JEE Mathematics/ Lakshmikanta Satapathy/ Theory of Probability part 8/ Partition of sample space, Theorem of total probability, Bayes' theorem with examples

1. Physics Helpline
L K Satapathy
Probability Theory 8

2. Physics Helpline
L K Satapathy
Partition of a Sample Space
A set of events constitutes the partition of a sample
space S , if
1 2, , . . . , nE E E
(i) They are pair wise disjoint :
(ii) They are exhaustive :
(iii) They have non-zero probabilities :
; ; , 1, 2, 3 . . .i jE E i j i j n   
1 2 . . . nE E E S   
( ) 0 1,2, . . .iP E i n  
Example -1 : Consider a non empty event E and its complement E
corresponding to a sample space S. They form a partition of S , since
( ) ; ( ) , ( ) ( ) 0 ( ) 0i E E ii E E S iii P E and P E       
Probability Theory 8

3. Physics Helpline
L K Satapathy
Example – 2 : Consider two non-empty events E and F corresponding
to a sample space S , as shown in the Venn diagram.
E
F
S
EF
EF
EF EF
Consider the events :(i) E  F (ii) E  F
(iii) E  F (iv) E  F
We observe that these four events
(i) are pair wise disjoint
(ii) are exhaustive
(iii) have non-zero probabilities
 Events E  F , E  F , E  F and E  F form a partition of S
It may be noted that , partition of a sample space is not unique.
There can be more than one partitions of the same sample space.
Probability Theory 8

4. Physics Helpline
L K Satapathy
Theorem of Total Probability :
If the set of events constitute a partition of the
sample space S. and A be any other event associated with S , then
1 2, , . . . , nE E E
1 1 2 2( ) ( ) ( ) ( ) ( ) . . . ( ) ( )n nP A P E P A E P E P A E P E P A E   
1
1
( ) ( )
n
i
i
P E P A E

 
Proof : Since constitute a partition of S , we have1 2, , . . . , nE E E
1 2( ) . . . ni S E E E   
1 2( ) ( ) [ ( . . . )]nP A P A S P A E E E      
( ) ,i jii E E i j   
 For event A , we have
1 2( ) [( ) ( ) . . . ( )] . . . (1)nP A P A E A E A E       
Probability Theory 8

5. Physics Helpline
L K Satapathy
AE2
S
AE1 AEn
A
E1 E2 En
The events & the sample space are
shown in the diagram. The events
(AE1) , (AE2) - - - (AEn) are
pair-wise disjoint .
1 2(1) ( ) ( ) ( ) . . . ( ) . . . (2)nP A P A E P A E P A E       
1 1 2 2( ) ( ) ( ) ( ) ( ) . . . ( ) ( )n nP A P E P A E P E P A E P E P A E    
From Multiplication Theorem , we have
( ) ( ) ( ) , 1,2, . . . , , [ ( ) 0]i i i iP A E P E P A E i n as P E    
1
1
( ) ( )
n
i
i
P E P A E

 
Probability Theory 8

6. Physics Helpline
L K Satapathy
1
( ) ( )
( ) , 1,2, . . . ,
( ) ( )
i i
i n
j j
j
P E P A E
P E A for any i n
P E P A E

 

Bayes’ Theorem : [ Formula for the probability of causes ]
If be non-empty events constituting a partition of S
& A be any other event of non-zero probability associated with S , then
1 2, , . . . , nE E E
Proof : From conditional probability , we have
( ) ( )
( )
( ) ( )
i i
i
P E A P A E
P E A
P A P A
 
 
( ) ( )
( )
i iP E P A E
P A

By multiplication theorem of probability
1
( ) ( )
( ) ( )
i i
n
j j
j
P E P A E
P E P A E



By theorem of total probability
gives the probability of a particular (cause)
after event A has occurred
( )iP E A iE
Probability Theory 8

7. Physics Helpline
L K Satapathy
Ans : Let the events
Example : A bag contains 4 red and 4 black balls. Another bag
contains 2 red and 6 black balls. One of the two bags is selected at
random and a ball is drawn from the bag which is found to be red. Find
the probability that the ball is drawn from the first bag.
Choosing bag 2 = E2
 Probability of drawing a red ball from bag 1
Drawing a red ball = A
Choosing bag 1 = E1 1
1( )
2
P E 
2
1( )
2
P E 
Bag 1 contains 4 red balls in a total of 8 balls
1
4 1( )
8 2
P A E  
Bag 2 contains 2 red balls in a total of 8 balls
 Probability of drawing a red ball from bag 2 2
2 1( )
8 4
P A E  
Probability Theory 8

8. Physics Helpline
L K Satapathy
1 1
1
1 1 2 2
( ) ( )
( )
( ) ( ) ( ) ( )
P E P A E
P E A
P E P A E P E P A E


Probability of drawing a ball from bag1 , given that it is red = P(E1/A)
1 1
2 2
1 1 1 1
2 2 2 4


  
 Using Bayes’ Theorem , we get
1
24 [ ]
1 1 3
4 8
Ans 

Probability Theory 8

9. Physics Helpline
L K Satapathy
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