DECISION MAKING UNDER UNCERTAINTY PROBABILITY It is a numerical measure of the likelihood that an event will occur. It is the quantification of uncertainty. Business decisions are often based on an analysis of uncertainty such as: i. What are the “chances” that sales will increase if we decrease prices? ii. What is the “likelihood” that a new assembly method will increase productivity? iii. How “likely” is it that the project will be completed on time? iv. What are the “odds” in favour of a new investment being profitable? If probabilities were available, we could determine the likelihood of each of the events occurring. Probability values are always assigned on a scale from 0 to 1. A probability near 0 indicates that an event is unlikely to occur; a probability near 1 indicates that an event is almost certain to occur. Other probabilities between 0 and 1 represent varying degrees of likelihood that an event will occur. EXPERIMENTS AND THE SAMPLE SPACE An experiment is any process that generates well-defined outcomes. Experiment Experimental Outcomes Toss a coin Select a part for inspection Conduct a sales call Roll a die Play a football Head, Tail Defective, Non- defective Purchase, No purchase 1, 2, 3, 4, 5, 6 Win, Lose, Draw 1
Transcript
DECISION MAKING UNDER UNCERTAINTY
PROBABILITY
It is a numerical measure of the likelihood that an event will occur. It is the quantification of uncertainty.
Business decisions are often based on an analysis of uncertainty such as:
i. What are the “chances” that sales will increase if we decrease prices?ii. What is the “likelihood” that a new assembly method will increase productivity?iii. How “likely” is it that the project will be completed on time?iv. What are the “odds” in favour of a new investment being profitable?
If probabilities were available, we could determine the likelihood of each of the events occurring.
Probability values are always assigned on a scale from 0 to 1. A probability near 0 indicates that an event is unlikely to occur; a probability near 1 indicates that an event is almost certain to occur. Other probabilities between 0 and 1 represent varying degrees of likelihood that an event will occur.
EXPERIMENTS AND THE SAMPLE SPACEAn experiment is any process that generates well-defined outcomes.
Experiment Experimental OutcomesToss a coin
Select a part for inspection
Conduct a sales call
Roll a die
Play a football game
Head, Tail
Defective, Non-defective
Purchase, No purchase
1, 2, 3, 4, 5, 6
Win, Lose, Draw
Sample SpaceIt is the set of all possible experimental outcomes.
Sample PointAny one particular experimental outcome and is an element of the sample space.
Examples:1. Consider the experiment of tossing a coin:
The experimental outcomes are: a Head or a TailIf S denotes the sample space then the sample space and sample points for the coin-tossing experiment: S = {Tail, Head}
2. Consider selection of a part for inspection, the sample space with sample points are:
S= {Defective, Non-defective}
3. Consider rolling a die, the sample space with sample points are:
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S= {1, 2, 3, 4, 5, 6}EVENTS AND THEIR PROBBLITIESAn event is a collection of sample points (experimental outcomes). Capital letters are used to denote events. Consider an event, A, the probability of the event A is denoted as:
P(A) read “the probability of the event A”.
0 ≤ P(A) ≤ 1
OBJECTIVE AND SUBJECTIVE PROBABILITYWhere the probability of an event is based on past data, and the circumstances are repeatable by test, the probability is known as statistical or objective probability. There are 2 types of such a probability:
Classical probability Relative frequency probability
Classical ProbabilityThis is a type of probability assigned to equally likely outcomes of an experiment. That is, when the assumption of equally likely outcomes is used as the basis for assigning probabilities the approach is referred to as the classical method. If an experiment has n possible outcomes that are assumed to be equally likely, the probability of each outcome is 1/n.
RELATIVE FREQUENCY METHODThis approach is used to assign probabilities in situations when an event contains more than one sample point or element.
Examples1. What is the probability of drawing an Ace from a shuffled pack of cards?
Solution:In a pack of cards there are 4 Aces. The total number of cards is 52.
P(Drawing an Ace)
2. Suppose in the test market evaluation of a product, 400 potential customers were contacted: 100 actually purchased the product, but 300 did not.(i) Find the probability of a customer contacted making a purchase.(ii) Find the probability of a customer contacted making no purchase.
Solution
(i) P(A customer making purchase)
(ii) P(A customer making no purchase)
Subjective ProbabilityThe classical and relative frequency methods cannot be applied to all situations where probability assignments are desired. Subjective probabilities are quantifications of personal judgment, experience and expertise. For example, the sales manager considers that there is 40% chance of obtaining the order for which the firm has just quoted. Clearly, this value cannot be tested by repeated trials.
Subjective probability expresses a person’s “degree of belief”, it is personal. Different people can be expected to assign different probabilities to the same event.
SOME BASIC RELATIONSHIPS OF PROBABILITY
Complement of an EventFor the event A, its complement, denoted Ac, ~A or A', is the event consisting of all sample points in the sample space (S) that are not in A.
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AcEventA
Sample Space
The shaded region depicts the complement of A, denoted Ac.In any probability application, event A and its complement, Ac, must satisfy the condition:
P(A) + P(Ac) = 1P(A) = 1 – P(Ac)
Examples1. A sales manager after reviewing sales reports, states that 80% of new customer contacts result
in no sale. Determine the probability that a contact may produce a sale.
SolutionLet A denote the event of a sale and Ac denote the event of no sale.→ P(Ac) = 0.80 P(A) = 1 - P(Ac) = 1 – 0.80 = 0.20
2. A purchasing agent states a 0.90 probability that a supplier will send a shipment that is free of defective parts. Find the probability that the shipment will contain some defective parts.
SolutionLet D denote the event of a defective part and D' be the event of non-defective part.→ P(D') = 0.90 P(D) = 1 - P(D') = 1 – 0.90 = 0.10
RULES OF PROBABILITY
Mutually Exclusive EventsTwo or more events are said to be mutually exclusive when they cannot happen together or at the same time. This means that the events do not have sample points or elements in common. For example, in the classification of people, the events “male” and “female” are mutually exclusive. If a person is “female” this automatically excludes the possibility of being “male”.
Exhaustive EventsTwo events, A and B, are said to be exhaustive if AUB = S (sample space) so that P(AUB) = 1.
DefinitionIf the two events, A and B, are mutually exclusive and exhaustive, then:
ExampleIn the throw of a die:Let A be the event of getting an even number and B, the event of getting an odd number.
The sample space, S = {1, 2, 3, 4, 5, 6}A = {2, 4, 6} and B = {1, 3, 5} A U B = S = {1, 2, 3, 4, 5, 6}
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Independent EventsTwo or more events are said to be independent if the occurrence or non-occurrence of any one event does not affect the occurrence or non-occurrence of the other(s). For example, the outcome of any throw of a die is independent of the outcome of any preceding or succeeding throws.
The Union of Events For the 2 events A and B, the union of A and B is the event containing all sample points belonging to A or B or both. It is denoted AUB.
The shaded region depicts AUB
The Intersection of EventsThe intersection of events A and B is the event containing the sample points (elements) belonging to both A and B. It is denoted as A∩B.
The shaded region depicts A∩B.
THE ADDITION RULE/LAW (OR)
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Event A
Sample Space S
Event B
Event A Event
B
Sample Space S
The addition rule/law provides a way to compute the probability of event A or B or both occurring. It is used to compute the probability of the union of two (2) events, A B. The general addition rule or law states that:
If the 2 events, A and B, are mutually exclusive (cannot happen or occur together) then A and B do not have any intersection (i.e. A∩B = 0), and the addition rule reduces to
Examples
1. Of the 200 students taking Management Accounting and Control examination paper (a two-section examination paper), 160 passed the Management Accounting section and 140 passed the Quantitative Techniques section; 124 students passed both. After reviewing the passes, the Examinations Committee decided to pass any student who passed at least one of the two sections. What is the probability of a student passing in this examination paper?
SolutionLet A = the event of passing the Management Accounting Section
B = the event of passing the Quantitative Technique Section
We want the probability that a student passed the Management Accounting, Quantitative Technique or passed both, i.e. P(A B).
The results indicate an 88% chance of a student passing the examination.
2. In a study involving the television viewing habits of married couples, it was reported that 30% of the husbands and 20% of the wives were regular viewers of a particular Friday evening programme. For 12% of the couples in the study, both husband and wife are regular viewers of the programme. What is the probability that at least one member of a married couple is a regular viewer of the programme?
SolutionLet H = the event of Husband is a regular viewer. W = the event of wife is a regular viewer.
3. What is the probability of throwing a 3 or a 6 with a throw of a die?
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Solution Let A = the event of throwing a 3 B = the event of throwing a 6
ExerciseSuppose that a sample space has five equally likely experimental outcomes: E1, E2, E3, E4, and E5.
Let A = {E1, E2} B = {E3, E4} C = {E2, E3, E5}
Find:(a) P(A), P(B) and P(C).(b) P(AUB). Are A and B mutually exclusive?(c) Ac, Cc, P(Ac), and P(Cc).(d) AUBc and P(AUBc)(e) P(BUC).
CONDITIONAL PROBABILITIES
In conditional probabilities, we determine the probability of one event when another event known to have occurred is important.
Given the 2 events, A and B which are related, the probability of the event A given the condition that event B has occurred is denoted:
which is read “the probability of A given B”.
With the 2 events A and B, the general definitions of conditional probability for A given B and for B given A are respectively:
Example 1
For 2 events A and B, P(A) = 0.5, P(B) = 0.60 and P(A∩B) = 0.40.
(a) Find P(A/B)
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(b) Find P(B/A)(c) Are A and B independent? Why or why not?
Solution
Given that P(A) = 0.5, P(B) = 0.6, and P(A∩B) = 0.4
(a)
(b)
=
(c) No because P(A) and P(B)
Example 2
Promotional status of male and female officers in the Police Service:
The Police Service consists of 1200 officers: 960 men and 240 women. Over the past 2 years, 324 officers have been promoted. The table below shows the specific breakdown of promotions for male and female officers:
After reviewing the promotional record, a committee of female officers filed a discrimination case on the basis that only 36 female officers had received promotions during the past 2 years. The police administration argued that the relatively low number of promotions for female officers is due not to discrimination but to the fact that few female officers are in the service. Using conditional probability to evaluate the discrimination charge, draw a conclusion.
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Note
Independent events: if 2 events, A and B are independent then
P(B/A) = P(B)Or P(A/B) = P(A)
If they are not independent then P(B/A) P(B) or P(A/B) P(A)
Solution:
Let M = event an officer is a man W = event an officer is a woman B = event an officer is promoted
Probability that an officer is a man and is promoted
Probability that an officer is a man and is not promoted
Probability that an officer is a woman and is promoted
Probability that an officer is a woman and is not promoted
Probability that an officer is a man
Probability that an officer is a woman
Probability that an officer is a promoted
Probability that an officer is a not promoted
The conditional probability values strongly support the argument presented by the female officers.
JOINT PROBABILITY TABLE
Joint probability is the probability of the intersection of 2 events. The Joint Probability Table provides a summary of probability information.
Example
The Joint Probability Table for Police Officer Promotions:
The Multiplication Law is used to find the probability of the intersection of 2 events. It is derived from the definition of conditional probability. The multiplication law states that:
Note:
From the conditional probability,
Multiplying through by P(B), we have
Also,
Multiplying through by P(A) we have,
Example:
Suppose that a newspaper circulation department knows that 84% of its customers subscribe to the Daily edition of the paper. In addition, the department knows that the conditional probability that a customer who already holds a Daily subscription also subscribes to the Sunday edition is 0.75. What is the probability that a customer subscribes to both the Daily and Sunday editions of the newspaper?
Solution
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Marginal ProbabilitiesJoint Probabilities
Let D = the event that a customer subscribes to the Daily edition. S = the event that a customer subscribes to the Sunday edition
63% of the newspaper’s customers take both the Daily and Sunday editions.
INDEPENDENT EVENTS
The general multiplication law states that:
=
But if A and B are independent events, then
Therefore
Examples
(1) A service station manager knows from past experience that 40% of her customers use credit card when purchasing gasoline. What is the probability that the next 2 customers purchasing gasoline will both use credit cards?
Solution
Let A = the event that the 1st customer uses credit a card. B = the event that the 2nd customer uses a credit card.
To find
Note: A and B are 2 independent events.
(2) What is the probability of throwing a 3 and a 6 with two throws of a die?
Solution
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Note: Also, P(D∩S) = P(D/S).P(S)But no information in the Question on P(D/S) and P(S), so we can’t use this.
(3) From past experience it is known that a machine is set up correctly on 90% of occasions. If the machine is set up correctly, then 95% of good parts are expected but if the machine is not set up correctly then the probability of a good part is only 30%. On a particular day the machine is set up and the first component produced and found to be good. What is the probability that the machine is set up correctly?
Solution Let C = the event that the machine is set up correctly.
G = the event that good parts are produced.
P(C) = 0.9 => P(C') = 1 – 0.9 = 0.1
P(G/C) = 0.95P(G/C') = 0.3
To find P(C/G)
(4) There are 100 students in a first year college intake. 36 are male and are studying accounting, 9 are male and are not studying accounting, 42 are female and studying accounting, 13 are female and are not studying accounting.
(i) Calculate the probability that a student is studying accounting given that he is a male.
(ii) If a student is studying accounting, calculate the probability that he is a male
The probability of a student studying accounting given that he is a male is 0.80.
(ii) To find P(M/A)
The probability of a student being male given that he is studying accounting is 0.462.
BAYES’ THEOREM
Probabilities may be revised when new information is obtained. The Bayes’ theorem is applied when we have initial or Prior probability estimates for specific events of interest. Then, from sources such as a sample, a special report, or a product test, we obtain some additional information about the events. With this new information we update the prior probability values by calculating revised probabilities, referred to as Posterior probabilities.
Definition
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PRIOR PROBABILITIES
NEW INFORMATION
APPLICATION OF BAYES’ THEOREM
POSTERIOR PROBABILITIES
With prior probabilities P(A1), P(A2), …, P(An) and the appropriate conditional probabilities P(B/A1), P(B/A2), …, P(B/An), the posterior probabilities of events A1, A2, …, An are:
But
and
Therefore the general Bayes’ theorem then becomes:
Example
A company has three production sections S1, S2 and S3 which contribute 40%, 35% and 25% respectively to total output. The following percentages of faulty units have been observed:
S1 2% (0.02)S2 3% (0.03)S3 4% (0.04)
There is a final check before output is dispatched. Calculate the probability that a unit found faulty at this check has come from Section 1 (S1).
SolutionLet S1 = the event that a unit chosen is from Section 1. S2 = the event that a unit chosen is from Section 2. S3 = the event that a unit chosen is from Section 3. F = the event that a unit chosen is faulty.
To be calculated using the Bayes’ theorem.Exercise1. Calculate
(i) P(S2/F)(ii) P(S3/F)
2. A manufacturing firm receives shipments of parts from two different suppliers: 1 and 2. Currently, 65% of the parts purchased by the company are from supplier 1 and the remaining 35% are from supplier 2. The quality of the purchased parts varies with the source of supply. Based on historical data, the conditional probabilities of receiving good and bad parts from the two suppliers are as follows:
Supplier Good Parts Bad Parts 1 0.98 0.02 2 0.95 0.05
(a) Find the probability that a part selected is bad.(b) Find the probability that a part selected is good.
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(c) Now, suppose that the parts from the two suppliers are used in the firm’s manufacturing process and that a bad part causes the machine to break down:
(i) What is the probability that it came from supplier 1?(ii) What is the probability that it came from supplier 2?
Note
Prior Probabilities:P(S1) = 0.65P(S2) = 0.35
New information = Conditional Probabilities:P(G/S1) = 0.98 P(B/S1) = 0.02P(G/S2) = 0.95 P(B/S2) = 0.05
1. Rev. Thomas Bayes (1702 – 1761), a Presbyterian minister, is credited with the original work leading to the version of Bayes’ Theorem in use today.
2. Bayes’ theorem is applicable when the events for which we want to compute, posterior probabilities, are mutually exclusive and their union is the entire sample space.
THE TABULAR APPROACH
The tabular approach is helpful in conducting the Bayes’ theorem calculations simultaneously for all events Ai.
The computations involve the following steps:
STEP I: Prepare 3 columns: Column 1 – The mutually exclusive events for which posterior probabilities are desired Column 2 – The prior probabilities for the events Column 3 – The conditional probabilities of the new information given each event.
STEP II: In column 4 compute the joint probabilities for each event and the new information B by using the multiplication law – i.e. multiply the prior probabilities in column 2 by the corresponding conditional probabilities in column 3:
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STEP III: Sum the joint probabilities in column 4 to obtain the probability of the new information, P(B).
STEP IV: In column 5, compute the posterior probabilities by using the basic relationship of conditional probability:
Note: The joint probabilities appear in column 4, whereas P(B) is the sum of the column 4 values.
(1) The prior probabilities of events A1, A2, and A3 are P(A1) = 0.20, P(A2) = 0.50 and P(A3) = 0.30. The conditional probability of event B given A1, A2, A3 are:
(a) Compute (b) Apply Bayes’ theorem to compute the posterior probability P(A2/B).(c) Use the tabular approach to applying Bayes’ theorem to compute: P(A1/B), P(A2/B)
and P(A3/B)
(2) The Wayne Manufacturing Company purchases a certain part from suppliers A, B and C. Supplier A supplies 60% of the parts, B 30% and C 10%. The quality of parts varies among the suppliers, with A, B, and C parts having 0.25%, 1% and 2% defective rates, respectively. The parts are used in one of the company’s major products.
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(a) What is the probability that the Company’s major product is assembled with a defective part? Use the tabular approach to applying Bayes’ theorem to solve.
(b) When a defective part is found which supplier is the likely source?
Solution
EVENT P(Ai) P(D/Ai) P(Ai∩D) P(Ai/D)Supplier A 0.60 0.0025 0.0015 0.23 B 0.30 0.0100 0.0030 0.46 C 0.10 0.0200 0.0020 0.31
P(D) = 0.0065 1.00
(a) P(D) = 0.0065(b) B is most likely supplier if a defective part is found.
Note:
An example using the form of Bayes’ theorem.
P(B/A) =
40 workers in a large office can be classified as follows:
Clerical Grade Admin. GradeMale 4 6Female 22 8
(a) If a worker is selected at random from the office, what is the probability that the worker will be in the administrative grade:
(b) If the worker selected is female, what is the probability that she will be in the administrative grade?
(c) If the worker selected is male, what is the probability that he will be in the administrative grade?
(d) If the worker selected is in the administrative grade, what is the probability of the worker being male?
Solution
Let M = event of selecting a male F = event of selecting a female C = event of selecting a clerical grade A = event of selecting an admin. Grade
(a)
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(b)
(c)
(d)
DECISION TREE
It is often helpful to visualize a problem involving probability on a decision tree. A decision tree provides a graphical representation of the decision-making process. Although decision trees can be used in any situation, they are particularly helpful when solving problems involving conditional probability.
Example
A bag contains 5 blue marbles, 3 yellow marbles and 2 white marbles. 2 marbles are then drawn from the bag, the first marble not being replaced before the second marble is drawn. What is the probability that:
(a) both marbles are the same colour?(b) both marbles are a different colour?(c) at least one marble is blue?
Solution
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Using a decision tree:
Let B = event of drawing a blue marble Y = event of drawing a yellow marble W = event of drawing a white marble
(a) P(both marbles are of the same colour) = P(BB) or P(YY) or (WW) = 2/9 + 1/15+1/45 = 14/45
(b) P(both marbles are of different colours) = P(BY) or P(BW) or P(YB) or P(YW) or P(WB) or P(WY)
= 1/6+1/9+1/6+1/15+1/9+1/15 = 1/3+2/9+2/15
Or P(both marbles are of different colours) = 1 – P(both marbles are of the same colour) = 1 – 14/45 = 31/45
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(c) P(at least one marble is blue) = P(BB) or P(RY) or P(BW) or P(YB) or P(WB) = 2/9 + 1/6 + 1/9 + 1/6 +1/9 = 70/90 = 7/9
Exercise
(1) A box of 20 components contains six defective components. It three components are drawn consecutively from the box, what is the probability of all three components being good (non-defective).
(a) if there is replacement(b) if there is no replacement.
Note:Decision tree is also used to solve problems that involve the use of the Bayes’ theorem.
EXPECTED VALUE
The expected value (EV) of a particular event which has different possible outcomes is a weighted average of the values (pay-offs) associated with each possible outcome. The probabilities of each outcome are used as the weights and these are multiplied by the respective pay-offs of each outcome.
For instance, for the 2 possible outcomes having pay-offs X1 and X2 with the probabilities of each outcome P1 and P2, the expected value EV(X) is
EV(X) = P1X1 + P2X2
Example
If there is 60 percent chance of earning GH¢1000 and 40% chance of earning GH¢5000 from an investment then
The expected value (EV) of a particular course of action over n possible outcomes can be defined as:
EV =
Where Pi = probability of ith outcomes Xi = value of ith outcome
and
Example
A fair coin is tossed and, for a GH¢1 stake, the gambler is promised GH¢1.80 if the result is a head, but only GH¢0.10 if the result is a tail, what is the expected value to the gambler of each throw?
Solution
Let P1 = probability of head, (0.5) P2 = probability of tail, (0.5) X1 = the pay-off of a head, (GH¢1.80) X2 = the pay-off of a tail, (GH¢0.10)
The expected value (EV) to the gambler of each throw is GH¢0.95. Therefore, for a GH¢1.00 stake, the gambler can expect to lose GH¢0.05 or GP 5.
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EXPECTED VALUE AND DECISION TREE
The idea of expected value (EV) may be related to the decision tree analysis. The sequence of decisions and outcomes is represented graphically as the branches of the ‘tree’. At every point (node) that a decision must be made or an outcome must occur, the tree branches out further until all the possible outcomes have been displayed.
Boxes (Decision Nodes)
They are usually used to indicate situations where the decision maker consciously selects a particular course of action (strategy) and where the outcome of that action is “certain”. Branches coming out of these boxes simply indicate the alternative decisions or strategies which might be taken, each of which has a probability of 1 once selected.
Circles (Chance Nodes)
They are usually used to indicate situations reflecting a “state of nature” i.e. situations whose outcomes are not entirely under the conscious control of the decision-maker. Branches coming out of such circles show the various possibilities which might occur, together with estimates of their probability of occurrences.
Pay-offs
They are the valuations placed at the end of particular branches emanating from chance nodes. They are the values which management allocates to that event or outcome should it actually occur, and are often denoted by a straight (vertical) line at the end of the branch.
Example
A decision maker must decide between a strategy of increasing price or keeping price unchanged. If this latter strategy is pursued, only one profit outcome is possible which is evaluated at GH¢200,000. If the former strategy (increasing price) is pursued then the decision maker intends to support the policy change by an active advertising campaign. However, the outcome of such an advertising campaign is not entirely under the control of the decision maker. He estimates that there is a 60% chance of the campaign being a success and a 40% chance of it being a failure.
The firm estimates profits of GH¢800,000 should the advertising campaign (allied to a price increase) be successful, but losses of GH¢600,000 should it fail. If the aim of the firm is to maximize expected profit, what decision will the decision make take?
Solution
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Backward Induction:
Which branch the firm should choose in order to maximize the expected profit can easily be determined. The process of solving this problem is called backward induction.
Procedure for Solving a Decision Tree
1. Start with the final branches of the decision tree, and evaluate each event node (chance node) and each decision node, as follows:
For an event node, compute the EMV of the node by computing the weighted average of the EMV of each branch weighted by its probability. Write this EMV above the event/chance node.
For a decision node, compute the EMV of the node by choosing that branch emanating from the node with the best EMV value. Write this number above the decision node and cross off those branches emanating from the node with inferior EMV values by drawing a double line through them.
2. The decision tree is solved when all nodes have been evaluated
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3. The EMV of the optical decision strategy is the EMV computed for the starting branch of the tree.
Firms Optimal Decision Strategy
The firm should increase its price and accompany it with an advertising campaign. The EMV of this strategy is GH¢240,000
Example
Mary is organizing a special outdoors show which will take place on August 15. The earnings from the show will depend heavily on the weather. If it rains on August 15, the show will lose GH¢20,000; if it is sunny on August 15, the show will earn GH 15,000. Historically, the likelihood of it raining on any given day in mid- August is 27%. Suppose that today is July 31. Mary has the option of cancelling the show by the end of the day on July 31, but if she does so, she will then lose her GH¢1,000 deposit on the facilities.
What is Mary’s optimal decision strategy?
Solution
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Optimal Decision Strategy
Mary should not cancel the show When it rains she loses GH¢20,000 but when it does not rain she will earn GH¢15,000 The EMV of the decision strategy is GH¢5,550.
OPTIMIZATION OF LEVELS OF ACTIVITY UNDER CONDITIONS OF UNCERTAINTY
Expected value principles can be used to calculate the maximum stock or profit level when demand is subject to random variations over a period.
Pay-off
In decision analysis, we refer to the consequence resulting from a specific combination of a decision alternative and a state of nature as a pay-off.
Pay-off Table
A table showing pay-off for all combinations of decision alternatives and states of nature is a pay-off table.
Pay-offs can be expressed in terms of profit, costs, time, distance or any other measure appropriate for the decision problem being analyzed.
Example
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A distributor buys perishable articles for GH¢2 per item and sells them at GH¢5. Demand per day is uncertain and items unsold at the end of the day represent a write off because of perishability. If he understocks he loses profit he could have made.
A 300 day record of past activity is as follows:
Daily Demand (units) No. of Days Probability (P)10 30 0.111 60 0.212 120 0.413 90 0.3
300 1.0
Required:
(a) Prepare a conditional and expected profit table.(b) What level of stock (using the expected value approach) should be held from day to day to
maximize profit?
Solution
We have to calculate the conditional profit (CP) and Expected Profit (EP)
If demand = 10 and stock = 10
CP = (5x10) – (2x10) = 50 – 20 = 30EP = CP x probability of the demand = 30 x 0.1 = 3
(b) The optimum stock position, given the pattern of demand, is to stock 12 units per day. This is because stocking 12 units per day has the highest expected value.
ALTERNATIVE DECISION RULES
Apart from the expected value approach, there are other alternative rules we can use:
Maximax RuleIt chooses the best of the best and it is applied by people who do not fear risk. It is an optimistic rule and maximizes the maximum that can be gained.
Maximin ruleIt chooses the “best of the worst” and it is applied by people who fear risk. This is a cautious decision rule based on maximizing the minimum loss that can occur.
Minimax Rule (Minimum Regret Criterion)It minimizes the maximum opportunity losses and it is applied by people who are neither risk-takers nor risk-avoiders. This decision seeks to” minimize the maximum regret” that there would be from choosing a particular strategy. We usually construct a regret table based on the pay-off table. The regret is the opportunity loss from taking one decision given that a certain contingency occurs.
Example
The following pay-off table shows potential profits and losses which are expected to use from launching various products in three markets conditions:
Pay-off Table in GH¢’000s
Demand Conditions/Product
Boom Steady RecessionConditions State
Product A +8 1 -10Product B -2 +6 +12Product C +16 0 -26
The probabilities are Boom 0.6, steady 0.3 and Recession 0.1.
Product A = (0.6x8) + (0.3x1)+(0.1x-10) = 4.1Product B = (0.6x-2)+(0.3x6)+(0.1x12) = 1.8Product C = (0.6x16)+(0.3x00+(0.1x-26) = 7.0
Using the expected value rule, the rankings are: C, A, B
(b) The Maximax Rule (Best of the Best)Best of Product A = +8Best of Product B = +12Best of Product C = +16
The Ranking using the maximax rule is C, B, A
(c) The maximum rule: (Best of the worst)Worst of Product A = -10Worst of Product B = -2Worst of Product C = -26
The ranking using the maximum rule is B,A,C
(d) Minimax Regret Rule:Minimise the maximum regret we have to prepare the regret opportunity lost table:
Regret Table in GH¢’000
Boom Condition Steady Condition Recession MaxProduct A
8 5 22 22
Product B
18 0 0 18
Product C
0 6 38 38
Note
The regret is the opportunity loss from taking one decision given that a certain contingency occurs.
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Set the best position (in the pay-off table) under any state (Boom, steady and Recession) to zero and calculate the amount of short fall these is by not being at the position.
Boom
The best =+16 = set to zero in opportunity If A is produced, we lose the +16 and gain 8= -16+8 = -8 = Opportunity loss (Regret) = 8
If B is produced, we lose the +16 and lose another 2(i.e. -2)
= -16-2 = -18 = Opportunity loss = 8
Steady
Best = +6 = Set to zero in opportunity loss table
If Product A produced, we lose 6 and gain 1. Opportunity loss = 6-0=6
Recession
Best = +12 = set of zero in opportunity loss table
If Product A is produced, we lose 12 and lose further 10.
Opportunity loss = 12+10=22
If Product C is produced, we lose 12 and lose further 26Opportunity loss = 12+26 = 38
Exercise
(1) How many ways can we select three elements from a set of eight elements.(a) where the selected elements are ranked in order?(b) where the selected elements are not ranked in order?
(2) A batch of 20 video recorders contains four defectives. A sample of five videos is to be selected from the batch.(a) How many possible samples are there?(b) if all the samples were to contain only non-defective videos, how many ways can this
occur?
(3) Five students (A, B, C, D and E) have been elected to the student union.(a) If three students are to be selected for a committee, how many different committees
are possible?(b) If the three members are to become chairperson, treasurer and secretary, how many
different line ups from the five students are possible?
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PERMUTATIONS AND COMBINATIONS
Factorial Notations
The exclamation sign in English is used to denote the factorial of a number. The factorial of the number n, denoted n!, where n is a whole number and n ≥ 0 is:
In general, if there are n items and r are to be placed in order, the number of different ways (permutations) in which this can be done is:
Example
Suppose you are given 10 desirable qualities for a new product and are asked to place 4 of these qualities in order of merit, i.e. 1st , 2nd ,3rd, 4th, in how many different ways can this be done?
Solution
The ordering of the 4 qualities selected does matter here therefore we use permutations
n = 10, v = 4
= 5,040
COMBINATIONS
In general, if there are n items and r are to be selected irrespective of order, the number of ways (combinations) in which this can be done is:
Example
Suppose we want to select 4 desirable qualities of a new product out of 10 desirable qualities given (irrespective of the order of merits), in how many ways can this be done?
Solution
Order is not important therefore it is a combination problem
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=
PROBABILITY DISTRIBUTION
Random Variable
A random variable is a numerical description of the outcome of an experiment. For example, if we consider the experiment of selling automobiles for one day at a particular dealership, we could describe the experimental outcomes in terms of the number of cars sold. In this case x = number of cars sold, x is called a random variable.
Examples of Random Variables
Experiment Random Variable(x) Possible Values for the Random Variable
1. Make 100 sales calls Total no. of sales 0, 1, 2,…,1002. Inspect a shipment of 70 radios No. of Defective radios 0, 1, 2,…, 703. Build a new library Percentage of project
completed after 6 months 0 ≤ x ≤ 100
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4. Operate a restaurant No. of customers entering in one day 0, 1, 2, …
A random variable may be classified as either discrete or continuous, depending on the numerical values it can have.
Discrete Random Variable
A random variable that may take on only a finite or infinite sequence (e.g. 1, 2, 3, …) of values.
Examples:
The number of units sold The number of defects observed The number of customers that enter a bank during one day of operation
Probability Distribution of a Discrete Random Variable
It is a distribution (a list) of the possible values of a random variable (x) with their corresponding probabilities [f(x)]. In the development of a discrete probability distribution, two conditions must always be satisfied:
1) f(x) ≥ 0
2) ∑f(x) = 1
Note: f(x) denotes the probability function and provides the probability that the random variable x takes on a specific value.
Example
The sales records of an automobile firm over the past year that the firm was opened for business for 300 days are:
Sales Volume(No. of cars sold)
No. of Days
0 541 1172 723 424 125 3
300
Required:Construct a probability distribution for the number of cars sold per day.
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Solution
Let x = a random variable representing the no. of cars sold per day. f(x) = the probability function of x
Probability Distribution for the Number of Cars Sold Per Day
The expected value of a discrete random variable is a weighted average of all possible values of the random variable, where the weights are the probabilities associated with the values
E(x) = μ = ∑xf(x)
Where E(x) and μ are used to denote the expected value (the mean) of a random variable.
Variance of a Random Variable:
The variance of a discrete random variable is given as:
The standard deviation is given as:
Example
Calculate the expected value (mean), variance and standard deviation for cars sold per day for the automobile sales.
1. The following Table shows a partial probability distribution for the MRA Company’s projected profits (in thousands of Ghana cedis) for the first year of operation (the negative value denotes a loss):
x f(x)-100
0
50
100
150
200
0.10
0.20
0.30
0.25
0.10
-
(a) Find the missing value of f(200). What is your interpretation of this value?(b) What is the probability that MRA will be profitable?(c) What is the probability that MRA will make at least GH¢100,000 profit?(d) Compute the expected profit [E(x)], variance (σ2), and standard deviation (σ).
2. The demand for a product of Banbo Industries varies greatly from month to month. Based on the past 2 years of data, the following probability distribution shows the company’s monthly demand:
x f(x)300
400
500
600
0.20
0.30
0.35
0.15
(a) If the company places monthly orders equal to the expected value of the monthly demand, what should Banbo’s monthly order quantity be for this product?
(b) Assume that each unit demanded generates GH¢70 in revenue and that each unit ordered costs GH¢50. How much will the company gain or lose in a month if it places an order based on your answer to part (a) and the actual demand for the item is 300 units?
(c) What are the variance and the standard deviation for the number of units demanded?
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BINOMAL PROBABILITY DISTRIBUTION
The Binomial Distribution can be used to describe the likely outcome of events for discrete variables which:
(a) have only 2 possible outcomes(b) are independent – i.e. the outcome of one trial does not affect the outcome of any other trial.(c) The probabilities of the 2 outcomes do not change from one trial to the next.
An important discrete random variable associated with the binomial experiment is the number of successful outcomes in trials.
If we let x denote the value of this random variables, then x can take the values 0,1,2,3,…,n, depending on the number of successes observed in the n trials. The probability distribution associated with this random variable is called the binomial probability distribution.
The binomial probability function is:
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x = 0, 1, …, n
where
n = no. of trialsp = probability of success on one trialx = no. of successes in n trialsf(x) = probability of x successes in n trials.
Note:n!= n(n-1)(n-2) …(2)(1)
Expected Value and Variance for the Binomial Distribution
For the special case of a binomial probability distribution, the expected value of the random variable is given by:
E(x) = µ = np
The variance is
Var(x) = σ2 = np(1-p)
The standard deviation is:
Example
Consider the experiment of customers entering the NICK Clothing Store. Based on experience, the store manager estimates that the probability of a customer making a purchase is 0.30. For the next three customers who enter the store, what is the probability that
(a) no customer makes a purchase?(b) exactly one customer makes a purchase?(c) exactly two customers make a purchase?(d) exactly three customers make a purchase?(e) calculate the expected number of customers making a purchase.(f) calculate the variance and standard deviation for the number of customers making a purchase
Solution
1) The experiment can be described as a sequence of 3 identical trials, one trial for each of the 3 customers who will enter the store.
2) Two outcomes – the customer makes a purchase (success) or the customer does not make a purchase(failure) – are possible for each trial.
3) The probabilities of the purchase (0.30) and no purchase (0.70) outcomes are assumed to be the same for all customers.
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4) The purchase decision of each customer is independent of the purchase decision of the other customers.
Let x = random variable of the no. of customers making a purchase i.e. no. of successes in the 3 trials.
The requirements of binomial probability distribution are satisfied.n=3, p=0.30, 1-p = 0.70
f(x) =
(a)
= 1(1)(0.343) = 0.343
(b) f(1) =
=
= 3(0.3)(0.49) = 0.44
(c)
=
= 3(0.09)(0.7)= 0.189
(d)
(e) E(x) = µ = np = 3(0.3) = 0.9
(f) Var(x) = σ2 = np(1-p) = 3(0.3)(0.7)
= 0.63
(g) = = 0.79
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Exercise: ASW pages 85 - 88
POISSON PROBABILITY DISTRIBUTIONThe Poisson probability distribution is a discrete probability distribution that deals with the number of occurrences of an event over a specified interval of time or space. For example the random variable of interest might be
the number of arrivals at a car wash in 1 hour the number of repairs needed in 10 miles of highway the number of leaks in 100 miles of pipeline.
The following 2 assumptions are satisfied:
(a) The probability of an occurrence of the event is the same for any two intervals of equal length.
(b) The occurrence or nonoccurrence of the event in any interval is independent of the occurrence or nonoccurrence in any other interval.
The probability function of the Poisson random variable is given as:
for
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where
λ = mean or average number of occurrences in an intervale = 2.71828x = number of occurrences in an interval
f(x) = probability of x occurrences in an interval
Note that x is a discrete random variable with no upper limit (infinite sequence of values: 0, 1, 2, …)
Example 1
Suppose that we are interested in the number of arrivals at the modern car park in Accra during a 15-minute period on weekday mornings. The Poisson probability function is assumed. The analysis of historical data shows that the average number of cars arriving during a 15-minute interval of time is 10. Find the probability of 5 arrivals.
Solution
for x = 0, 1, 2, …
λ = 10, e =2.71828, x = 5
Therefore the probability that 5 cars will arrive in the 15-minute interval of time is 0.0378.
NoteWhen working with the Poisson probability distribution, you need to be sure that λ is the mean number of occurrences for the desired interval. For instance, suppose that you know that 30 calls come into a switch board every 15 minutes.
Rate = 2 calls per minute
To compute the number of calls that come over a 5-minute period;λ = (2 calls per minute) x (5 minutes) = 10
To compute the number of calls that come over a 10-minute period;λ = (2 calls per minute) x (10 minutes) = 20
Example 2
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Suppose that we are concerned with the occurrence of major defects in a section of highway one month after resurfacing. If it is assumed that the probability of a defect is the same for any two intervals of equal length, and that the occurrence or nonoccurrence of a defect in any one interval is independent of the occurrence or nonoccurrence in any other interval. Suppose that major defects occur at the average rate of two per mile. Find the probability that
(i) no major defects will occur in a particular 3-mile section of the highway.(ii) at least one major defect will occur in the 3-mile section of the highway.
Solution
(a) Rate of occurrence of major defects = 2 per mileFor the 3-mile section, λ = (2 defects per mile) x (3 mile) = 6 (i.e. the expected number of major defects over the 3-mile section)
x = 0
(b) The probability of at least one major defect in the 3-mile section is1 – 0.0025 = 0.9975.
CONTINUOUS RANDOM VARIABLES
Random variables such as weight, time, and temperature that may take on any value in a certain interval or collection of intervals are continuous random variables.
NORMAL PROBABILITY DISTRIBUTION
This is perhaps the most important probability distribution used to describe a continuous random variable. It is applicable in a great many practical problem situations and its probability density function has the form of the bell-shaped curve
f(x)
0 X
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Characteristics of the Normal Distribution 1. Shape: It is bell-shaped
µ
2. Symmetry: It is symmetric about the mean.
3. Area: Total area under the normal curve is 1.
4. Random variable: It is continuous and takes all real values.
STANDARD NORMAL DISTRIBUTION
It is a random variable that has a normal distribution with mean of 0 and a standard deviation of 1. The letter z is used to designate this particular normal random variable.
The formula used to convert any normal random variable x with mean u and standard deviation 0 to the standard normal distribution is:
Where z is a measure of the number of standard deviations that x is from μ. The conversion to the z value allows us to use the standard normal distribution to compute probabilities for any normal distribution.
f(Z)
σ = 1
0.5 0.5
-Z μ = 0 +Z
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The total area above the mean (μ = 0) is 0.5 and the total area bellow the mean is 0.5 (as shown in the above graph)
The Standard Normal Distribution Table and the appropriate Z values are used to find the probability
Examples
1. For the standard normal random variable z, compute the following probabilities:
2. For the standard normal random variable z, find for each situation:
(a) The area between 0 and z is 0.4750.(b) The area between 0 and z is 0.2291.(c) The area to the right of z is 0.1314.(d) The area to the left of z is 0.6700.
3. For the standard normal random variable z, find z for each situation:
a) The area to the left of z is 0.2119.b) The area between - z and z is 0.9030.c) The area between - z and z is 0.2052.d) The area to the left of z is 0.9948.e) The area to the right of z is 0.6915.
4. The demand for a new product is estimated to be normally distributed with µ = 200 and σ = 40. Let x be the number of units demanded and find the following probabilities:
(a) P(180 ≤ x ≤ 220)(b) P(x ≥ 250)(c) P(x ≤ 100)(d) P(225 ≤ x ≤ 250)
5. Grear Tire Company just developed a new steel-belted radial tire that would be sold through a national chain of stores. Grear’s management has some probability information concerning the number of miles the tires will last.
From factual road tests with the tires, Grear’s engineering group estimates the mean tire mileage at µ = 36,500 miles and the standard deviation at σ =5,000 miles. In addition, the data collected indicate that a normal distribution is a reasonable assumption.
(a) What percentage of the tires, then, can be expected to last more than 40,000 miles? (In other words, what is the probability that the tire mileage will exceed 40,000?)
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(b) Grear is considering a guarantee that will provide a discount on a new set of tires if the mileage on the original tires doesn’t exceed the mileage stated on the guarantee. What should the guarantee mileage be if Grear wants no more than 10% of the tires to be eligible for the discount?
Solution
(a) μ= 36,500, σ = 5,000, x =40,000
From table, the area between the mean (Z=0) and Z=0.70 is 0.2580). Total area above the mean is 0.5000. Therefore, the area above Z= 0.70 is 0.5000 - 0.2580 = 0.2420.
Thus about 24.2% of the tires manufactured by Grear can be expected to last more than 40,000 miles.
(b) 40% of the area must be between the mean and the unknown guarantee mileage
f(Z)
-Z -1.28 μ = 0 +Z
Z(0.4000) = 1.28 but because this is below the mean , Z is negative(-1.28)
=
x = 36,500 – 1.28(5000) = 30,100
The guarantee of 30,100 miles will meet the requirement that approximately 10% of the tires will be eligible for the discount.
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Question 6 (Nov. 2008)
(a) State two (2) characteristics of the normal distribution. (2 marks)
(b) For the past four years, the sales of fairly used car engines at Abossey Okai have been as follows:
An engine is sold on average for GH¢350. Assume that the monthly sales of the engines are normally distributed.
Required:(i) What is the probability that between 200 and 400 engines will be sold in a month?(ii) What is the probability that more than 500 engines will be sold in a month?(iii) What is the maximum expected revenue per month of the sales of engines if the
probability of sales per month is 0.4?Solution
(b) Let X be he number of engines sold per month at Abossey Okai.
X P(X) XP(X) X2P(X)200
150
450
300
250
0.1875
0.3125
0.125
0.125
0.250
37.50
46.875
56.25
37.50
62.50
7,500.00
7,031.25
25,312.50
11,250.00
15,625.00
240.625 66,718.75
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f(Z)
-Z -0.43 0 1.70 +Z
From Tables, P(200 ≤ X ≤ 400) = 0.1664 + 0.4554 = 0.6218
f(Z)
-Z 0 2.76 +Z
From Tables, P(X > 500) = 0.029
(iii) Let W be the maximum sales per month that gives a probability of 0.4.
Then P(X ≤ W) = 0.4
f(Z)
0.4
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-Z 0 +Z
From Tables, Z = - 0.25
Hence, the maximum expected revenue per month = GH¢350 x 217 = GH¢75,950
7. Weights of bags of potatoes are normally distributed with mean 5 and standard deviation 0.2 pounds. The potatoes are delivered to a supermarket, 200 bags at a time.
(a) What is the probability that a random bag will weigh more than 5.5 pounds?(b) How many bags, from a single delivery, would be expected to weigh more than 5.5
pounds?8. The time taken to complete a job of a particular type is known to be normally distributed with
mean 6.4 hours and standard deviation 1.2 hours. What is the probability that a randomly selected job of this type takes:
(a) less than 7 hours?(b) Less than 6 hours?(c) Between 6 and 7 hours?
9. Trading volume on the Ghana Stock Exchange has been growing in recent years. For the first two weeks in January 2008, the average daily volume was 646 thousand shares. The probability distribution of daily volume is approximately normal with standard deviation of about 100 thousand shares.
(a) What is the probability that the trading volume will be less than 400 thousand shares?(b) What percentage of the time does the trading volume exceed 800 thousand shares?(c) If the exchange wants to issue a press release on the top 5% of trading days, what volume
