Physics Helpline / JEE Mathematics / Lakshmikanta Satapathy / Axioms of Probability, Mutually exclusive and Complementary events and Illustrations of Probability

1. Physics Helpline
L K Satapathy
Probability Theory 5

2. Physics Helpline
L K Satapathy
Probability of an Event :
Let S be the sample space associated with a random experiment
E be a subset of S representing an event
The number of elements of S = n(S)
The number of elements of E = n(E)
Then Probability of the event E is
( )
( )
( )
number of elements of E n E
P E
number of elements of S n S
 
 It follows that ,
( ) 0( ) 0 ( )
( ) ( )
n
P minimum
n S n S

   
( )
( ) 1 ( )
( )
n S
P S maximum
n S
 
Probability Theory 5

3. Physics Helpline
L K Satapathy
Axiomatic definition of Probability
Probability P(E) of an event E corresponding to the sample space S
of a random experiment is a Real valued function , whose domain is
the power set of S (since E is a subset of S) and Range is the closed
interval [0,1] (since minimum and maximum values of P(E) are 0 and
1 respectively)
(i) For any event E , P(E)  0
(iii) If E and F are mutually exclusive events , P(E  F) = P(E) + P(F)
(ii) P(S) = 1
Axioms or Rules
Probability Theory 5

4. Physics Helpline
L K Satapathy
To prove (iii) :
Set theory  n(E  F) = n(E) + n(F) – n(E  F)
If E and F are disjoint , n(E  F) = 0
 n(E  F) = n(E) + n(F)
( ) ( ) ( )
( ) ( ) ( )
n E F n E n F
n S n S n S

  
( ) ( ) ( )P E F P E P F   
P(E  ) = P(E) + P() [ since E and  are disjoint ]
 P(E ) = P(E) + P() [ since E   = E ]  P() = 0
To prove that P() = 0 :
Probability Theory 5

5. Physics Helpline
L K Satapathy
Probability of complementary events
Consider an event E of a random experiment and its complement E.
 The two events E and E are mutually exclusive
We know that E  E =  and E  E = S
 P( E  E) = P(S)
 P( E ) + P( E) = P(S) [since P( E  E) = P( E ) + P( E)
 P( E ) + P( E) = 1 [ since P( S ) =1 ]
 P ( E) = 1 – P(E)
Probability Theory 5

6. Physics Helpline
L K Satapathy
Illustration 1
Three coins are tossed simultaneously. Then find the probability of
getting exactly one head or two heads.
Let E be the event of getting exactly one head or two heads
Sample space S = { HHH , HHT , HTH , HTT , THH , THT , TTH , TTT }
 E = { HHT , HTH , HTT , THH , THT , TTH }
 n(S) = 8 and n(E) = 6
( ) 6 3( )
( ) 8 4
n E
P E
n S
   
Answer
Probability Theory 5

7. Physics Helpline
L K Satapathy
Illustration 2
Two dice are thrown simultaneously. Then find the probability of
getting a prime number as the sum
Sample space S = { 11 , 12 , . . . , 16 , 21 , 22 , . . . , 26 ,
31 , 32 , . . . , 36 , 41 , 42 , . . . , 46 ,
51 , 52 , . . . , 56 , 61 , 62 , . . . , 66 }
 n(S) = 36 and n(E) = 15
( ) 15 5( )
( ) 36 12
n E
P E
n S
   
Answer
Sum is a prime number  Sum = 2 , 3 , 5 , 7 or 11
Sum 2 = {11} , Sum 3 = {12 , 21} , Sum 5 = {14 , 41 , 23 , 32}
Sum 7 = {16 , 61 , 25 , 52 , 34 , 43} , Sum 11 = {56 , 65}
 E = { 11, 12, 21, 14, 41, 23, 32, 16, 61, 25, 52, 34, 43, 56, 65 }
Probability Theory 5

8. Physics Helpline
L K Satapathy
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