event of getting at least two defective parts. This implies that E will contain two defectives, and three defectives. Looking at the sample space diagram above, E ={GDD, DGD, DDG, DDD}. It is easy to see that E is a part of S and commonly called as a subset of S. Hence an event is always a subset of the sample space. Definition of Probability Probability of an event A is defined as the ratio of two numbers m and n. In symbols m P (A) = n where m= number of ways that are favorable to the occurrence of A and n= the total number of outcomes of the experiment (all possible outcomes) Please note that P (A) is always > = 0 and always < = 1. P (A) is a pure number. Diagram Explaining Three Extreme Values of Probability The range with in which probability of an event lies can be best understood by the following diagram. The glass shows three stages-Empty, half-full, and full to explain the properties of probability. 100% Chance or Certainty (50% Chance) Equally Likely (0% Chance) Impossibility 2) Types of ProbabilityClassical Classical Probability is same as the one we have used while defining probability in the beginning. This is called a priori probability because you can compute the answer in advance without actually having to perform an experiment. Consider for example of getting a King from a pack of cards. You can compute the answer in advance as 4/52. Even though classical probability comprehensively defines probability, its use in business analysis for assessing uncertainty is seriously handicapped because you cannot compute probability in advance with out performing trials (experiments), or observing the behavior of the random variable. Here, random variable means a variable that takes many values with corresponding probabilities. 2) Types of Probability Relative Frequency "What is the chance that our company will achieve a sale of more than 15000 tons of specialty paper?" "What is the chance that there will be no stock out problem this year?" The answers to these questions cannot be obtained in advance with out performing experiments or using the past data. The tool organization deploys is a typical frequency distribution of past data expressed as a Relative Frequency that was already covered earlier. For example, your company can use the past data on sales for assessing the probability of achieving a particular target. All you need to do is to convert the past sales figures into a well-structured frequency distribution containing target range in every class and the number of times that was achieved in every class. If you now express relative frequency either in proportion or in percentages to the total frequency, you can easily compute the probability of achieving particular target sales. 2) Types of ProbabilitySubjective When you cannot get probability worked out either based on classical approach or based on relative frequency of past data, the only option available to you is to use Subjective Probability. Assigning subjective probabilities to various events is based on some past experience, personal opinion, and an analysis of the business situation. Survey of Experts' Opinion for knowing the chances of certain key events affecting business is an example of subjective probability. 3)Mutually Exclusive Events Two events A and B are said to be mutually exclusive if the occurrence of A precludes the occurrence of B. For example, from a well shuffled pack of cards, if you pick up one card at random and would like to know whether it is a King or a Queen. The selected card will be either a King or a Queen. It cannot be both a King and a Queen. If King occurs, Queen will not occur and Queen occurs, King will not occur. 4) Independent Events Two events A and B are said to be independent if the occurrence of A is in no way influenced by the occurrence of B. Likewise occurrence of B is in no way influenced by the occurrence of A. For example, when you toss a coin twice, getting head in the second trial is in no way influenced by what happened in the first toss. In the first toss you might have got a head or a tail. In other words, the outcome of the second toss has nothing to do with what happened in the first toss. 5) Rules for Computing Probability 1) Addition Rule -Mutually Exclusive Events P(A B) P(A) P(B) This rule says that the probability of the union of A and B is determined by adding the probability of the events A and B. B Here the symbol A is called A union B meaning A occurs, or B occurs or both A and B simultaneously occur. When A and B are mutually exclusive, A and B cannot simultaneously occur. 5) Rules for Computing Probability 2) Addition Rule –Events are not Mutually Exclusive P(A B) P(A) P(B) P(A B) This rule says that the probability of the union of A and B is determined by adding the probability of the events A and B and then subtracting the probability of the intersection of the events A and B. B The symbol A is called A intersection B meaning both A and B simultaneously occur. Example for Addition Rules From a pack of well-shuffled cards, a card is picked up at random. 1) What is the probability that the selected card is a King or a Queen? 2) What is the probability that the selected card is a King or a Diamond? Solution to part 1) Look at the Diagram: Event A Event B King 4 Numbers Queen 4 Numbers Let A = getting a King Let B = getting a Queen There are 4 kings and there are 4 Queens. The events are clearly mutually exclusive. Applying the formula P(A B) P(A) P(B) = 4/52 +4/52 = 8/52 =2/13 Solution to part 2) Look at the Diagram: There are totally 52 cards in a pack out of which 4 are Kings and 13 are Diamonds. Let A= getting a King and B= getting a Diamond. The two events here are not mutually exclusive because you can have a card, which is both a King and a Diamond called King Diamond. P(K P(K) D) P(D) P(K D) = 4/52+13/52-1/52= 16/52=4/13 King and Diamond King Diamond Multiplication Rule 3) Independent Events P(A P(A).P(B) B) This rule says when the two events A and B are independent, the probability of the simultaneous occurrence of A and B (also known as probability of intersection of A and B) equals the product of the probability of A and the probability of B. Of course this rule can be extended to more than two events. Multiplication Rule 3) Independent Events-Example Example: The probability that you will get an A grade in Quantitative Methods is 0.7. The probability that you will get an A grade in Marketing is 0.5. Assuming these two courses are independent, compute the probability that you will get an A grade in both these subjects. Solution: Let A = getting A grade in Quantitative Methods Let B =getting A grade in Marketing It is given that A and B are independent. P(A P(A).P(B) B) = 0.7.0.5 = 0.35. Multiplication Rule 4) Events are not independent P(A B) P(A).P(B/A ) This rule says that the probability of the intersection of the events A and B equals the product of the probability of A and the probability of B given that A has happened or known to you. This is symbolized in the second term of the above expression as P (B/A). P(B/A) is called the conditional probability of B given the fact that A has happened. We can also write P(A B) P(B).P(A/B) if B has already happened. Multiplication Rule 4) Events are not independent-Example From a pack of cards, 2 cards are drawn in succession one after the other. After every draw, the selected card is not replaced. What is the probability that in both the draws you will get Spades? Solution: Let A = getting Spade in the first draw Let B = getting spade in the second draw. The cards are not replaced. This situation requires the use of conditional probability. P(A) = 13/52 (There are 13 Spades and 52 cards in a pack) P(B/A)=12/51(There are 12 Spades and 51 cards because the first card selected is not replaced after the first draw) P(A B) P(A).P(B/A ) = (13/52).(12/51)=156/2652= 1/17. 6) Marginal Probability Contingency table consists of rows and columns of two attributes at different levels with frequencies or numbers in each of the cells. It is a matrix of frequencies assigned to rows and columns. The term marginal is used to indicate that the probabilities are calculated using a contingency table (also called joint probability table). Marginal probability means the probability of a simple event like P (A) or P(B). Joint probability is the simultaneous occurrence of events. Probability that both A and B occur. P(A is an example of joint probability. B) Marginal Probability-Example A survey involving 200 families was conducted. Information regarding family income and whether the family buys a car are given in the following table. Family Income below Rs 4 lakhs /year Income of Rs. 4 lakhs and above Total Buyer of Car 38 42 80 Non-Buyer 82 38 120 120 80 200 Total a) What is the probability that a randomly selected family is a buyer of the car? b) What is the probability that a randomly selected family is both a buyer of car and belonging to income of Rs. 4 lakhs and above? c) A family selected at random is found to be belonging to income of Rs 4 lakhs and above. What is the probability that this family is buyer of car? Solution a) What is the probability that a randomly selected family is a buyer of the Car? Look at the contingency table above. There are 80 families in total are buyers of a car. There are 200 families in the survey. Hence the probability that a randomly chosen family is a buyer of the car =80/200 =0.40. Note this is a case of marginal probability of a family buying a car. b) What is the probability that a randomly selected family is both a buyer of car and belonging to income of Rs. 4 lakhs and above? Again look at the table above. Both a buyer of car and an income of Rs 4 lakhs and above = 42 families. Total families in the survey =200. Hence the probability that a randomly selected family is both a buyer of car and belonging to income of Rs. 4 lakhs and above =42/200 =0.21. Note this is a case of joint probability (buyer and Rs. 4 lakhs and above) c) A family selected at random is found to be belonging to income of Rs 4 lakhs and above. What is the probability that this family is buyer of car? A family is selected and found to be having an income of Rs 4 lakhs and above. In this category there are totally 80 families out of which 42 are buyers. Hence the probability =42/80 =0.525. Note this is a case of conditional probability of buyer given income is Rs. 4 lakhs and above. 7) Probability Tree Probability concepts are more effectively understood using a probability tree as the basis. The branches of the trees represent the probabilities of associated events. The tree can solve intriguing problems including Bayes’ Theorem of posterior analysis. Posterior analysis is a process by which the probabilities are revised based on new or additional information. Bayes’ Theorem is essentially a case of conditional probability. Let us look at an example to comprehend the power of the probability tree including Bayes’ Method. 7) Probability Tree-Example There are three Machines designated as A, B, and C producing the same item. The output of A, B, and C are 40%, 35% , and 25% respectively. A produces 5 % defectives, B produces 10% defectives and C produces 12% defectives. Draw the probability tree and then answer the following: a. If an item is selected from the total output at random, what is the probability that it is defective? b. If an item is selected from the total output at random, what is the probability that it is non-defective (good piece)? c. If an item is selected and found to be defective. What is the chance that it is produced from A? Produced from B? Produced from C? 7) Probability Tree-Example 7) Solution a. If an item is selected from the total output at random, what is the probability that it is defective? Look at the tree. Defective occurs in three places. A and Defective, B and Defective, C and Defective. Adding these three joint probabilities, namely 0.02+0.035+0.03 you get the answer. The answer is 0.085 or 8.5%. That is P(D) =0.085(Marginal Probability of D) b. If an item is selected from the total output at random, what is the probability that it is non-defective (good piece)? Again look at the tree. There are three branches in which you find good pieces. A and Good, B and Good, C and Good. Adding these three joint probabilities, you get the answer. That is adding 0.38+0.315+0.22=0.915 or 91.5%. This is the answer.(Note: P(G) = 0.915 is the marginal probability of G) 7) Solution Cont c. An item is selected and found to be defective. What is the chance that it is produced from A? Produced from B? Produced from C? This part involves the application of Bayes’ Theorem of computing conditional probability. From the tree, computing these probabilities is very simple. Let us first take Machine A. You want P(A/D) = = 0.02/0.085 = 0.23529. Likewise for Machine B, P(B/D) = =0.035/0.085 = 0.41177 For Machine C, P(C/D) = = 0.03/0.085 = 0.35294 P(C D) P(D) P(B D) P(D) P(A D) P(D) ...
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This note was uploaded on 02/24/2012 for the course BUSINESS 281 taught by Professor Gray during the Spring '12 term at Florida State College.

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