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1
Section 2.8
Some Probability Laws
©
John J Currano, 01/26/2010
2
We have already seen:
The
Multiplicative Law of Probability
.
P
(
A
∩
B
) =
P
(
B
)
P
(
A

B
)
provided
P
(
B
)
≠
0
P
(
A
∩
B
) =
P
(
A
)
P
(
B

A
)
provided
P
(
A
)
≠
0
P
(
A
∩
B
) =
P
(
A
)
P
(
B
)
if
A
and
B
are independent
.
Countable Additivity
. If {
A
k
 k
∈
K
} is a
countable collection of mutually exclusive
events, then
If
A
and
B
are mutually exclusive
, then
P
(
A
∪
B
) =
P
(
A
) +
P
(
B
).
If
A
,
B
,
C
are mut. exclusive
,
P
(
A
∪
B
∪
C
)=
P
(
A
)+
P
(
B
P
(
C
).
Etc.
∑
∈
=
⎟
⎠
⎞
⎜
⎝
⎛
K
k
k
∈
K
P
(
A
k
).
A
k
P
U
3
The
Multiplicative Law of Probability
generalizes to three
or more events.
Assume all have nonzero probability.
Then:
P
(
A
∩
B
) =
P
(
A
)
P
(
B

A
)
P
(
A
∩
B
∩
C
) =
P
(
A
)
P
(
B

A
)
P
(
C

A
∩
B
)
P
(
A
∩
B
∩
C
∩
D
) =
P
(
A
)
P
(
B

A
)
P
(
C

A
∩
B
)
P
(
D

A
∩
B
∩
C
)
etc.
4
()
∑
=
⎟
⎠
⎞
⎜
⎝
⎛
∈
∈
K
k
k
K
k
k
A
P
A
P
U
Consequences:
If
A
and
B
are events:
P
(
A
)
=
P
(
A – B
) +
P
(
A
∩
B
)
P
(
B
)
=
P
(
A
∩
B
) +
P
(
B – A
)
P
(
A
∪
B
)
=
P
(
A – B
) +
P
(
A
∩
B
) +
P
(
B – A
)
Sketch of Proof
:
The results follow by countable additivity.
As can be seen from the picture,
A – B
,
A
∩
B
, and
B – A
a
re pairwise disjoint (mutually exclusive) and
A
=
(
A – B
)
∪
(
A
∩
B
),
B
=
(
A
∩
B
)
∪
(
B – A
),
(
A
∪
B
)
=
(
A – B
)
∪
(
A
∩
B
)
∪
(
B – A
).
Countable Additivity
If {
A
k
} countable and mutually exclusive, then
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Consequences:
If
A
and
B
are events:
P
(
A
)
=
P
(
A – B
) +
P
(
A
∩
B
)
P
(
B
)
=
P
(
A
∩
B
) +
P
(
B – A
)
P
(
A
∪
B
)
=
P
(
A – B
) +
P
(
A
∩
B
) +
P
(
B – A
)
Additive Law of Probability
P
(
A
∪
B
) =
P
(
A
) +
P
(
B
) –
P
(
A
∩
B
).
NonDecreasing
If
A
⊆
B
, then
P
(
A
)
≤
P
(
B
).
 Note that
B
=
A
∪
(
B – A
), so
P
(
B
) =
P
(
A
) +
P
(
B – A
)
≥
P
(
A
)
Complement
:
P
(
A
) = 1 –
P
(
A
)
 Note that
S = A
∪
A
so
1 =
P
(
S
) =
P
(
A
) +
P
(
A
)
()
∑
=
⎟
⎠
⎞
⎜
⎝
⎛
∈
∈
K
k
k
K
k
k
A
P
A
P
U
Countable Additivity
If {
A
k
} countable and mutually exclusive, then
6
Additive Law of Probability
:
P
(
A
∪
B
) =
P
(
A
) +
P
(
B
) –
P
(
A
∩
B
).
P
(
A
) =
P
(
A – B
) +
P
(
A
∩
B
)
P
(
B
) =
P
(
A
∩
B
) +
P
(
B – A
)
P
(
A
∪
B
) =
P
(
A – B
) +
P
(
A
∩
B
) +
P
(
B – A
)
Complement
:
P
(
A
) = 1 –
P
(
A
)
Show:
1.
P
(
B – A
) =
P
(
B
)
–P
(
A
∩
B
)
2.
If
A
⊆
B
, then
P
(
B – A
) =
P
(
B
)
(
A
)
3.
P
(
∅
) = 0
7
InclusionExclusion
P
(
A
∪
B
∪
C
)=
P
(
A
) +
P
(
B
) +
P
(
C
)
–
P
(
A
∩
B
) –
P
(
A
∩
C
) –
P
(
B
∩
C
)
+
P
(
A
∩
B
∩
C
)
Sketch of proof
:
A
∪
B
∪
C = A
∪
(
B
∪
C
)
−
associative law
A
∩
(
B
∪
C
) = (
A
∩
B
)
∪
(
A
∩
C
)
−
distributive law
P
(
A
∪
(
B
∪
C
)) =
P
(
A
) +
P
(
B
∪
C
)
–
P
(
A
∩
(
B
∪
C
))
=
P
(
A
) +
P
(
B
∪
C
) –
P
((
A
∩
B
)
∪
(
A
∩
C
))
Additive Law of Probability
:
P
(
A
∪
B
) =
P
(
A
) +
P
(
B
) –
P
(
A
∩
B
)
Text, p. 58
8
∑
=
⎟
⎠
⎞
⎜
⎝
⎛
∈
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This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.
 Fall '10
 Any
 Statistics, Probability

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