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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) ProbabilityMeaning 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 nondefective. 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 = NonDefective 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
stagesEmpty, halffull, 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 wellstructured 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 wellshuffled 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+13/521/52= 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 EventsExample
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 independentExample
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 ProbabilityExample
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 NonBuyer 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 TreeExample
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 nondefective (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 TreeExample 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 nondefective (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.
 Spring '12
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