JEE Mathematics/ Lakshmikanta Satapathy/Questions on reverse probability solved using the concept of total probability and Bayes' theorem with complete explanation

1. Physics Helpline
L K Satapathy
Probability QA 6

2. Physics Helpline
L K Satapathy
QA Probability - 6
Q1: In a set of 10 coins , 2 coins are with heads on both sides. A coin is selected at
random from this set and tossed five times. If all the five times , the result was heads ,
find the probability that the selected coin had heads on both sides.
1
2 1( )
10 5
P E  
Ans : Let us define the events as follows:
E1 = Selecting a coin having heads on both sides
E2 = Selecting a coin not having heads on both sides
A = Getting a head in all the 5 tosses
We are required to find the probability of event E1 given that event A has
already occurred which is equal to P(E1 /A)
There are 2 coins having heads on both sides in a total of 10 coins.
2
8 4( )
10 5
P E  
There are 8 coins not having heads on both sides in a total of 10 coins.

3. Physics Helpline
L K Satapathy
QA Probability - 6
In 1 toss of a coin having heads on both sides, probability of getting a head = 1.
 Probability of getting a head in 5 successive tosses is
P(A / E1) = Probability of getting a head in all the five tosses
when the selected coin has heads on both sides
P(A / E2) = Probability of getting a head in all the five tosses
when the selected coin does not have heads on both sides
5
1( ) (1) 1P A E  
The 5 successive tosses of the coin are independent events.
In 1 toss of a coin not having heads on both sides, probability of getting a head = 1/2.
 Probability of getting a head in 5 successive tosses is
 
5
2
1 1( )
2 32
P A E  

4. Physics Helpline
L K Satapathy
QA Probability - 6
1( ) 1P A E 
2
1( )
32
P A E 
1
1( )
5
P E 
2
4( )
5
P E 
We have obtained :
Using Bayes’ Theorem ,
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


1 11
5 5
1 4 1 1 11
5 5 32 5 40

 
   
1
1 40 85
9 5 9 9
4
[
0
]Ans   

5. Physics Helpline
L K Satapathy
QA Probability - 6
Q2: There are three coins. One is a two-headed coin (having head on both faces) ,
another is a biased coin that comes up heads 75% of the times and the third is also a
biased coin that comes up tails 40% of the times. One of the coins is chosen at random
and tossed. If it shows heads , then what is the probability that it was the two headed
coin?
1 2 3
1( ) ( ) ( )
3
P E P E P E   
Ans : Let us define the events as follows :
E1 = choosing the two headed coin
E2 = choosing the coin which come up heads 75% of the times
E3 = choosing the coin which come up tails 40% of the times
A = getting a head
To find: Probability that the coin is two headed, given that it shows head = P(E1/A)
The coin is chosen at random.

6. Physics Helpline
L K Satapathy
QA Probability - 6
On selecting the 1st (two headed) coin ,
probability of getting a head is 100%
Using Bayes’ theorem:
1( ) 1P A E 
On selecting the 2nd coin ,
probability of getting a head is 75% 2
75 3( )
100 4
P A E  
On selecting the 3rd coin , probability of getting a tail
is 40%  Probability of getting a head is 60%
3
60 3( )
100 5
P A E  
1 1
1
1 1 2 2 3 3
( ) ( )
( )
( ) ( ) ( ) ( ) ( ) ( )
P E P A E
P E A
P E P A E P E P A E P E P A E

 
1 1
1 13
1 1 3 1 3 3 3 20 15 121 1
3 3 4 3 5 4 5 20

  
       
[ ]20
47
Ans

7. Physics Helpline
L K Satapathy
QA Probability - 6
Q3: A card is lost from a pack of 52 cards. From the remaining pack , two cards are
drawn , which are found to be both diamonds. Find the probability that the lost card
being diamond.
Ans : Let us define the events as follows :
E1 = the lost card is a diamond
E2 = the lost card is not a diamond
A = the two drawn cards are both diamond
We are required to find the probability that the lost card is a diamond , given that
the 2 drawn cards are diamond = P(E1/A)
There are 13 diamond cards in a total of 52 cards. 1
13 1( )
52 4
P E  
There are 39 non-diamond cards in a total of 52 cards. 2
39 3( )
52 4
P E  

8. Physics Helpline
L K Satapathy
QA Probability - 6
P(A/E1) = probability of drawing 2 diamond cards when the lost card is a diamond
To apply Bayes’ theorem , we need to find the following:
12
2
1 51
2
12 11 2 11 22( )
51 50 17 25 425
C
P A E
C
     
 
P(A/E2) = probability of drawing 2 diamond cards when the lost card is not a diamond
When the lost card is a diamond we have 12 diamond cards in a total of 51 cards
2 diamond cards can be chosen from 12 diamond cards in ways
2 cards can be chosen from a total of 51 cards in ways
12
2C
51
2C
13
2
2 51
2
13 12 13 2 26( )
51 50 17 25 425
C
P A E
C
     
 
When the lost card is non-diamond we have 13 diamond cards in a total of 51 cards
2 diamond cards can be chosen from 13 diamond cards in ways
2 cards can be chosen from a total of 51 cards in ways
13
2C
51
2C

9. Physics Helpline
L K Satapathy
QA Probability - 6
1
22( )
425
P A E 
2
26( )
425
P A E 
1
1( )
4
P E 
2
3( )
4
P E 
We have obtained :
Using Bayes’ Theorem ,
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


1 22
224 425
1 22 3 26 22 (3 26)
4 425 4 425

 
   
22 22 11
22 78 100 50
[ ]Ans  


10. Physics Helpline
L K Satapathy
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