Paper.15-
Quantitative Techniques for Management Decisions
Module10 Conditional Probabilitym Bayes, Theorem
Principal Investigator
Prof. S P Bansal
Vice Chancellor
Maharaja Agrasen
University, Baddi
YoginderVerma
Co-Principal Investigator Prof
Pro–Vice Chancellor
Central University of Himachal Pradesh. Kangra. H.P.
Paper Coordinator
Content Writer
Dr. Pankaj Madan
Dean, F M S, Gurukul Kangri viswavidalaya, Haridwar
Dr. Nidhi Handa
Assistt. Prof.
Department of
Mathematics & Statistics, Gurukul Kangri
viswavidhalaya, Haridwar
vishwavidayalaya.
1
ITEMS
DESCRIPTION OF MODULE
Subject Name
Management
Paper Name
QUANTITATIVE TECHNIQUES FOR MANAGEMENT DECISIONS
Module Title
Conditional Probability, Bayes Theorem
Module Id
Module No.-10
Pre- Requisites
Basic knowledge of what Probability is.
Objectives
To study the basic concept of conditional probability and the rules for its
computation.
Keywords
Compound event, Conditional Probability, Complementary events,
Independent events, Multiplication rule.
QUADRANT – 1
Module-10 Conditional Probability Bayes Theorem .
1. Introduction
2. Marginal Probability
2.
3.
4.
5.
6.
Conditional Probability
Multiplication Rule
Bayes Theorem
Problems
Summary
Introduction
Suppose all 100 employees of a company were asked whether they are in favour of or against
paying high salaries to chief executive officers of U.S. companies. Table 3 gives a two way
classification of the response of these 100 employees.
Table 1. Two way classifications of the responses of the employees
In favour
Against
Male
15
45
Female
4
36
2
Table 1 gives the distribution of 100 employees based on two variables or characteristics:
gender (male or female) and opinion (in favour or against). Such a table is called a
contingency table. In table 1, each box that contains a number is called a cell. Notice that
there are four cells in Table 1. Each cell gives frequencies for two characteristics. For
example, 15 employees in this group possess two characteristics: They are male and in favour
of paying high salaries to CEOs. We can interpret the number in other cells the same way. By
adding the row of totals and the column of totals to Table 1, we write Table 2.
Table 2 two way classifications of employees
In favour
Against
Total
Male
15
45
60
Female
4
36
40
Total
19
81
100
Suppose one employee is selected at random from these 100 employees. This employee may
be classified either on the basis of gender alone or on the basis of opinion. If only one
characteristic is considered at a time, the employees selected can be male, a female, in favour,
or against. The probability of each of these four characteristics events is called marginal
probability or simple probability. These probabilities are called marginal probabilities
because they are calculated by dividing the corresponding row margins (totals for the rows)
or column margins (totals for the columns by the grand total.
MARGINAL PROBABILITY:
Marginal probability is the probability of a single event without consideration of any other
event .Marginal probability is also called simple probability.
Table 3. Listing the marginal probabilities
In favour
(A)
Against
(B)
Total
Male (M)
15
45
60
Female (F)
4
36
40
Total
19
81
100
P(A)= 19/100
P(B)= 81/100
= .19
= .81
P(M)= .60
P(F)= .40
Now suppose that one employee is selected at random. Futhermore, assume that it is known
that this selected employee is a male. In other words, the event that the employees selected is
3
a male has already occurred. What is the probability that the employee selected (male) is in
favour of paying high salaries to CEOs?
This probability is called conditional probability i.e. P (in favour / male).
CONDITIONAL PROBABILITY:
The Probability that an event A will occur, knowing that a certain event B has already
occurred is called the conditional probability of A for given B and denoted by P(A / B) .
.
p(A / B) =
𝑝(𝐴 ⋂ 𝐵)
p (B / A) =
𝑝(𝐵)
and
𝑝(𝐴 ⋂ 𝐵)
𝑝(𝐴)
Example: The probability that a regular scheduled flight depart in time is 0.85 and the
probability that it arrives in time is 0.80. The probability that it arrives and departs in time
0.75. Find the probability that a plane arrives in time given that it departed on time.
Solution:
p (event of departure) = p (D) = 0.85
p (event of arrival)
= p (A) = 0.80
p (A ⋂ 𝐷) = 0.75.
Required probability is
P (A / D) =
𝑝 (𝐴 ⋂ 𝐷)
𝑝 (𝐷)
=
0.75
0.85
=
15
17
= 0.88
Bayés Theorem: Thomas Bayes a renowned British Mathematician introduced the concept
of inverse probability.
This theory finds the probability of an event by taking into account a given sample
information.
This is called inverse probability because it consists of finding the probability of a
probability.
This is called posterior probability as it is calculated after information is taken into
account.
Bayes theorem is based on the formula for conditional probability.
When A1 & A2 are two events which are mutually exclusive (i.e. the two events
cannot occur together) & exhaustive (i.e. the two events A1 & A2 are the entire experiment).
B is a sample event which intersects each of the A events as shown then part of B which is
within A1 shows the area “A1 & B” whereas the part of B within A2 represents the area.
4
INDEPENDENT EVENTS:
The occurrence of one event does not change the probability of the occurrence of other event
is called independent events
either P(A/B) = P(A) or P(B/A) = P(B)
Multiplication Rule:
The probability that events A and B happen together is called the joint probability of
events A and B can also be denoted by P (𝐴 ⋂ 𝐵) or P (AB).
The probability of event A1 given event B is
P (A1/B) =
𝑝(𝐴1 ⋂ 𝐵)
𝑝(𝐵)
also P (A/B) =
p(A2 ⋂ B)
p(B)
Where p (B) = p (𝐴1 ⋂ 𝐵) + p (𝐴2 ⋂ 𝐵)
∴ p (𝐴1 ⋂ 𝐵) = P(A1) x p (B/A1)
p (𝐴2 ⋂ 𝐵) = P(A2) x p (B/A2)
GENERALIZED FORM OF BAYE’S THEOREM :
Suppose the events A1, A2 ……𝐴𝑛 form a partition of a sample space S, that is, the
events Ai are mutually exclusive and their union is S. Now let B be any other event. Then.
B = 𝑆 ⋂ 𝐵 = (𝐴1 ⋃ 𝐴2 ⋃ … … . ⋃ 𝐴𝑛 ) ⋂ 𝐵
= (𝐴1 ⋂ 𝐵) ⋃(𝐴2 ⋂ 𝐵) ⋃ … … . . ⋃(𝐴𝑛 ⋂ 𝐵)
Where the 𝐴𝑖 ∩ 𝐵 are also mutually exclusive.
Accordingly
P(B)=𝑃(𝐴1 ∩ 𝐵) + 𝑃(𝐴2 ∩ 𝐵) … … … … . . +𝑃(𝐴𝑛 ∩ 𝐵)
Thus by multiplication theorem
P(B) = p(A1) p(B/A1) + P(A2) P(B/A2) +……p(An) P(B/An)
------ [1]
On the other hand, for any i, the conditional probability of Ai given B is defined by
P(Ai/B) =
5
𝑃(𝐴𝑖 ⋂ 𝐵)
𝑃(𝐵)
Here we use (1) to replace P(B) and use P(𝐴𝑖 ⋂ 𝐵) = P(Ai)P(B/Ai) to replace P(A𝑖 ∩
𝐵)
P(𝐴𝑖/𝐵) =
OR
𝑃(𝐴𝑖)𝑃(𝐵/𝐴𝑖)
𝑃(𝐴1)𝑃(𝐵/𝐴1)+𝑃(𝐴2)𝑃(𝐵/𝐴2)…………….+𝑃(𝐴𝑛)𝑃(𝐵/𝐴𝑛).
P(Ai/B) =
𝑃(𝐴𝑖)𝑝(𝐵/𝐴𝑖)
∑𝑛
𝑖=1 𝑃(𝐵/𝐴𝑖)
[2]
To summarise it if an event B can occur only in combination with one of the n
mutually exclusive, & exhaustive events A1, A2……. An then by actual observation, B is
found to have occurred, then probability that it was preceded by a particular event A1 is
defined by (2) . This is Baye’s theorem.
Suppose the probabilities of A1, A2 ……………An as well as conditional probabilities
p(A1), ………. ,P(An) are known and we are interested in finding the conditional probability
p(A1/B) from the given information. This is possible with the Baye’s Theorem as follows
P(Ai/B) =
=
𝑝(𝐴𝑖 ⋂ 𝐵)
𝑝(𝐵)
𝑝(𝐴1 )𝑝(𝐵/𝐴1 )
𝑝(𝐴1 ) 𝑝(𝐵/𝐴1)……..+ 𝑝 (𝐴𝑛 ) 𝑃(𝐵/𝐴𝑛 )
Example : An item is manufactured by three machines M1, M2 M3. Out of the total
manufactured during a specified production period, 50% are manufactured on M1, 30% on
M2 and 20% of M3. It is also known that 2% of the item produced by M1 and M2 are
defective, while 3% of those manufactured by M3 are defective. All the items are put into one
bin. From the bin, one item is drawn at random and is found to be defective. What is the
probability that it was made on M1, M2 or M3.
Answer :
Let
B
:
the item is defective.
A1
:
the item was made on M1.
A2
:
the item was made on M2.
A3
:
the item was made on M3.
It an item is drawn at random, we know that a prior probability.
P(A1) = 0.5, P(A2) = 0.3, P(A3) = 0.2
We also know P(B/A1) = 0.02. P(B/A2) = 0.02, P(B/A3) = 0.3. We are interested in
finding the posterior probability, i.e. having known that the item is defective, what is the
probability that it was manufactured on M1, M2 or M3.
6
𝑀1
,P(B/𝐴1 )=0.02
P(𝐴1 )=0.5
P(𝐴2 )=0.3 P(B/𝐴2 )=0.02
𝑀2
P(B/𝐴3 )=0.03
P(𝐴2 )=0.2
𝑀3
P(𝐴1 ) P(B/𝐴1 )=(0.5)(.02) =0.01
Now, the probability that the item was manufactured on
M1
=
=
M2
=
M3
=
𝐽𝑜𝑖𝑛𝑡 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑏𝑟𝑎𝑛𝑐ℎ
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑗𝑜𝑖𝑛𝑡 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑏𝑟𝑎𝑛𝑐ℎ𝑒𝑠
0.01
0.01 + 0.006 + 0.006
0.006
0.022
0.006
0.022
=
0.01
0.022
= 0.454
= 0.2727
= 0.2727
Example.2 A company has two plants to manufacture, sctoors plant I manufacture 70% of
the sctoors and plant II manufactures 30%. At plant I, 80% of sctoors are good
standard quality and at plant II, 90% of sctoors are rated standard quality. A sctoor is
picted up at random and is found to be of standard quality. What is the chance that it
has come from plant I or plant II.
Ans : Let
A1 = The event of drawing a sctoor produced by plant I.
A2 =
The event of drawing a scooter produced by plant II
E =
Event of drawing a standard quality scooter either by plant I
or plant II.
P(A1) = 70% = 0.7
P(A2) = 30% = 0.3
P(E / A1) = 80% = 0.8
P(E / A2) = 0.9
P (E ⋂ 𝐴1 ) = P(E / A1) x P(A1) = 0.8 x 0.7 = 0.56
P(E ⋂ 𝐴2 ) = P(E / A2) x P(A2) = 0.9 x 0.3 = 0.27
The probability that a standard quality scooter is chosen at random is manufactured by
plant I is given by the Bay’s theorem as :
𝐸
P (A1 / E) =
=
7
𝑝(𝐴 ) ×𝑝(𝐴1 )
1
𝐸
𝐸
𝑝(𝐴 ) × 𝑝(𝐴1 )+ 𝑝(𝐴 ) × 𝑝(𝐴2 )
1
2
0.56
0.56 + 0.27
=
56
83
Similarly, we can obtain P(A2/E)
= 0.424
Example 3. – A box contains 20 cassettes, 4 of which are defective. If 2 cassettes are selected
at random (without replacement) from this box, what is the probability that both are
defective?
Solution: Let us define the following events for this experiment.
G1
=
event that the first cassette selected is good
D1
=
event that the first cassette selected is defective
G2
=
event that the second cassette selected is good
D2
=
event that the second cassette selected is defective.
We are to calculate the joint probability of D1 and D2, which is given by
P(D1 and D2) = P(D1) P(D2 I D1)
As we know, there are 4 defective cassettes in 20. Consequently, the probability of selecting a
defective cassette at the first selection is
P(D1) = 4/20
To calculate the probability P(D2/ D1), we know that the first cassette selected is defective
because D1 has already occurred. Because the selections are made without replacement, there
are 19 total cassettes and 3 of them are defective at the time of the second selection.
Therefore,
P(D2 / D1)= 3/19
Hence, the required probability is
P(D1 and D2) = P(D1) P(D2 / D1) = (4/20) (3/19) = .032
The tree diagram in Figure shows the selection procedure and the final four outcomes
for this experiment along with their probabilities. The joint probability of D1 and D2 is
highlighted in the tree diagram.
Conditional probability was discussed already . It is obvious from the formula for
joint probability that we know the probability of an even A and the joint probability of
events A and B i.e. P (𝐴 ⋂ 𝐵) , then we can calculate the conditional probability of B given A
i.e. P(B/A)..
8
First Slection
Second Slection
G1
G 2/ 9
15/1
G1
/20
16
D
4/
1
20
Final out comes
P(G1 and G2 ) = (16/20) (15/19) = .632
D
2/
4/1 G1
9
P(G1 and D2 ) = (16/20) (4/19)
/D 1
G2 9
1
1 6/
D
2/
3/1 D1
9
P(D1 and G1 ) = (4/20) (16/19) = .168
P(D1 and D2 ) = (4/20) (3/19) =0 .032
Fig.I
Example 4.Compute the conditional probability P (in favour / male) for the data on 100
employees given in table.
Solution – The probability P (in favour/male) is the conditional probability that a randomly
selected employee is in favour given that this employee is a male. It is known that the venet
“male” has already occurred. Based on the information that the employee selected is a male,
we can infer that the employee seleted must be one of the 60 males and, hence, must belong
to the first row of table 3. Therefore, we are concerned only with the first row of that table.
Male
In Favor
Against
Total
15
45
60
Male who are in favour
Total number of males
The required conditional probability is calculated as follows.
P(in favour/male) =
𝑁𝑢𝑚𝑏𝑒𝑟𝑠 𝑜𝑓 𝑚𝑎𝑙𝑒𝑠 𝑤ℎ𝑜 𝑎𝑟𝑒 𝑖𝑛 𝑓𝑎𝑣𝑜𝑟
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑎𝑙𝑒𝑠
=
15
60
= .25
As we can observe from this computation of conditional probability the total number of males
(the event that has already occurred) is written in the denominator and the number of males
who are in favour (the event whose probability we are to find) is written in the numerator.
Note that we are considering the row of the event that has already occurred. The tree diagram
in figure II as follows :
9
we are to find
probability of this event
This has
already occurred
Male
Fe
m
40 ale
/10
0
a le
s /M
r
u
o
Fav 5 /60
1
Ag
a in
s
4 5/ t/Ma
le
60
required probability
le
ma
/Fe
s
r
ou
Fav 4/4 0
Ag
ain
s
36 t/Fe
/40 m
a le
Fig II
Summary
The probability of an event is subject to the condition that another event has already
occurred is conditional probability. This type of probability opens many spheres for
researchers to make inferences/decisions. It involves independent verses dependent
probability, mutually exclusive verses mutually non-exclusive events along with
complementary events. Bayes’ Theorem involves such type of problem e.g. the
following table involves two-way classification of all scientists & engineers (Male &
Female)
Scientists
Engineers
Male
680
527
Female
120
108
Find one person selected is an engineer / a scientist given a person is female/ a
male given the person is scientist etc.Here conditional probability and dependent
probabilities are involved.
10