Probability: Definition
Probability: Definition
Probability: Definition
Probability
Definition:
An experiment is any well-defined operation or procedure that results in one of two or
more possible outcomes. An outcome is particular result of an experiment.
Counting Techniques:
(a)
(b)
(c)
(d)
Tree Diagram,
Multiplication Rule,
Permutation, and
Combination.
Pr
n!
(n r )!
Where r n
Example:
How many three-digit numbers can be formed from the digits 1, 2, 4, 5 and 9 without
replacement?
Solution:
n = 5 and r = 3
n
Pr
n!
5!
60 ways
(n r )! (5 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:
n
Pr (n) r
Example:
Arrange the license plate with 3 alphabets and 3 digits with replacement.
Solution:
n1 n Pr 26 P3 26 3 17576
n2 n Pr 10 P3 10 3 1000
n1 n2 17576 1000 17576000 ways
Situation 3:
For n non-distinct items out of which n1 are of one kind, n2 are of another kind,
, nk are of another, and n1 + n2 + + nk = n, the number of
permutations of all n items is:
n
n!
n1 ! n 2 ! n3 ! ............ n k !
Example:
Find the possible permutations of 7558.
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Solution:
n = 4 (7558),
n1 = 1 (one 7),
n2 = 2 (two 5), and
n3 = 1 (one 8).
4
P1, 2,1
4!
12 ways
1! 4! 1!
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)!
Cr
n!
r! (n r )!
Example:
In how many ways five students can be selected from a group of 12 students?
Solution:
n = 12; r = 5
n
C r 12 C 5
12!
792 ways
5! (12 5)!
Possibility Space,
Event,
Complementary Event,
Mutually Exclusive Events,
Composite Events, and
Joint Events.
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 = 2n = 24 = 16 events
Possible Events of two tossed coins:
E1 = { }
E2 = {HH}
Null Event
Simple Events
Compound Events
Sure Event
S
E5 =
{TT}
Event E5
E5
{HH,
HT, TH}
Event
(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.
Non-mutually exclusive events are the vice versa of the above definition and are
also known as over-lapping events.
S
A
For example, in case of two tossed coins, E6 and E7 are not mutually exclusive
events, because both events have an outcome HH in common. However, E6 and
E11 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 E6 and E7 is as follows:
E6 U E7 = {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 AB.
For example, the joint event of E6 and E7 is as follows:
E6 E7 = {HH}
Take another example, the joint event of E2 and E15 is as follows:
E2 E15 = { }
Probability Theory:
1.
(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:
number of outcomes in the event A
number of outcomes in the possibility space S
n( A)
P ( A)
n( S )
n( A) 1
0.25
n( S ) 4
(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
m
P ( A) lim
n
m
n
m
is considered as an estimate of the actual probability of event A
n
The ratio
and normally this estimate is called the probability of the event A and is written
as:
P ( A)
m
n
m
63
0.175
n 360
Example:
n( A ) 1
n( S )
8
Since P ( A ) 1 P ( A)
or P ( A) 1 P ( A )
therefore P ( A) 1
1 7
0.875
8 8
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 non-occurrence 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:
P( B / A)
P( A B)
(dependent event)
P ( A)
and
P( A / B )
P( A B)
P( B )
(independent events)
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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:
P( S / B)
13
0.5
26
Male
70
120
190
Female
150
160
310
Total
220
280
500
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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; P( A)
B = student selected is a female; P( B)
220
500
310
500
(ii)
150
220
0. 3
500 220 500 10
500 500
310 190
P ( A B ) P ( A ) P ( B / A ), where P ( B / A )
120
280
0.24
500 280 500
(i)
(ii)
Solution:
(i)
it is either Ace or King:
Let A = Ace; P( A)
B = King; P ( B)
4
52
4
52
4
4
8
2
0.154
52 52 52 13
(ii)
4
52
13
D = Diamond; P( D)
52
Let C = Queen; P (C )
4 1
1
52 4 52
4 13 1
16
4
0.308
52 52 52 52 13
Bayes Theorem:
1. The Bayes Theorem is based on conditions probabilities. It calculates
probabilities of the causes that may have produced an observed event.
2. Given A1, A2, , Ai, ., An mutually exclusive events, whose union
is the entire possibility space S, and let B an arbitrary event in S, such that P(B)
O, then:
P( B Ai )
P( B)
P( B / Ai ) P( A)
P( B / Ai ) P( Ai )
P( Ai / B)
Where i = 1, 2, 3, 4, ., n
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