Digital Signal Processing Reference
In-Depth Information
Table 1.1
Outcomes of the space
S
in Example 1.5.1
s
1
¼
(1,1)*
s
7
¼
(2,1)
s
13
¼
(3,1)*
s
19
¼
(4,1)
s
¼ (5,1)
*
s
31
¼
(6,1)
25
s
2
¼
(1,2)
s
8
¼
(2,2)*
s
14
¼
(3,2)
s
¼ (4,2)
*
s
26
¼
(5,2)
s
32
¼
(6,2)*
20
s
3
¼
(1,3)*
s
9
¼
(2,3)
s
¼ (3,3)
*
s
21
¼
(4,3)
s
27
¼
(5,3)*
s
33
¼
(6,3)
15
s
4
¼
(1,4)
s
¼ (2,4)
*
s
16
¼
(3,4)
s
22
¼
(4,4)*
s
28
¼
(5,4)
s
34
¼
(6,4)*
10
s
¼ (1,5)
*
s
11
¼
(2,5)
s
17
¼
(3,5)*
s
23
¼
(4,5)
s
29
¼
(5,5)*
s
35
¼
(6,5)
5
s
6
¼
(1,6)
s
12
¼
(2,6)*
s
18
¼
(3,6)
s
24
¼
(4,6)*
s
30
¼
(5,6)
s
36
¼
(6,6)*
From (
1.34
) the probability of
A
is given as:
PfAg¼n
A
=N ¼
5
=
36
:
(1.55)
The outcomes of the event
B
are denoted with an asterisk in Table
1.1
. The total
number of outcomes
n
B
is 18:
B ¼
s
1
; s
3
; s
5
; s
8
; s
10
; s
12
; s
13
; s
15
; s
17
; s
20
; s
22
; s
24
; s
25
; s
27
; s
29
; s
32
; s
34
; s
36
g
:
f
(1.56)
Using (
1.51
), we find the desired conditional probability:
PfAjBg¼n
A;B
=n
B
¼
5
=
18
:
(1.57)
1.6 Total Probability and Bayes' Rule
1.6.1 Total Probability
Consider two events,
A
and
B
, which are not mutually exclusive, as shown in Fig.
1.9
.
We wish to obtain information regarding the occurrence of
A
by exploring two
cases:
B
would have occurred (event (A
\
B)).
B
would not have occurred (event (
A \ B
)).
The events (
A\B
) and (
A \ B
) are mutually exclusive, resulting in:
PfAg¼PfA \ BgþPfA \ Bg:
(1.58)
Using (
1.43
) we can express (
1.58
) in terms of conditional probabilities:
PfAg¼PfAjBgPfBgþPfAjBgPfBg:
(1.59)
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