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• For n = 2, the result holds because by Eq. (10.26) we have:

PA( ∪ A ) = P A( ) + P A( ) − P A( ∩ A )
1 2 1 2 1 2
and since any probability is a non-negative number, this leads to
the inequality:

PA( ∪ A ) ≤ P A( ) + P A( )
1 2 1 2

• Assume that the theorem is true for (n – 1) events, then we can write:

 n  n
≤
P U A k ∑ P A( k )
 k=2  k=2

• Using associativity, Eq. (10.26), the result for (n – 1) events, and the
non-negativity of the probability, we can write:

 n    n   n    n 
P A ∪
P A ∩
P A +
k
k
P U A k =   1 U A  = ( 1 ) P U A k −   1 U A  

 k=1    k=1   k=2    k=2 


n
n
k ∑
P A ∩
≤ PA ( ) + ∑ P A −   1 U A  ≤ n PA( )
)
(

1 k   k
k=2    k=2  k =1
which is the desired result.
In-Class Exercises
Pb. 10.8 Show that if the events A , A , …, A are such that:
2
1
n
A ⊂ A ⊂ … ⊂ A
1 2 n
then:
 n 
P U A k = P A( n )
 k=1 

Pb. 10.9 Show that if the events A , A , …, A are such that:
1
n
2
A ⊃ A ⊃… ⊃ A
1 2 n
© 2001 by CRC Press LLC