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Physics Helpline
L K Satapathy
Probability QA 9
Physics Helpline
L K Satapathy
Question: Let two fair six faced dice A and B be thrown simultaneously. If E1 is
the event that die A shows up four , E2 is the event that die B shows up two and E3
is the event that the sum of the numbers on both dice is odd , then which of the
following options is not true ?
(a) E1 and E2 are independent (b) E2 and E3 are independent
(c) E1 and E3 are independent (d) E1 , E2 and E3 are independent
Answer :
Probability QA 9
Rule : Two events A & B are independent if P(AB) = P(A).P(B)
For three events A , B & C P(AB C) = P(A).P(B).P(C)
[Refer Probability Theory 7]
For two dice , S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }
( ) 36n S 
Physics Helpline
L K Satapathy
Probability QA 9
E1 ( A shows 4) = { (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) }
1
1 1
( ) 6 1( ) 6 ( )
( ) 36 6
n E
n E P E
n S
     
E2 ( B shows 2) = { (1,2),(2,2),(3,2),(4,2),(5,2),(6,2)}
2
2 2
( ) 6 1( ) 6 ( )
( ) 36 6
n E
n E P E
n S
     
E3 ( Sum is odd) =
3
3 3
( ) 18 1( ) 18 ( )
( ) 36 2
n E
n E P E
n S
     
{(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(3,6),
(4,1),(4,3),(4,5),(5,2),(5,4),(5,6),(6,1),(6,3),(6,5) }
Physics Helpline
L K Satapathy
Probability QA 9
(E1  E2) ( A shows 4 & B shows 2) = {(4,2)}
1 2 1 2
1( ) 1 ( ) . . . (1)
36
n E E P E E     
(E1  E3) ( A shows 4 & Sum is odd) = {(4,1),(4,3),(4,5)}
1 3 1 3
3 1( ) 3 ( ) . . . (2)
36 12
n E E P E E      
(E2  E3) ( B shows 2 & Sum is odd) = {(1,2),(3,2),(5,2)}
2 3 2 3
3 1( ) 3 ( ) . . . (3)
36 12
n E E P E E      
(E1E2  E3) ( A shows 4 , B shows 2 & Sum is odd) = 
1 2 3 1 2 3( ) 0 ( ) 0 . . . (4)n E E E P E E E       
[ It may be noted that (E1  E2) & (E3) are mutually exclusive ]
Physics Helpline
L K Satapathy
Probability QA 9
1 2 2 3 1 3 1 2 3
1 1 1( ) ( ) ( ) ( ) 0
36 12 12
P E E P E E P E E P E E E        
Correct option = (d)
1 2 3
1 1 1( ) ( ) ( )
6 6 2
P E P E P E  
 E1 , E2 and E3 are not independent  (d) is not true
Results obtained :
1 2 1 2 1 2
1 1 1( ) ( ). ( ) ( ) &
6 6 36
a P E P E P E E E E are independent     
2 3 2 3 2 3
1 1 1( ) ( ). ( ) ( ) &
6 2 12
b P E P E P E E E E are independent     
1 3 1 3 1 3
1 1 1( ) ( ). ( ) ( ) &
6 2 12
c P E P E P E E E E are independent     
1 2 3 1 2 3
1 1 1 1( ) ( ). ( ). ( ) ( )
6 6 2 72
d P E P E P E P E E E      
Physics Helpline
L K Satapathy
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Probability QA 9/ Independent events

  • 1. Physics Helpline L K Satapathy Probability QA 9
  • 2. Physics Helpline L K Satapathy Question: Let two fair six faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four , E2 is the event that die B shows up two and E3 is the event that the sum of the numbers on both dice is odd , then which of the following options is not true ? (a) E1 and E2 are independent (b) E2 and E3 are independent (c) E1 and E3 are independent (d) E1 , E2 and E3 are independent Answer : Probability QA 9 Rule : Two events A & B are independent if P(AB) = P(A).P(B) For three events A , B & C P(AB C) = P(A).P(B).P(C) [Refer Probability Theory 7] For two dice , S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), (3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6), (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) } ( ) 36n S 
  • 3. Physics Helpline L K Satapathy Probability QA 9 E1 ( A shows 4) = { (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) } 1 1 1 ( ) 6 1( ) 6 ( ) ( ) 36 6 n E n E P E n S       E2 ( B shows 2) = { (1,2),(2,2),(3,2),(4,2),(5,2),(6,2)} 2 2 2 ( ) 6 1( ) 6 ( ) ( ) 36 6 n E n E P E n S       E3 ( Sum is odd) = 3 3 3 ( ) 18 1( ) 18 ( ) ( ) 36 2 n E n E P E n S       {(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(3,6), (4,1),(4,3),(4,5),(5,2),(5,4),(5,6),(6,1),(6,3),(6,5) }
  • 4. Physics Helpline L K Satapathy Probability QA 9 (E1  E2) ( A shows 4 & B shows 2) = {(4,2)} 1 2 1 2 1( ) 1 ( ) . . . (1) 36 n E E P E E      (E1  E3) ( A shows 4 & Sum is odd) = {(4,1),(4,3),(4,5)} 1 3 1 3 3 1( ) 3 ( ) . . . (2) 36 12 n E E P E E       (E2  E3) ( B shows 2 & Sum is odd) = {(1,2),(3,2),(5,2)} 2 3 2 3 3 1( ) 3 ( ) . . . (3) 36 12 n E E P E E       (E1E2  E3) ( A shows 4 , B shows 2 & Sum is odd) =  1 2 3 1 2 3( ) 0 ( ) 0 . . . (4)n E E E P E E E        [ It may be noted that (E1  E2) & (E3) are mutually exclusive ]
  • 5. Physics Helpline L K Satapathy Probability QA 9 1 2 2 3 1 3 1 2 3 1 1 1( ) ( ) ( ) ( ) 0 36 12 12 P E E P E E P E E P E E E         Correct option = (d) 1 2 3 1 1 1( ) ( ) ( ) 6 6 2 P E P E P E    E1 , E2 and E3 are not independent  (d) is not true Results obtained : 1 2 1 2 1 2 1 1 1( ) ( ). ( ) ( ) & 6 6 36 a P E P E P E E E E are independent      2 3 2 3 2 3 1 1 1( ) ( ). ( ) ( ) & 6 2 12 b P E P E P E E E E are independent      1 3 1 3 1 3 1 1 1( ) ( ). ( ) ( ) & 6 2 12 c P E P E P E E E E are independent      1 2 3 1 2 3 1 1 1 1( ) ( ). ( ). ( ) ( ) 6 6 2 72 d P E P E P E P E E E      
  • 6. Physics Helpline L K Satapathy For More details: www.physics-helpline.com Subscribe our channel: youtube.com/physics-helpline Follow us on Facebook and Twitter: facebook.com/physics-helpline twitter.com/physics-helpline