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196
Solutions to Exercises
20. [BB] We use induction on
n,
the strong form. For
n
=
1, the result is clear.
If
n
=
2,
P(AI
U
A2)
=
P(AI)
+
P(A2)

P(AI
n
A2)
~
P(AI)
+
P(A2)'
Now assume
n
>
2 and the result holds for all
integers
k
with 1
~
k
<
n.
We have
P(AI
U
A2
U .
.. U
An)
= P((AI
U
A2
U· U
AnI)
U
An)
~
P(AI
U
A2
U .
.. U
AnI)
+
P(An)
~
(P(AI)
+
P(A2)
+ .
.. +
P(An I)
+
P(An)
By the Principle of Mathematical Induction, the result follows.
using the
k
=
2 case
using the
k
=
n
 1 case.
21. We use induction on
n,
the strong form. For
n
=
1, the result is clear.
If
n
=
2,
P(AI
n
A2)
=
P(AI)
+
P(A2)

P(AI
U
A2)
~
P(AI)
+
P(A2)

1. Now assume
n
>
2
and the result holds for
all integers
k
with 1
~
k
<
n.
We have
P(AI
n
A2
n .
..
nAn)
=
P((AI
n
A2
n· n
AnI} nAn)
~
P(AI
n
A2
n .
..
nAnI)
+
P(An)
 1
~
(P(AI)
+
P(A2)
+ .
.. +
P(An I)

(n

2)
+
P(An)
 1
=
P(AI)
+
P(A2)
+ .
.. +
P(An)

(n

1).
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory, Integers

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