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THEOREM-1:
Probability of the impossible event is zero i.e. p(Φ) = 0.
PROOF:
Impossible event contains no sample point and hence the certain event S and the impossible event Φ are mutually exclusive.
Hence S U Φ = S , [where S= Sample space ]
∴ P( S U Φ ) = P( S )
⇒ P( S ) + P(Φ ) = P( S )
⇒ P( Φ ) = 0 [Proved]
REMARK:
It may be noted P( A )=0, does not imply that A is necessarily an empty set. In practice, probability ' 0 ' is assigned to the events which are so rare for example , let us consider the random tossing of a coin. The event that the coin will stand erect on its edge, is assigned the probability 0.
THEOREM -2:
Probability of the complementary event Ā of A is given by P( Ā ) =1-P( A ).
PROOF:
We know that A and Ā are dis-joint events.
Moreover, A U Ā = S
⇒ P( A U Ā ) = P( S )
⇒ P( A ) + P( Ā ) = 1
⇒ P( Ā ) = 1 - P( A ) [Proved]
THEOREM - 3:
For any event A, 0 ≤ P( A ) ≤ 1.
PROOF:
Clearly for any event A ,
we have Φ ⊆ A⊆ S
⇒ | Φ | ⊆ | A | ⊆ | S |
⇒ | Φ |/ | S | ≤ | A |/ | S | ≤ | S |/ | S |
⇒ 0 ≤ P( A ) ≤ 1 [Proved]
ADDITION RULE:
★ For any two events A and B
P( A ⋃ B ) = P( A ) + P( B ) - P( A ⋂ B ).
★ If A and B are two mutually exclusive events then,
P( A ⋃ B ) = P( A ) + P( B ) [ ∵ P( A ⋂ B ) = 0 ]
★ For any three events A , B , C ,
P( A ⋃ B ⋃ C )=P( A ) + P( B ) + P( C ) - P( A ⋂ B ) - P(B ⋂ C ) - P(C ⋂ A ) + P(A ⋂ B ⋂ C)
PROOF:
Let us take A ⋃ B = D
P( A ⋃ B ⋃ C ) = P( D ⋃ C ).
= P( D ) + P( C ) - P ( D ∩ C )
= P( A ⋃ B ) + P( C ) - P(( A ⋃ B ) ∩ C ))
= P( A ⋃ B ) + P( C ) - [ P(( A ⋂ C ) ⋃ ( B ⋂ C )) ]
= P( A ) + P( B ) + P( C ) -P( A ⋂ B ) - [ P( A ⋂ C ) + P( B ⋂ C ) - P((A ⋂ B ) ⋂ ( B ⋂ C)) ].
= P( A ) + P( B ) + P( C ) -P( A ⋂ B ) - P(B ⋂ C ) - P(C ⋂ A ) + P(A ⋂ B ⋂ C)
[Proved]
★ If A , B , C are the pair-wise mutually exclusive events , then
P( A ⋃ B ⋃ C )=P( A ) + P( B ) + P( C ) .
Example:
A dice is thrown is thrown twice , find the
i) P( getting total is 8)
ii) P( getting total is 10)
iii) P( total is 8 and 10 )
iv) P ( total is 8 or 10 )
v) P( total is an even number)
vi) P( total is not 8)
vii) P( tota is an even number but not 8 )
Answer- Since the dice is thrown twice , the sample space ( S ) = | 36 |
i) Let A is the event of getting total is 8
A={ (2,6) , (3,5) , (4,4) , (5,3) , (6,2) } , | A | = 5
so, P( A )= | A | / | S | = 5/36.
ii) Let B is the event of getting total is 10
B={ (4,6) , (5,5) , (6,4) } , | B |=3
so, P( B )= | B | / | S | = 3/36= 1/12
iii) P( getting total is 8 and 10)
= P( A ⋂ B ) = P(Φ ) = 0
iv)P( getting total is 8 pr 10)
= P( A ⋃ B ) = P( A ) + P( B ) - P( A ⋂ B )
= 5/36 + 3/36 - 0 = 8/36=2/9
v) Let C be the event of total is an even no.
P( C ) = 18/ 36 = 1/2
vi) P( getting total is not 8)
= 1 - P( A ) = 1- (5/36) = 31/ 36
vii) P( getting total is an even no. but not 8)
= P(getting total is an even no.) -
P(getting total is 8)
= (18/36) - (5/36) = 13/36.
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