Probability theory

1. INSTITUTE OF
BUSINESS
MANAGEMENT
STATISTICAL
ANALYSIS

2. PROFESSOR
SHAH MOHD.

3. GROUP NAME:- FERRERO FACTION
GROUP NO:- 3
 GROUP MEMBERS:-
PARUL SINGH
SHIVANI SINGH
MOHD ATIF
MOHD BAIZAD MALIK
MOHD NADEEM

4. PROBABILITY
THEORY

5. 1. Preface
2. Probability meaning
3. Probability theory
4. Definition of Probability
5. Importance of Probability
6. Types of Probability
7. Addition Probability Model
8. Multiplication Probability
Model
9. Conditional Probability
10. Baye’s Theorem
11. Bibliography
Page No.
6
8
9
10
12
13
17
27
36
46
61

7. FATHER OF PROBABILITY
GIROLAMO
CARDANO

8. PROBABILITY
Probability is the
ratio of
favourable
events to the
total number of
equally likely
events.
Probability is an
attitude of mind
towards
uncertain

9.  Credit of development of probability theory goes to gamblers,
and people who bets on horses, who has started discussion with
famous mathematician of that age, after disappointed from
goddess fortune famous scientists as Gallelio, Pascal, Fermett
used their power to solve these problems.
 This undeveloped idea is developed by the scientists as-
Laplace, Gauss, Yuler, James Bernoulli.

10. THE THEORY OF
PROBABILITY is of interest not only to
card and dice players who were its God fathers but also to
all men of action, heads of industries or heads of armies
whose success depends on decision.

11. DEFINITION OF
PROBABILITY
Mathematical
Definition
Statistical Definition
An event happen a times and
does not happen b times
and all ways are equally
likely then probability of
happening of an event will
be (a/a+b) and
probability of not
happening of an event will
be (b/a+b).
Probability is calculated on
the basis of available data
or frequencies or pre-
experiences.
P = r/n
r= Relative frequency
n= Number of the items

12. Importance of
Probability Theory
BASIS OF
STATISTICAL
LAWS
IMPORTANCE
IN GAMES OF
CHANCE
USE IN
SAMPLING
SPECIFIC
IMPORTANCE
IN INSURANCE
BUSINESS
USE IN
ECONOMICS
AND BUSINESS
DECISIONS
BASIS OF
TESTS OF
HYPOTHESIS
AND TEST OF
SIGNIFICANCE

13. TYPES OF
PROBABILITY
THEORETICAL
RELATIVE
SUBJECTIVE

14. Theoretical Probability
We assume that all n possible outcomes of a
particular experiment are equally likely, and we
assign a probability of 1/n to each possible
outcomes.
Example:- The theoretical probability of rolling a 3
on a regular 6 sided die is 1/6.

15. Relative Probability
 We conduct an experiment many, many times. Then we say:-
 The probability of an event A= How many times A occurs
How many trials
 Relative probability is based on observation or actual measurements.
 Example:- A die is rolled 100 times. The number 3 is rolled 12 times. The
relative probability of rolling a 3 is 12/100.

16. Subjective
Probability
 These are values (between 0 and 1 or 0 and 100%) assigned
by individuals based on how likely they think events are to
occur.
 Example:- The probability of my being asked on a date for
this weekend is 10%

17. ADDITON PROBABILITY
THEOREM
(An Important Theorem of Probability)
 “The literal meaning of addition theorem is to add
the individual probabilities of two or more events.”

18. The addition theorem in the probability
concept is the process of determination of the
probability that either ‘A’ or event ‘B’ occur or
both occur. The notation between two events
‘A’ and ‘B’ the addition is denoted as ‘U’ and
pronounced as union.
 Then, probability of occurrence of at least one
of these two events is given by the sum of the
individual probabilities.

19. Proof-
Let the event A can occur in p ways
and B in q ways then the number of ways
in which either event can happen is p+q. If
the total number of possibilities is n, then
by definition of probability,

20. = (p+q)/n = p/n + q/n = P(A)+P(B)
P(A or B) = P(A∪B) = P(A)+ P(B)
•The probability of either A or B
occurs
favourable no. of cases
Total no. of cases

21. • Addition theorem probability can
be defined and proved as
follows:-
•Where,
•P(A)= Probability of occurrence of event ‘A’
•P(B)= Probability of occurrence of event ‘B’
•P(A∪B)= Probability of occurrence of event ‘A’ or event ‘B’
•P(A∩B)= Probability of occurrence of event ‘A’ or event ‘B’

22. Case I –
When event are mutually exclusive
Let ‘A’ and ‘B’ are subsets of a finite non empty set ‘S’ then
according to the addition rule-
P(A∪B)= P(A)+P(B)-P(A).P(B)
on dividing both sides by P(S), we get-
P(A∪B)/P(S)=
P(A)/P(S)+P(B)/P(S)P(AUB)/P(S)………....(1) eq.

23. Case I I –
When event are not mutually exclusive
If the event ‘A’ and ‘B’ correspond to the two events ‘A’ and ‘B’ of
a random experiment and if the set ‘S’ corresponds to the sample
space ‘S’ of the experiment then the equation (1) become-
P(AUB)= P(A)+P(B)-P(A).P(B)

24. •These are event A∪ B refers to the
meaning that either event ‘A’ or event ‘B’
occurs.
•So we can find the logic behind the
addition probability theorem both may
occur simultaneously.

25. For example:-
The probability of getting spot 2 in a throw of a single dice is 1/6,
the probability of getting spot 4 is also 1/6, If it is asked that what is
the probability of getting 2 or 4, it will be{1/6+1/6=2/6}=1/3 on the
basis of addition theorem.
Hint:
while doing a question it should be kept in mind that addition theorem
will be applicable in all those problems in which either the word ‘or’ has been used
explicitly or the word ‘or’ is used in analysis of the question

26. LIMITATION:-
•Addition theorem is applicable only when following two
conditions are fulfilled- (a)Events are mutually exclusive
and (b)They all are related to same set.
•It should be noted that if two or more events are not
mutually exclusive completely, the addition theorem has
to be modified. Suppose, there are two events A and B
and in some occurrence they are not mutually exclusive,
then modified formula will be as follows:-
• P(AorB)= P(A)+P(B)-P(A&B)

27. Multiplication Law of Probability
Multiplication law in probability applies to
combination of events. When the events
have to occur together then we make use of
the multiplication law of probability.

28. Now two cases arise: whether
the events are
independent or dependent.
The events are said to
be independent only
when the occurrence of
one event does not
change the probability
of occurrence of any
other event with it.
The given events
are said to be
dependent when the
occurrence of one
event changes the
probability of
occurrence of any

29. Rules
The multiplication rule states that:
“The probability of occurrence of given two events or in other words the probability of intersection of two given events
is equal to the product obtained by finding the product of the probability of occurrence of both events.”
This implies that if A and B are two events given then:
P (A and B) = P (A ∩ B) = P (A) * P (B)
This rule is applicable in all the cases, that is, when events are independent or
dependent.
In case when we have dependent events we have to be very careful in determining
the probability of the second event after the occurrence of first event.

30. Let us see an example to understand this.
 We have two events of drawing two candies
from a box one by one but with replacement. It
is clear that here the given events are completely
independent and thus we can multiply the
probabilities of the given two events for finding
out the probability of combined events.
Now, suppose the candies are taken from the
box without putting the first one back. It is clear
that the events are dependent and thus we need
to find the conditional probability for finding
the probability of occurrence of combined
event.
In such case the multiplication rule is modified as:
P (A and B) = P (A ∩ B) = P
(A) * P (B|A)
Here, P (B|A) is the probability
of occurrence of the second
event B when the first event A
has already occurred.

31. Problems
Let us see some examples on the multiplication law of probability.
Example 1:
A bag contains 3 pink candies and 7 green candies. Two candies are taken out
from the bag with replacement. Find the probability that both candies are pink.
Solution:
Let A = event that first candy is pink and B = event that second candy is pink.
→ P (A) = 3/10 …(i)
Since the candies are taken out with replacement, this implies that the given events A and B are
independent.
→ P (B|A) = P (B) = 3/10 …(ii)
Hence by the multiplication law we get,
P (A ∩ B) = P (A) * P (B|A)
→ P (A ∩ B) = 3/10 * 3/10 [using (i) and (ii)]
= 9/100 = 0.09

32. Example 2:
A bag has 4 white cards and 5 blue cards. We draw two cards
from the bag one by one without replacement. Find the
probability of getting both cards white.
Solution:
Let A = event that first card is white and B = event that second card is white.
From question, P (A) = 4/9.
Now P (B) = P (B|A) because the events given are dependent on each other.
→ P (B) = 3/8.
So, P (A ∩ B) = 4/9 * 3/8 = 1/6.

33. COMBINED USE OF ADDITION AND
MULTIPLICATIONTHEOREM
Question : A speaks truth in 80% cases, B in 90% cases. In What percentage of
cases are they likely to contradict each other In starting the same fact.
Solution :
Let (A) and (B) denote the probability that A and B speak the truth. Then,
P(A) = 80/100 = 4/5 P(A) = 1 - P(A) = 1 - 4/5 = 1/5
P(B) = 90/100 = 9/10 P(B) = 1 - P(B) = 1 - 9/10 = 1/10
They will contradict each other only when one of them speaks the truth and the
other speaks a lie.

34. Thus, there are two possibilities:
1. A Speaks the truth and B tells a lie.
2. B Speaks the truth and A tells a lie.
Since, the events are independent, so by using the multiplication
theorem, we have:
1. Probability in the 1st case = 4/5 *1/10 =4/50
2. Probability in the 2nd case = 9/10 * 1/5 = 9/50
Since, these cases are mutually exclusive, so by using the addition
theorem. We have the required probability
= 4/50 + 9/50 = 13/50 = 26%

35. In a random experiment, if A and B are two events, then the probability of
occurrence of event A when event B has already occurred and P(B) ≠ 0, is called the
conditional probability and it is denoted by P(AB)
P(A/B) = Number of outcomes favourable to A which are also favourable to B
Number of outcomes favourable to B
P(A/B) = P(A ∩ B) , P(B) ≠ 0
P(B)
Similarly, P(B/A) = P(A ∩ B) , P(A) ≠ 0
P(A)
MULTIPLICATION THEOREMIN CASE OF
CONDITIONAL PROBABILITY

36. These are the probabilities calculated on
the basis that something has already
happened.

37.  The probability that I will pay my electricity
bill given that have just been paid.
 The probability that my students will turn
upto class given that it is a sunny day.

38. CONDITIONAL
PROBABILITY
Unconditional
Probability
Conditional
Probability

39. A fair dice is about to be tossed. The
probability that it lands with 5 showing up is
1/6 this is UNCONDITIONAL PROBABILITY,

40. The probability that it lands with 5 showing up,
given that it lands with an odd number showing up, is
1/3 this is a CONDITIONAL PROBABILITY.

41. P(A/B)
P(A/B)=P(A∩B) , P(B)≠0
P(B)
- P(A/B)=Probability of occurrence of event A given
that event B has already occurred P(B)≠0

42. - P(B/A)=P(A∩B), P(A)≠0
P(A)
- P(B/A)=Probability of occurrence event B
given that the event A has already occurred.
P(A)≠0

43. •The conditional probability of an event A
given that B has occurred lies between 0&1.
* P(A′/B)=1-P(A/B)

44. •P(E∩F)=P(E/F)×P(F)
•If E,F,&G are independent given that an event H has
occurred, then
•P(E∩F∩G/H)=P(E/H)×P(F/H)×P(G/H)

45. Example of Conditional
Probability
Table shows the result of a class survey:-
Find P (wash the dishes/male)
Did you wash the dishes last night?
YES NO
The condition male limits the sample space to 15 possible outcomes. Out of 15
males 7 did the dishes.
Therefore, P(washes the dishes/male)= 7/15
FEMALE 7 6 13 FEMALES
MALE 7 8 15 MALES

47. Bayes’ theorem (also known as
Bayes’rule or Bayes’law)
Is a result in probability theory that relates conditional probabilities.
If A and B denote two events,
P(A|B) denotes the conditional probability of A occurring, given that
B occurs.
The two conditional probabilities P(A|B) and P(B|A) are in general
different.
Bayes theorem gives a relation between P(A|B) and P(B|A).
An important application of Bayes’ theorem is that it gives a rule
how to update or revise the strengths of evidence-based beliefs in
light of new evidence
a posteriori.

48. As a formal theorem
Bayes’ theorem is valid in all interpretations of
probability. However, it plays a central role in the debate
around the foundations of statistics: frequentist and
Bayesian interpretations disagree about the kinds of
things to which probabilities should be assigned in
applications.

49. Whereas frequentists assign probabilities to
random events according to their frequencies of
occurrence or to subsets of populations as
proportions of the whole,
Bayesians assign probabilities to propositions
that are uncertain.
A consequence is that Bayesians have more
frequent occasion to use Bayes’ theorem.
The articles on Bayesian probability and
frequentist probability discuss these debates at
greater
length.

50. Suppose we have estimated
prior probabilities for events
we are concerned with, and
then obtain new information.
We would like to a sound
method to computed revised
or posterior probabilities.
Bayes’ theorem gives us a
way to do this.

51.  The probability of two events A and B happening, P(AB), is the
probability
 of A, P(A), times the P(A B) = P(A)P(B|A) (1)
 probability of B given that A has occurred, P(B|A). (1)
 On the other hand, the probability of A and B is also equal to the
probability
 of B times the probability of A given B.
 P(A B) = P(B)P(A|B) (2)
 Equating the two yields:
 P(B)P(A|B) = P(A)P(B|A) (3)
 and thus
 P(A|B) = P(A)P(B|A)
 P(B) (4)

 This equation, known as Bayes Theorem is the basis of statistical
inference.

52. STATEMENT OF BAYES’
THEOREM
 Bayes’ theorem relates the conditional and marginal
probabilities of stochastic
 events A and B:
 P(A|B) = P(B|A) P(A)
 P(B)
 Each term in Bayes’ theorem has a conventional
name:

53.  • P(A) is the prior probability or marginal probability
of A. It is ”prior” in the sense that it does not take
into account any information about B.
 • P(A|B) is the conditional probability of A, given B. It
is also called the posterior probability because it is
derived from or depends upon the specified value of
B.
 • P(B|A) is the conditional probability of B given A.
 • P(B) is the prior or marginal probability of B, and
acts as a normalizing constant.

54. Bayes Theorem (Bayes’ Rule)
 Bayes' theorem. Let A1, A2, ... , An be a set of mutually exclusive
events that together form the sample space S. Let B be any event
from the same sample space, such that P(B) > 0. Then,
P( Ak | B ) = P( Ak ∩ B )
 P( A1 ∩ B ) + P( A2 ∩ B ) + . . . + P( An ∩ B )

Note: Invoking the fact that P( Ak ∩ B ) = P( Ak )P( B | Ak ), Baye's
theorem can also be expressed as
P( Ak | B ) = P( Ak ) P( B | Ak )
 P( A1 ) P( B | A1 ) + P( A2 ) P( B | A2 ) + . . . + P( An ) P( B | An )

55. When to Apply Bayes' Theorem
 Part of the challenge in applying Bayes' theorem involves recognizing
the types of problems that warrant its use. You should consider Bayes'
theorem when the following conditions exist.:
 The sample space is partitioned into a set of mutually exclusive
events { A1, A2, . . . , An }.
 Within the sample space, there exists an event B, for which P(B)
> 0.
 The analytical goal is to compute a conditional probability of the
form: P( Ak | B ).
 You know at least one of the two sets of probabilities described
below.
 P( Ak ∩ B ) for each Ak
 P( Ak ) and P( B | Ak ) for each Ak

56. Example:-
3 boxes contain 6 red, 4 black, 5 red, 5
black and 4 red, 6 black balls
respectively. One of the box is selected
at random and a ball is drawn from it. If
the ball drawn is red, find the probability
that if it is drawn from the first bag.

57. Solution:-
Let A, B, C, & D, be the events
defined as follows:-
A= 1st box is choosen
B= 2nd box is choosen
C= 3rd box is choosen
D= 4th box is choosen

58. Since there are three boxes and one of the
three boxes is choosen at random,
therefore:-
P(A)=P(B)=P(C)= 1/3
If A has already occurred, then first box has
been chosen which contains 6 red and 4
black balls. The probability of drawing a red
ball from it is 6/10.

59. So,
P(D/A)=6/10
Similarly, P(D/B)=5/10 and P(D/C)=4/10
We are required to find P(A/D) i.e., given that the ball drawn is red,
what is the probability that it is drawn from the first box.
BY BAYE’S RULE:-
P(A).P(D/A)
P(A/D) = P(A).P(D/A)+P(B).P(D/B)+P(C).P(D/C)
1/3 * 6/10
(1/3 * 6/10)+(1/3 * 5/10)+(1/3 * 4/10)
=2/5

60. Probability Revision using Bayes’
Theorem
Prior
Probabilities
New
Information
Application of
Bayes’
Theorem
Posterior
Probabilities

61. BIBLIOGRAPHY
 Reference taken from the book BUSINESS STATISTICS
by:- Prof. R. P. Varshney.
 Reference from Google.
 Wikipedia
 Other educational sites.