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NOTE In solving certain category of probability problems, it is often conve-
nient to solve for P(A) by computing the probability of its complement and
then applying the above relation.
Pb. 10.7 Show that if A , A , …, A are mutually exclusive events, then:
1
2
n
PA( ∪ A ∪…∪ A ) = PA( ) + PA( ) +…+ PA( )
1 2 n 1 2 n
(Hint: Use mathematical induction and Eq. (10.15).)

10.3 Addition Laws for Probabilities
We start by reminding the reader of the key results of elementary set theory:

• The Commutative law states that:

A ∩ B = B ∩ A (10.16)

A ∪ B = B ∪ A (10.17)

• The Distributive laws are written as:

A ∩ ( B ∪ C =) ( A ∩ B ∪) ( A ∩ C) (10.18)

A ∪ ( B ∩ C =) ( A ∪ B ∩) ( A ∪ C) (10.19)

• The Associative laws are written as:

(A ∪ ) B ∪ C = A ∪ (B ∪ C ) = A ∪ ∪ C (10.20)
B

(A ∩ ) B ∩ C = A ∩ (B ∩ C ) = A ∩ ∩ C (10.21)
B
• De Morgan’s laws are

(A ∪ ) B = A ∩ B (10.22)

(A ∩ ) B = A ∪ B (10.23)

© 2001 by CRC Press LLC