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# Probability (Lec 1) (Probability and Distributions) Engineering Mathematics Notes | EduRev

## Engineering Mathematics : Probability (Lec 1) (Probability and Distributions) Engineering Mathematics Notes | EduRev

``` Page 1

NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   1

MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem

1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let  ?? ,F,??  be a probability space and let  ?? ?? : ?? ???  be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ???  using Theorem of Total Probability .  _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
Page 2

NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   1

MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem

1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let  ?? ,F,??  be a probability space and let  ?? ?? : ?? ???  be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ???  using Theorem of Total Probability .  _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   2

(ii) In Bayes’ theorem the probabilities ?? ?? ?? ,?? ??? , are referred to as prior
probabilities and the probabilities ?? ?? ?? |?? ,?? ??? , are referred to as posterior
probabilities. _
To see an application of Bayes’ theorem let us revisit Example 3.4.
Example 3.5
Urn ?? 1
contains 4 white and 6 black balls and urn ?? 2
contains 6 white and 4 black balls.
A fair die is cast and urn ?? 1
is selected if the upper face of die shows five or six dots.
Otherwise urn ?? 2
is selected. A ball is drawn at random from the selected urn.
(i) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 1
;
(ii) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 2
.
Solution. Define the events:
?? : drawn ball is white;

?? 1
: urn ?? 1
is selected
?? 2
: urn ?? 2
is selected
mutually exclusive & exhaustive events
(i) We have
?? ?? 1
|??  =
?? ?? |?? 1
?? ?? 1

?? ?? |?? 1
?? ?? 1
+?? ?? |?? 2
?? ?? 2

=
4
10
×
2
6
4
10
×
2
6
+
6
10
×
4
6

=
1
4
·
(ii) Since ?? 1
and ?? 2
are mutually exclusive and ?? ?? 1
??? 2
|?? =?? ?? |?? = 1,we
have
?? ?? 2
|?? = 1-?? ?? 1
|??
=
3
4
· _
In the above example
?? ?? 1
|?? =
1
4
<
1
3
=?? ?? 1
,
Page 3

NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   1

MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem

1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let  ?? ,F,??  be a probability space and let  ?? ?? : ?? ???  be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ???  using Theorem of Total Probability .  _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   2

(ii) In Bayes’ theorem the probabilities ?? ?? ?? ,?? ??? , are referred to as prior
probabilities and the probabilities ?? ?? ?? |?? ,?? ??? , are referred to as posterior
probabilities. _
To see an application of Bayes’ theorem let us revisit Example 3.4.
Example 3.5
Urn ?? 1
contains 4 white and 6 black balls and urn ?? 2
contains 6 white and 4 black balls.
A fair die is cast and urn ?? 1
is selected if the upper face of die shows five or six dots.
Otherwise urn ?? 2
is selected. A ball is drawn at random from the selected urn.
(i) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 1
;
(ii) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 2
.
Solution. Define the events:
?? : drawn ball is white;

?? 1
: urn ?? 1
is selected
?? 2
: urn ?? 2
is selected
mutually exclusive & exhaustive events
(i) We have
?? ?? 1
|??  =
?? ?? |?? 1
?? ?? 1

?? ?? |?? 1
?? ?? 1
+?? ?? |?? 2
?? ?? 2

=
4
10
×
2
6
4
10
×
2
6
+
6
10
×
4
6

=
1
4
·
(ii) Since ?? 1
and ?? 2
are mutually exclusive and ?? ?? 1
??? 2
|?? =?? ?? |?? = 1,we
have
?? ?? 2
|?? = 1-?? ?? 1
|??
=
3
4
· _
In the above example
?? ?? 1
|?? =
1
4
<
1
3
=?? ?? 1
,
NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   3

and                                       ?? ?? 2
|?? =
3
4
>
2
3
=?? ?? 2
,
i.e.,
(i) the probability of occurrence of event ?? 1
decreases in the presence of the
information that the outcome will be an element of ?? ;
(ii) the probability of occurrence of event ?? 2
increases in the presence of information
that the outcome will be an element of ?? .

These phenomena are related to the concept of association defined in the sequel.
Note that
?? ?? 1
|?? <?? ?? 1
??? ?? 1
n?? <?? ?? 1
?? (?? ),
and
?? ?? 2
|?? >?? ?? 2
??? ?? 2
n?? >?? ?? 2
?? (?? ).

Definition 3.2

Let  ?? ,F,??  be a probability space and let ?? and ?? be two events. Events ?? and ?? are
said to be

(i) negatively associated if ?? ?? n?? <?? ?? ?? ?? ;
(ii) positively associated if ?? ?? n?? >?? ?? ?? ?? ;
(iii) independent  if ?? ?? n?? =?? ?? ?? ?? . _
Remark 3.3
(i) If ?? ?? = 0 then ?? ?? n?? = 0 =?? ?? ?? ?? ,??? ?F, i.e., if  ?? ?? = 0 then
any event ?? ?F and ?? are independent;
(ii) If ?? ?? > 0  then ?? and ?? are independent if, and only if, ?? ?? |?? =?? (?? ),
i.e., if ?? ?? > 0, then events ?? and ?? are independent if, and only if, the
availability of the information that event ?? has occurred does not alter the
probability of occurrence of event ?? . _
Now we define the concept of independence for arbitrary collection of events.

Page 4

NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   1

MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem

1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let  ?? ,F,??  be a probability space and let  ?? ?? : ?? ???  be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ???  using Theorem of Total Probability .  _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   2

(ii) In Bayes’ theorem the probabilities ?? ?? ?? ,?? ??? , are referred to as prior
probabilities and the probabilities ?? ?? ?? |?? ,?? ??? , are referred to as posterior
probabilities. _
To see an application of Bayes’ theorem let us revisit Example 3.4.
Example 3.5
Urn ?? 1
contains 4 white and 6 black balls and urn ?? 2
contains 6 white and 4 black balls.
A fair die is cast and urn ?? 1
is selected if the upper face of die shows five or six dots.
Otherwise urn ?? 2
is selected. A ball is drawn at random from the selected urn.
(i) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 1
;
(ii) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 2
.
Solution. Define the events:
?? : drawn ball is white;

?? 1
: urn ?? 1
is selected
?? 2
: urn ?? 2
is selected
mutually exclusive & exhaustive events
(i) We have
?? ?? 1
|??  =
?? ?? |?? 1
?? ?? 1

?? ?? |?? 1
?? ?? 1
+?? ?? |?? 2
?? ?? 2

=
4
10
×
2
6
4
10
×
2
6
+
6
10
×
4
6

=
1
4
·
(ii) Since ?? 1
and ?? 2
are mutually exclusive and ?? ?? 1
??? 2
|?? =?? ?? |?? = 1,we
have
?? ?? 2
|?? = 1-?? ?? 1
|??
=
3
4
· _
In the above example
?? ?? 1
|?? =
1
4
<
1
3
=?? ?? 1
,
NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   3

and                                       ?? ?? 2
|?? =
3
4
>
2
3
=?? ?? 2
,
i.e.,
(i) the probability of occurrence of event ?? 1
decreases in the presence of the
information that the outcome will be an element of ?? ;
(ii) the probability of occurrence of event ?? 2
increases in the presence of information
that the outcome will be an element of ?? .

These phenomena are related to the concept of association defined in the sequel.
Note that
?? ?? 1
|?? <?? ?? 1
??? ?? 1
n?? <?? ?? 1
?? (?? ),
and
?? ?? 2
|?? >?? ?? 2
??? ?? 2
n?? >?? ?? 2
?? (?? ).

Definition 3.2

Let  ?? ,F,??  be a probability space and let ?? and ?? be two events. Events ?? and ?? are
said to be

(i) negatively associated if ?? ?? n?? <?? ?? ?? ?? ;
(ii) positively associated if ?? ?? n?? >?? ?? ?? ?? ;
(iii) independent  if ?? ?? n?? =?? ?? ?? ?? . _
Remark 3.3
(i) If ?? ?? = 0 then ?? ?? n?? = 0 =?? ?? ?? ?? ,??? ?F, i.e., if  ?? ?? = 0 then
any event ?? ?F and ?? are independent;
(ii) If ?? ?? > 0  then ?? and ?? are independent if, and only if, ?? ?? |?? =?? (?? ),
i.e., if ?? ?? > 0, then events ?? and ?? are independent if, and only if, the
availability of the information that event ?? has occurred does not alter the
probability of occurrence of event ?? . _
Now we define the concept of independence for arbitrary collection of events.

NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   4

Definition 3.3

Let  ?? ,F,??  be a probability space. Let ?? ?R be an index set and let  ?? ?? :?? ???  be a
collection of events in F.

(i) Events  ?? ?? :?? ???  are said to be pairwise independent if any pair of events ?? ??
and ?? ?? ,?? ??? in the collection  ?? ?? :?? ???  are independent. i.e., if ?? ?? ?? n
?? ?? =?? ?? ?? ?? ?? ?? , whenever ?? ,?? ??? and ?? ??? ;

(ii) Let ?? = 1, 2,… , n , for some ?? ?N, so that  ?? ?? :?? ??? = ?? 1
,… ,?? ??  is a
finite collection of events in F. Events ?? 1
,… ,?? ?? are said to be independent if,
for any sub collection  ?? ?? 1
,… ,?? ?? ??  of  ?? 1
,… ,?? ??   ?? = 2,3,… ,??
??  ?? ?? ?? ?? ?? =1
= ?? ?? ?? =1
?? ?? ?? .                                                                (3.6)

(iii) Let ?? ?R be an arbitrary index set. Events  ?? ?? :?? ???  are said to be
independent if any finite sub collection of events in  ?? ?? :?? ???  forms a
collection of independent events.  _

Remark 3.4
(i) To verify that?? events  ?? 1
,… ,?? ?? ?F are independent one must verify
2
?? -?? - 1 =
?? ??
?? ?? =2
conditions in (3.6). For example, to conclude that
three events ?? 1
,?? 2
and ?? 3
are independent, the following 4  = 2
3
- 3- 1
conditions must be verified:
?? ?? 1
n?? 2
=?? ?? 1
?? ?? 2
;
?? ?? 1
n?? 3
=?? ?? 1
?? ?? 3
;
?? ?? 2
n?? 3
=?? ?? 2
?? ?? 3
;
?? ?? 1
n?? 2
n?? 3
=?? ?? 1
?? ?? 2
?? ?? 3
.

(ii) If events ?? 1
,… ,?? ?? are independent then, for any permutation  ?? 1
,… ,?? ??  of
1,… ,?? , the events ?? ?? 1
,… ,?? ?? ?? are also independent. Thus the notion of
independence is symmetric in the events involved.

(iv) Events in any subcollection of independent events are independent. In
particular independence of a collection of events implies their pairwise
independence.  _

Page 5

NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   1

MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem

1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let  ?? ,F,??  be a probability space and let  ?? ?? : ?? ???  be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? (?? )

=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ???  using Theorem of Total Probability .  _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   2

(ii) In Bayes’ theorem the probabilities ?? ?? ?? ,?? ??? , are referred to as prior
probabilities and the probabilities ?? ?? ?? |?? ,?? ??? , are referred to as posterior
probabilities. _
To see an application of Bayes’ theorem let us revisit Example 3.4.
Example 3.5
Urn ?? 1
contains 4 white and 6 black balls and urn ?? 2
contains 6 white and 4 black balls.
A fair die is cast and urn ?? 1
is selected if the upper face of die shows five or six dots.
Otherwise urn ?? 2
is selected. A ball is drawn at random from the selected urn.
(i) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 1
;
(ii) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 2
.
Solution. Define the events:
?? : drawn ball is white;

?? 1
: urn ?? 1
is selected
?? 2
: urn ?? 2
is selected
mutually exclusive & exhaustive events
(i) We have
?? ?? 1
|??  =
?? ?? |?? 1
?? ?? 1

?? ?? |?? 1
?? ?? 1
+?? ?? |?? 2
?? ?? 2

=
4
10
×
2
6
4
10
×
2
6
+
6
10
×
4
6

=
1
4
·
(ii) Since ?? 1
and ?? 2
are mutually exclusive and ?? ?? 1
??? 2
|?? =?? ?? |?? = 1,we
have
?? ?? 2
|?? = 1-?? ?? 1
|??
=
3
4
· _
In the above example
?? ?? 1
|?? =
1
4
<
1
3
=?? ?? 1
,
NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   3

and                                       ?? ?? 2
|?? =
3
4
>
2
3
=?? ?? 2
,
i.e.,
(i) the probability of occurrence of event ?? 1
decreases in the presence of the
information that the outcome will be an element of ?? ;
(ii) the probability of occurrence of event ?? 2
increases in the presence of information
that the outcome will be an element of ?? .

These phenomena are related to the concept of association defined in the sequel.
Note that
?? ?? 1
|?? <?? ?? 1
??? ?? 1
n?? <?? ?? 1
?? (?? ),
and
?? ?? 2
|?? >?? ?? 2
??? ?? 2
n?? >?? ?? 2
?? (?? ).

Definition 3.2

Let  ?? ,F,??  be a probability space and let ?? and ?? be two events. Events ?? and ?? are
said to be

(i) negatively associated if ?? ?? n?? <?? ?? ?? ?? ;
(ii) positively associated if ?? ?? n?? >?? ?? ?? ?? ;
(iii) independent  if ?? ?? n?? =?? ?? ?? ?? . _
Remark 3.3
(i) If ?? ?? = 0 then ?? ?? n?? = 0 =?? ?? ?? ?? ,??? ?F, i.e., if  ?? ?? = 0 then
any event ?? ?F and ?? are independent;
(ii) If ?? ?? > 0  then ?? and ?? are independent if, and only if, ?? ?? |?? =?? (?? ),
i.e., if ?? ?? > 0, then events ?? and ?? are independent if, and only if, the
availability of the information that event ?? has occurred does not alter the
probability of occurrence of event ?? . _
Now we define the concept of independence for arbitrary collection of events.

NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   4

Definition 3.3

Let  ?? ,F,??  be a probability space. Let ?? ?R be an index set and let  ?? ?? :?? ???  be a
collection of events in F.

(i) Events  ?? ?? :?? ???  are said to be pairwise independent if any pair of events ?? ??
and ?? ?? ,?? ??? in the collection  ?? ?? :?? ???  are independent. i.e., if ?? ?? ?? n
?? ?? =?? ?? ?? ?? ?? ?? , whenever ?? ,?? ??? and ?? ??? ;

(ii) Let ?? = 1, 2,… , n , for some ?? ?N, so that  ?? ?? :?? ??? = ?? 1
,… ,?? ??  is a
finite collection of events in F. Events ?? 1
,… ,?? ?? are said to be independent if,
for any sub collection  ?? ?? 1
,… ,?? ?? ??  of  ?? 1
,… ,?? ??   ?? = 2,3,… ,??
??  ?? ?? ?? ?? ?? =1
= ?? ?? ?? =1
?? ?? ?? .                                                                (3.6)

(iii) Let ?? ?R be an arbitrary index set. Events  ?? ?? :?? ???  are said to be
independent if any finite sub collection of events in  ?? ?? :?? ???  forms a
collection of independent events.  _

Remark 3.4
(i) To verify that?? events  ?? 1
,… ,?? ?? ?F are independent one must verify
2
?? -?? - 1 =
?? ??
?? ?? =2
conditions in (3.6). For example, to conclude that
three events ?? 1
,?? 2
and ?? 3
are independent, the following 4  = 2
3
- 3- 1
conditions must be verified:
?? ?? 1
n?? 2
=?? ?? 1
?? ?? 2
;
?? ?? 1
n?? 3
=?? ?? 1
?? ?? 3
;
?? ?? 2
n?? 3
=?? ?? 2
?? ?? 3
;
?? ?? 1
n?? 2
n?? 3
=?? ?? 1
?? ?? 2
?? ?? 3
.

(ii) If events ?? 1
,… ,?? ?? are independent then, for any permutation  ?? 1
,… ,?? ??  of
1,… ,?? , the events ?? ?? 1
,… ,?? ?? ?? are also independent. Thus the notion of
independence is symmetric in the events involved.

(iv) Events in any subcollection of independent events are independent. In
particular independence of a collection of events implies their pairwise
independence.  _

NPTEL- Probability and Distributions

Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur                                   5

The following example illustrates that, in general, pairwise independence of a collection
of events may not imply their independence.

Example 3.6

Let ?? = 1, 2, 3, 4  and let F =?? ?? , the power set of ?? . Consider the probability space
?? ,F, P , where ??  ??  =
1
4
,?? = 1, 2, 3, 4 . Let ?? = 1, 4 , ?? = 2, 4  and  ?? = 3, 4 .
Then,
?? ?? =?? ?? =?? ?? =
1
2
,
?? ?? n?? =?? ?? n?? =?? ?? n?? =?? {4} =
1
4
,
and                     ?? ?? n?? n?? =??  4  =
1
4
·
Clearly,
?? ?? n?? =?? ?? ?? ?? ;?? ?? n?? =?? ?? ?? (?? ), and ?? ?? n?? =?? ?? ?? (?? ),
i.e., ?? ,?? and ?? are pairwise independent.
However,
?? ?? n?? n?? =
1
4
??? ?? ?? ?? ?? (?? ).
Thus  ?? ,?? and ?? are not independent. _

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