Definition:
An experiment is any welldefined
operation or procedure that results in one of two or more possible outcomes.
An outcome is particular result of an experiment.
Counting Techniques:
(a) Tree Diagram,
(b) Multiplication Rule,
(c) Permutation, and
(d)
Combination.
(a)
Tree Diagram:
Counting the number of possible outcomes of an experiment plays a major role in
probability theory. These possible
outcomes can be shown by the branches of a treelike diagram called ‘TreeDiagram’
or ‘Branch Diagram’.
(b)
Multiplication Rule:
If an operation of an experiment can be performed in n_{1} ways and if
for each of these ways another operation can be performed in n_{2} ways,
then the two operations can be performed together in n_{1} × n_{2}
ways, and the k^{th} in n_{k} ways, then all the k operations
can be performed together in n_{1} × n_{2} × n_{3} ×
…….. × n_{k} ways.
For example, a
coin and a die tossed together, the possible outcomes will be:
n_{1} = 2 (coin: two sides)
n_{2}
= 6 (die: 6 sides)
The two
operations can result in (n_{1} × n_{2}) 12 ways.
(c)
Permutations:
A permutation is a group of items with a certain ordered arrangement. For example: ABC, ACB, CAB, CBA and BCA are different
permutations. The rules of
permutation are different under each of the following four situations:
Situation
I:
(i) All the items are distinct,
(ii) Each item can occur only once in an arrangement, i.e., repetition not allowed or without replacement, and
(iii)
Each item can occupy any place in an arrangement.
In the above
situation, the number of permutations of n
items arranged r at a time, denoted by
is:
Where r
≤ n
Example:
How many threedigit numbers can be
formed from the digits 1, 2, 4, 5 and 9 without replacement?
Solution:
n = 5 and r
= 3
Situation
2:
(i) All the items are distinct,
(ii) An item can be repeated in an arrangement (i.e., repletion allowed or with replacement), and
(iii)
Each item can occupy any place in an arrangement.
In the above
situation, the number of permutations of n
items arranged r at a time is:
Example:
Arrange the license plate with 3
alphabets and 3 digits with replacement.
Solution:
Situation
3:
For n
nondistinct items out of which n_{1} are of one kind, n_{2} are
of another kind, ……, n_{k} are of another, and n_{1} + n_{2}
+ …………… + n_{k} = n, the number of permutations of all n items
is:
Example:
Find the possible permutations of 7558.
Solution:
n = 4 (7558),
n_{1} = 1 (one 7),
n_{2} = 2 (two 5), and
n_{3} = 1 (one 8).
Situation
4:
When
the items are arranged in a circle, two arrangements are not considered
different, unless corresponding items of both are preceded or followed by a
different item.
The number of
permutations of n distinct items arranged in a circle is:
= (n – 1)!
The above lines have different permutations:
The
above circles have same permutations.
Example:
Arrange 5 different trees in a circle:
Solution:
n = 5 trees
= (n – 1)!
= (5 – 1)! = 4! = 24 ways
(d)
Combinations:
A combination is a group of items without regard to the arrangement of items.
ABC and BCA are two different permutations but are same combinations of
three letters.
The number of
combination of n distinct items taken r at a time, denoted by
, is:
Example:
In how many ways five students can be
selected from a group of 12 students?
Solution:
n = 12; r = 5
Basic Concepts of Probability
Theory:
(a) Possibility Space,
(b) Event,
(c) Complementary Event,
(d) Mutually Exclusive Events,
(e) Composite Events, and
(f)
Joint Events.
(a)
Possibility Space:
Also known as ‘sample space’ or ‘outcome space’.
A possibility space is a set of all possible outcomes of an experiment
and is denoted by S.
If an
experiment consists of a toss of a fair die and the numbers are of interest, the
possibility space would be:
S = {1, 2, 3,
4, 5, 6}
If the interest
is whether the number is even or odd, the possibility space would be:
S = {even,
odd}
A possibility
space may be represented by a rectangle. The
number of outcomes in the possibility space is denoted by n(S).
(b)
Event:
A subset of a possibility space is called an event and is usually denoted by
first few capital letters A, B, C, ……… For example, a coin is tossed, the
sample space is S = {H, T} and the subset A = {H} is the event when a head
occurs.
Take another
example, two coins are tossed. The
sample space is S = {HH, HT, TH, TT} and the subset B = {HH, HT, TH} is the
event that at least one head appears when two coins are tossed.
An
event may be represented by a circle inside the rectangle of the possibility
space.
An event is
further divided into the following:
(i) Simple Event: is the subset if it contains only one outcome of the possibility space.
(ii) Compound Event: is the subset if it contains more than one outcomes of the possibility space.
(iii) Null Event: is a subset containing no outcomes. It is also called ‘impossible event’.
(iv)
Sure Event:
It is also called ‘certain event’. It is a subset containing all outcomes of the possibility
space.
Number
of possible events = 2^{n
}
Example:
Two coins are tossed. List all the
possible events or subsets of the possibility space.
Solution:
S = {HH, HT, TH, TT}
n(S) = 4
Number of possible events = 2^{n}
= 2^{4} = 16 events
Possible Events of two tossed coins:
(c)
Complementary Event:
For an event A, the complementary event is defined as a set of those outcomes of
the possibility space which are not in A. The
complementary event of A is written as
(not A).
For example, the complementary events of E_{5} = {TT} are . It is diagrammatically shown as below:
(d)
Mutually Exclusive Events:
Two events A and B are mutually exclusive if they have no outcomes in common and
therefore they cannot happen together. Mutually
exclusive events are also known as disjoint events.
Nonmutually exclusive events are the vice versa of the above definition and are also known as overlapping events.
For example, in
case of two tossed coins, E_{6} and E_{7} are not mutually
exclusive events, because both events have an outcome HH in common.
However, E_{6} and E_{11} are two mutually exclusive
events because no single outcome is common.
(e)
Composite Events:
For two events A and B, a composite event is defined as a set of outcome of
either A or B or both A and B and therefore at least one of two events must
occur. The composite event of A and
B is written as AUB, or A or B.
For example,
the composite event of E_{6} and E_{7} is as follows:
E_{6}
U E_{7} = {HH, HT, TH}
(f)
Joint Events:
For two events, A and B, a joint event is defined as a set of common outcomes of
A and B and therefore both the events must occur together.
The joint event of A and B is written as ‘A and B’ or
‘A∩B’.
For example,
the joint event of E_{6} and E_{7} is as follows:
E_{6}
∩ E_{7} = {HH}
Take another
example, the joint event of E_{2} and E_{15} is as follows:
E_{2}
∩ E_{15} = { }
(i.e., Null Event)
Probability Theory:
1. A probability is a numerical measure of the likelihood (or chance) that a particular event will occur.
2.
The probability of any event must satisfy the following two conditions.
(i) No probability is negative, P(Event) ≥ 0.
(ii)
No probability is greater than one, P(Event) ≤ 1.
3.
There are three different approaches to assign probabilities to the
events:
(a) Classical or Mathematical Approach
(b) Empirical or Relative Frequency Approach
(c)
Subjective Approach
(a)
Classical Approach:
It is the approach in which probabilities are assigned to the events before the
experiment is actually performed and therefore, such probabilities are also
called ‘a priori’ probabilities.
If the
possibility space of the experiment is finite, and if each outcome of the
possibility space is equally likely to occur, then the probability of event A:
It is also
referred to as ‘axiomatic function of probability’.
Example:
Two coins are tossed once, what is the
probability that two heads will appear?
Solution:
S = {HH, HT, TH, TT}
n(S) = 4
Event = Two head appear
i.e., n(A) = 1
Now the probability of event A is as
calculated below:
(b)
Relative Frequency Approach:
This approach is applied when the possibility spaces are infinite, or the
outcomes cannot be assumed equally likely.
If an
experiment is represented ‘n’ times under uniform conditions and if
‘m’ times the outcome of the experiment is in favour of an event A,
then the ratio
approaches the probability of the
event A as ‘n’ approaches infinity.
The ratio
is considered as an estimate of the
actual probability of event A and normally this estimate is called the
probability of the event A and is written as:
Since in this
approach probabilities are assigned after performing a large experiment,
therefore, they are also known as ‘a posteriori’ probabilities.
Example:
A die has been rolled 360 times and
‘Six’ has been observed 63 times. Estimate
the probability of occurring a ‘Six’ when the die is to be rolled once
again.
Solution:
m = 63
n = 360
(c)
Subjective Approach of Probability:
Subjective probability can be defined as the probability assigned to an event by
an individual, based on whatsoever evidence is available.
This evidence may be in the form of relative frequency of past
occurrences, or it may be just an educated guess.
Probability of Complementary
Events:
If A and
are complementary events in a
probability S, then:
Example:
A coin is tossed 3 times.
Find the probability of getting at least one tail.
Solution:
Number of possible outcomes in S = n(S) = 2^{3} = 8
Let A = At least one tail appears
Then = No tail appears = All heads appear
(the only outcome with all heads)
Independent and Dependent
Events:
1. When two events are given, the occurrence of the first event may or may not have an effect on the occurrence of the second event.
2. When the occurrence of one of the two events has no effect on the probability of the occurrence of the other event, the two events are called ‘Independent Events’.
3.
On the other hand, when the occurrence of first event has some effect on
the probability of occurrence of second event, the second event is said to be ‘Dependant’
on the first event.
Conditional Probability:
If there are two events A and B such
that the probability of event B depends on the occurrence or nonoccurrence of
the event A, then the probability of event B occurs when the event A occurs is
called the ‘conditions probability of event B given event A’ and is
written as:
and
If two events are independent then:
P(B/A)
= P(B)
(independent events)
Example:
A black card is drawn from an ordinary
deck of 52 playing cards. What is
the probability that it is of spade (♠) suit?
Solution:
Let B = Black card drawn
and S = Spade card drawn
Since the card drawn is black (event B
has occurred), and there are 13 spade cards in 26 black cards, therefore:
Multiplication Law of
Probability:
If there are two nonmutually exclusive
events A and B such that event B is dependent on event A, then the probability
of joint event (A and B) is given by:
P(A
and B) = P(A) × P(B/A)
or
P(A∩B) = P(A) × P(B/A) or P(A) + P(B) – P(AUB)
or it may
equivalently be written as:
P(A∩B)
= P(B) × P(A/B)
When two events A and B are independent:
P(A∩B)
= P(A) × P(B)
When two events A and B are mutually
exclusive, their joint probability is zero:
P(A∩B)
= 0
Example:
In a graduate college, there are 500
male and female students learning B.Sc and B.Com.
The break up is as follows:

Male 
Female 
Total 
B.Sc 
70 
150 
220 
B.Com 
120 
160 
280 
Total 
190 
310 
500 
What is the probability that a randomly
selected student is (i) a female B.Sc student, and (ii) a male B.Com student?
Solution:
(i)
a female B.Sc student:
Let A = student selected is learning B.Sc;
B = student selected is a female;
A∩B = female student learning B.Sc
P(A∩B) = P(A) × P(B/A), where
(ii)
a male B.Com student:
Let = student selected is not learning B.Sc;
= student selected is not a female;
Addition Law of Probability:
If A and B are two nonmutually
exclusive events, then the probability of the composite event (A or B) is given
by:
P(A
or B) = P(A) + P(B) – P(A and B)
or
P(AUB) = P(A) + P(B) – P(A∩B)
For two mutually exclusive events A and
B:
P(AUB)
= P(A) + P(B)
Example:
A card is drawn from a pack of 52 cards.
What is the probability that:
(i) it is either Ace or King
(ii)
it is either Queen or a Diamond
Solution:
(i)
it is either Ace or King:
(AUB) = the card drawn is either an Ace or a King
PAUB) = P(A) + P(B)
(ii) it is either Queen or a Diamond:
CUD = Card drawn is either a Queen or a Diamond
P(CUD) = P(C) + P(D) – P(C∩D); where P(C∩D) = P(C) × P(D/C) =
P(CUD)
Baye’s Theorem:
Where i
= 1, 2, 3, 4, ……., n